├── .gitignore ├── LICENSE ├── Makefile ├── README.md ├── errata.tex ├── lemmas.py ├── lemmas.tex ├── license.tex ├── macros.tex ├── notes.txt ├── prelude.tex ├── python ├── ex_8.1.2.py ├── ex_9.1.13.py ├── good_weak.py └── thrm_5.1.7b.py ├── sections ├── sec_1_3.tex ├── sec_1_4.tex ├── sec_5_1.tex ├── sec_5_2.tex ├── sec_6_1.tex ├── sec_6_2.tex ├── sec_6_3.tex ├── sec_6_4.tex ├── sec_6_5.tex ├── sec_6_6.tex ├── sec_7_1.tex ├── sec_7_2.tex ├── sec_8_1.tex ├── sec_8_2.tex └── sec_9_1.tex ├── shared ├── cor_aleph_wonlegt.tex ├── lem_aleph_infwo_succ.tex ├── lem_aleph_initle.tex ├── lem_aleph_initlimit.tex ├── lem_aleph_initltwa.tex ├── lem_aleph_wolege.tex ├── lem_aoc_notin.tex ├── lem_ord_expin_sups.tex └── lem_ord_lub.tex ├── solutions.tex └── theorems.tex /.gitignore: -------------------------------------------------------------------------------- 1 | *.aux 2 | *.log 3 | *.out 4 | *.pdf 5 | *.toc 6 | lemmas_content.tex 7 | -------------------------------------------------------------------------------- /LICENSE: -------------------------------------------------------------------------------- 1 | This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. 2 | To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/4.0/ or send a letter to Creative Commons, PO Box 1866, Mountain View, CA 94042, USA. 3 | -------------------------------------------------------------------------------- /Makefile: -------------------------------------------------------------------------------- 1 | ############################################################################### 2 | # FILE: Makefile 3 | # 4 | # Author: Dan Whitman 5 | ############################################################################## 6 | TARGET = solutions 7 | .PHONY: all lemmas clean 8 | 9 | #============================================================================== 10 | all: 11 | #============================================================================== 12 | pdflatex -halt-on-error $(TARGET).tex 13 | 14 | #============================================================================== 15 | lemmas: 16 | #============================================================================== 17 | ./lemmas.py 18 | pdflatex -halt-on-error lemmas.tex 19 | 20 | #============================================================================== 21 | theorems: 22 | #============================================================================== 23 | pdflatex -halt-on-error theorems.tex 24 | 25 | #============================================================================== 26 | errata: 27 | #============================================================================== 28 | pdflatex -halt-on-error errata.tex 29 | 30 | #============================================================================== 31 | clean: 32 | #============================================================================== 33 | rm -f *.aux *.log *.pdf *.out *.toc 34 | rm -f -r python/__pycache__ 35 | rm -f lemmas_content.tex 36 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | Introduction to Set Theory by Karel Hrbacek and Thomas Jech, 3rd Edition, Revised and Expanded 2 | ============================================================================================== 3 | 4 | Solutions Manual 5 | ---------------- 6 | 7 | The main solutions manual is `solutions.tex`. 8 | To build the LaTeX manual, simply run `make` to compile `solutions.pdf`. 9 | This may need to be run multiple times to get all the cross-references right. 10 | Unfortunately I only started typing up my solutions starting with Chapter 5, having kept them in a handwritten notebook prior to that. 11 | I do intend to eventually go back and start typing up my solutions to the earlier chapters as well to have a more complete manual. 12 | The `python` directory contains some quick and dirty Python scripts that were used to gain insight while working on some exercises. 13 | These are not really documented at all and so probably will not be of interest to anyone else. 14 | 15 | Other Documents 16 | --------------- 17 | 18 | When doing exercises it can be useful to see a list of lemmas that have been written as part of the solutions. 19 | You can then build a lemma list document by running `make lemmas`to compile `lemmas.pdf`. 20 | This may also need to be run multiple times to get all the cross-references right. 21 | 22 | I have also found it useful to work out more detailed proofs of some theorems in the text, for which the text usually provides proofs that gloss over the details. 23 | A document with these proofs can be built by running `make theorems` to compile `theorems.pdf`. 24 | This need to be run multiple times to get all the cross-references right. 25 | Many of these more detailed proofs also have their own lemmas, some of which are also used in the solutions manual or vice versa. 26 | Such lemmas resided in their own `.tex` files in the `shared` directory, so they can be included in both documents. 27 | 28 | Lastly, I have also been keeping a list of errata in the book. 29 | Run `make errata` to compile the `errata.pdf` document. 30 | 31 | Miscellaneous 32 | ------------- 33 | 34 | Please submit an issue if you discover any errors in any of the proofs or any typos. 35 | 36 | For an up-to-date PDF of the solutions manual, theorems document, errata list, and links to other resources for this text head over to [my website](https://math-study.net/set-theory/). 37 | Note that this is still a work in progress. 38 | -------------------------------------------------------------------------------- /errata.tex: -------------------------------------------------------------------------------- 1 | \documentclass{report} 2 | \usepackage{amsmath} 3 | \usepackage{amssymb} 4 | \usepackage[hidelinks]{hyperref} 5 | 6 | \input{macros} 7 | 8 | \title{ 9 | \booktitle \\ 10 | \ \\ 11 | Errata List 12 | \ \\ 13 | \ \\ 14 | Compiled by Dan Whitman 15 | } 16 | 17 | \begin{document} 18 | 19 | \maketitle 20 | 21 | \begin{enumerate} 22 | 23 | \item Page 62, Exercise 3.5.4. 24 | The hint should read ``Let $h(x) = A - x$; \ldots'' instead of ``Let $h(x) = B - x$; \ldots''. 25 | (Confirmed by Dr. Jech) 26 | 27 | \item Page 101, Exercise 5.2.3. 28 | This means that that the countable dense subset is dense in $P$ rather than simply dense with respect to itself. 29 | This has to be the case because, if were merely dense with respect to itself, then \emph{any} larger linearly ordered set containing the subset would have the same property so that it is impossible to put a bound on the size of that set (Confirmed) 30 | 31 | \item Page 114, Exercise 6.3.5 part (c). 32 | This should read ``\ldots, then $f[A] \in V_\w$.'' instead of ``\ldots, then $f[X] \in V_\w$.'' since $X$ has not been previously defined. 33 | (Unconfirmed) 34 | 35 | \item Page 140, bottom. 36 | The reference to Assumption~1.7 in Chapter~4 should really be Assumption~1.8. 37 | (Unconfirmed) 38 | 39 | \item Page 142, bottom. 40 | The reference to Theorem~4.4 in Chapter~7 should be Theorem~4.4 in Chapter~6. 41 | (Unconfirmed) 42 | 43 | \item Page 143 top. 44 | In the proof of Theorem~8.1.13 when showing that (c) implies (a) the sentence, ``Let $F$ be the system of all functions $f$ for which $\dom{f} \ss S$ and $f(X) \in X$ holds for any $X \in S$.'' should be, ``\ldots and $f(X) \in X$ holds for any $X \in \dom{f}$.'' 45 | This is because, if $\dom{f} \pss S$, then there is an $X \in S$ where $X \notin \dom{f}$ so that $f(X)$ is not defined. 46 | (Unconfirmed) 47 | 48 | \item Page 143, middle. 49 | In the proof of Theorem~8.1.14 there are two times in the first paragraph where statements are made for all or for some $\x \in A$. 50 | These should be $\x < \l$. 51 | (Unconfirmed) 52 | 53 | \item Page 158, bottom. 54 | Near the end of the first paragraph of the proof of Theorem~9.1.7 (K{\"o}nig's Theorem), the sentence, ``If $i_x \neq i_y = i$, then $a_i = d_i \notin A$ while $b_i = y \in A$.'' should be, ``If $i_x \neq i_y = i$, then $a_i = d_i \notin A_i$ while $b_i = y \in A_i$.'' 55 | (Unconfirmed) 56 | 57 | \item Page 160, Exercise 9.1.11. 58 | The special case mentioned in the hint should be $\parens{\k \cdot \l}^\mu = \k^\mu \cdot \l^\mu$. 59 | This is evidenced by the fact that $\parens{\k^\mu}^\l = \parens{\k^\l}^\mu$ does not make sense in the context of the problem (i.e. it is not a special case) as well as by the fact that it is not part of Theorem~5.1.7. 60 | (Unconfirmed) 61 | 62 | \end{enumerate} 63 | 64 | \end{document} 65 | -------------------------------------------------------------------------------- /lemmas.py: -------------------------------------------------------------------------------- 1 | #! /usr/bin/env python3 2 | """ 3 | Build lemmas document contents from other tex files. 4 | 5 | Run with -h for more information. 6 | """ 7 | from collections import namedtuple 8 | import argparse 9 | import os 10 | 11 | # Input directory 12 | idir = "sections" + os.sep 13 | 14 | # Output file name 15 | ofname = "lemmas_content.tex" 16 | 17 | # Argument parsing 18 | parser = argparse.ArgumentParser(description="Builds lemmas document content from\ 19 | section tex files as well as the theorems tex file. Output is written to " + ofname + ".") 20 | args = parser.parse_args() 21 | 22 | 23 | def get_arg(line, cmd): 24 | """ 25 | Retuns the argument if the Latex cmd 26 | is found. Otherwise None is returned. 27 | """ 28 | lcmd = "\\" + cmd + "{" 29 | if line.find(lcmd) != 0: 30 | return None 31 | j = line.find("}") 32 | if j < 0: 33 | return None 34 | return line[len(lcmd):j] 35 | 36 | 37 | sargs = ("lem", "cor", "defin", "thrm") 38 | counter_map = {"section": "chapter", 39 | "subsection": "section"} 40 | 41 | 42 | def process_file(fpath, ofile, headers=False, shared=None, defined_labels=[]): 43 | """ 44 | Processes latex file to extract statements 45 | 46 | fpath - Path to .tex file 47 | ofile - Output file object 48 | headers - Increment counters for header lines 49 | shared - Name of shared statement file we are 50 | reading (None indicates we are not in 51 | a shared statement) 52 | defined_labels - LaTeX labels that have already 53 | been defined so they are not 54 | defined again. This list be be 55 | added to. 56 | """ 57 | # Open up input file for reading 58 | with open(fpath, "r") as ifile: 59 | # Whether or not we are in a statement or comment block 60 | ins = False 61 | inc = False 62 | mdef = 0 63 | 64 | # Go through each line in the file 65 | for line in ifile: 66 | # We don't care about whitespace at the beginning or end of the line 67 | sl = line.strip() 68 | 69 | # Are we in a commented out block? 70 | if inc: 71 | if sl.find(r"\fi") == 0: 72 | inc = False 73 | continue 74 | 75 | # Are we in a statement block? 76 | if ins: 77 | # Pass the line through 78 | print(sl, file=ofile) 79 | 80 | # Are ending the statement? 81 | if get_arg(sl, "end") in sargs: 82 | # Print a blank line to leave some spaces 83 | print("", file=ofile) 84 | ins = False 85 | else: 86 | # No, only pass through certain lines pertaining to the statements 87 | earg = get_arg(sl, "exercise") 88 | iarg = get_arg(sl, "input") 89 | targ = get_arg(sl, "theorem") 90 | scarg = get_arg(sl, "setcounter") 91 | if (sl.find(r"\def") == 0 and sl.find("{") >= 0) or mdef > 0: 92 | # Include any macro definitions 93 | print(sl, file=ofile) 94 | if sl[-1] == "{": 95 | # Multiline def 96 | mdef += 1 97 | elif sl[0] == "}": 98 | # End multiline def 99 | mdef -= 1 100 | elif sl.find(r"\iffalse") == 0: 101 | # Start of comment block 102 | inc = True 103 | continue 104 | elif earg: 105 | # Exercise line, set counters 106 | e = int(earg) 107 | print(r"\setcounter{subsection}{" + str(e-1) + 108 | "}\stepcounter{subsection}", file=ofile) 109 | elif targ: 110 | # Theorem line, set counters 111 | t = int(targ) 112 | print(r"\setcounter{subsection}{" + str(t-1) + 113 | "}\stepcounter{subsection}", file=ofile) 114 | elif iarg and iarg.find("shared") == 0: 115 | # Input line for shared satement 116 | process_file(iarg + ".tex", ofile, 117 | shared=iarg.split("/")[-1], defined_labels=defined_labels) 118 | elif get_arg(sl, "begin") in sargs: 119 | # We are starting a statement, verify that no content is also on this line 120 | if sl[-1] != "}": 121 | raise ValueError( 122 | "Statement has content on the same line!") 123 | 124 | add = "" 125 | 126 | # Add shared tag if applicable 127 | if shared: 128 | add += "[Shared: " + \ 129 | shared.replace("_", r"\textunderscore ") + "] " 130 | 131 | # Determine the label tag (if there is one) 132 | lbl = "label" 133 | i = sl.find("\\" + lbl) 134 | if i >= 0: 135 | larg = get_arg(sl[i:], lbl) 136 | if larg: 137 | add += r"\{" + larg + r"\}" 138 | 139 | if larg in defined_labels: 140 | # Need to strip out the label to prevent compiler warnings, 141 | # because it has already been defined before. 142 | sl = sl.replace(f"\\{lbl}{{{larg}}}", "") 143 | else: 144 | defined_labels.append(larg) 145 | 146 | print(sl + add, file=ofile) 147 | 148 | ins = True 149 | elif scarg: 150 | # Set counter line 151 | print(sl.replace(scarg, counter_map[scarg]), file=ofile) 152 | elif headers: 153 | # Search for counter lines to increment counters 154 | for hdr in counter_map.keys(): 155 | if sl.find("\\" + hdr) == 0: 156 | print( 157 | r"\stepcounter{" + counter_map[hdr] + "}", file=ofile) 158 | 159 | 160 | # Section file record 161 | Section = namedtuple("Section", ("chap", "sec", "fname")) 162 | 163 | # List of already-defined item labels 164 | defined_labels = [] 165 | 166 | # Open up output file for writing 167 | with open(ofname, "w") as ofile: 168 | # Go through files in the current directory 169 | secs = [] 170 | for fname in sorted(os.listdir(idir)): 171 | # We only care about certain tex files 172 | (bname, ext) = os.path.splitext(fname) 173 | if ext != ".tex": 174 | continue 175 | if bname.find("sec_") != 0: 176 | continue 177 | 178 | secs.append(Section(*(int(s) 179 | for s in bname.split("_")[1:]), idir + fname)) 180 | 181 | # Sort sections and go through them 182 | print(r"\section{Solutions}", file=ofile) 183 | for sec in sorted(secs, key=lambda s: (s.chap, s.sec)): 184 | print("Processing " + sec.fname + "...") 185 | 186 | # Set counters 187 | print(r"\setcounter{chapter}{" + str(sec.chap) + "}", file=ofile) 188 | print(r"\setcounter{section}{" + str(sec.sec) + "}", file=ofile) 189 | 190 | # Process section file 191 | process_file(sec.fname, ofile, defined_labels=defined_labels) 192 | 193 | # Go through theorems 194 | print("Processing theorems...") 195 | print(r"\setcounter{chapter}{0}", file=ofile) 196 | print(r"\setcounter{section}{1}", file=ofile) 197 | print(r"\section{Theorems}", file=ofile) 198 | process_file("theorems.tex", ofile, headers=True, 199 | defined_labels=defined_labels) 200 | -------------------------------------------------------------------------------- /lemmas.tex: -------------------------------------------------------------------------------- 1 | \documentclass{report} 2 | \usepackage{amsmath} 3 | \usepackage{amssymb} 4 | \usepackage[hidelinks]{hyperref} 5 | \usepackage{color} 6 | \usepackage[margin=0.75in]{geometry} 7 | \usepackage{parskip} 8 | \usepackage{calc} 9 | 10 | \input{macros} 11 | 12 | \newtheorem{lem}{Lemma}[subsection] 13 | \newtheorem{defin}[lem]{Definition} 14 | \newtheorem{cor}[lem]{Corollary} 15 | \newtheorem{thrm}[lem]{Theorem} 16 | 17 | \begin{document} 18 | \input{lemmas_content} 19 | \end{document} 20 | -------------------------------------------------------------------------------- /license.tex: -------------------------------------------------------------------------------- 1 | \section*{Copyright} 2 | 3 | \textcopyright{} \the\year{} by Dan Whitman 4 | \newline 5 | \newline 6 | \def\liclink{http://creativecommons.org/licenses/by-nc-sa/4.0/} 7 | This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. 8 | To view a copy of this license, visit \href{\liclink}{\liclink} or send a letter to Creative Commons, PO Box 1866, Mountain View, CA 94042, USA. 9 | 10 | \begin{center} 11 | \ccbyncsa 12 | \end{center} 13 | -------------------------------------------------------------------------------- /macros.tex: -------------------------------------------------------------------------------- 1 | % Common sets 2 | \def\nats{{\boldsymbol{N}}} 3 | \def\ints{{\boldsymbol{Z}}} 4 | \def\rats{{\boldsymbol{Q}}} 5 | \def\prats{\rats^+} 6 | \def\reals{{\boldsymbol{R}}} 7 | \def\es{\varnothing} 8 | 9 | % Common variables/symbols 10 | \def\vphi{\varphi} 11 | \def\a{\alpha} 12 | \def\b{\beta} 13 | \def\g{\gamma} 14 | \def\d{\delta} 15 | \def\e{\varepsilon} 16 | \def\z{\zeta} 17 | \def\k{\kappa} 18 | \def\l{\lambda} 19 | \def\n{\nu} 20 | \def\r{\rho} 21 | \def\s{\sigma} 22 | \def\t{\tau} 23 | \def\x{\xi} 24 | \def\w{\omega} 25 | \def\W{\Omega} 26 | \def\al{\aleph} 27 | 28 | % Logic 29 | \def\bic{\leftrightarrow} 30 | \newcommand{\propP}[1]{\prop{P}(#1)} 31 | \def\lxor{\veebar} 32 | 33 | % For set-buider notation 34 | \def\where{\mid} 35 | 36 | % Other stuff 37 | \def\dom{\mathrm{dom}\,} 38 | \def\ran{\mathrm{ran}\,} 39 | 40 | % Cardinality shortcuts 41 | \def\cnats{{\al_0}} 42 | \def\ccont{{2^\cnats}} 43 | 44 | % Equivalence classes 45 | \newcommand\eclass[2]{\squares{#1}_{#2}} 46 | 47 | % Shortcuts that make writing easier 48 | \newcommand\parens[1]{\left( #1 \right)} 49 | \newcommand\squares[1]{\left[ #1 \right]} 50 | \newcommand\braces[1]{\left\{ #1 \right\}} 51 | \newcommand\angles[1]{\left\langle #1 \right\rangle} 52 | \newcommand\ceil[1]{\left\lceil #1 \right\rceil} 53 | \newcommand\floor[1]{\left\lfloor #1 \right\rfloor} 54 | \newcommand\abs[1]{\left| #1 \right|} 55 | \newcommand\dabs[1]{\left\| #1 \right\|} 56 | \newcommand\vect[1]{\mathrm{\mathbf{#1}}} 57 | \newcommand\conj[1]{\overline{#1}} 58 | \newcommand\pset[1]{\mathcal{P}\left(#1\right)} 59 | \newcommand\inv[1]{#1^{-1}} 60 | \newcommand\prop[1]{\mathbf{#1}} 61 | \def\rest{\restriction} 62 | \newcommand\tet[2]{{^{#1}#2}} 63 | 64 | % Families of set 65 | \def\famF{\mathcal{F}} 66 | 67 | % These are needed so that half-open intervals do not cause auto-indentation issues due to unmatched brackets 68 | \newcommand\clop[1]{[#1)} 69 | \newcommand\ilab[1]{#1)} 70 | 71 | % Other miscellaneous stuff 72 | \def\ss{\subseteq} 73 | \def\pss{\subset} 74 | \def\Seq{\mathrm{Seq}} 75 | \def\prece{\preccurlyeq} 76 | \def\sd{\bigtriangleup} 77 | 78 | % Environment shortcuts 79 | \newcommand\gath[1]{\begin{gather*} #1 \end{gather*}} 80 | \newcommand\ali[1]{\begin{align*} #1 \end{align*}} 81 | \newcommand\qproof[1]{\begin{proof} #1 \end{proof}} 82 | \newcommand\bicproof[2]{ 83 | $(\to)$ #1 84 | 85 | $(\leftarrow)$ #2 86 | } 87 | \newcommand\seteqproof[2]{ 88 | $(\ss)$ #1 89 | 90 | $(\supseteq)$ #2 91 | } 92 | 93 | % Environment for indenting nested paragraphs (useful in case trees) 94 | \newenvironment{indpar} 95 | { 96 | \begin{adjustwidth}{1cm}{} 97 | }{ 98 | \end{adjustwidth} 99 | } 100 | 101 | % Exercise, Theorem, and solution shortcuts 102 | \newcommand\exercise[2]{{ 103 | \renewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections 104 | \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering 105 | \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset 106 | \setcounter{question}{#1-1} % Manually set the question number since we always know the exercise number 107 | \question{#2} % The actual exam class question 108 | }} 109 | \newcommand\exerciseapp[3]{ 110 | \setqf{#2} 111 | \exercise{#1}{#3} 112 | \setqf{} 113 | } 114 | \newcommand\theorem[2]{{ 115 | \renewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections 116 | \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering 117 | \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset 118 | \setcounter{question}{#1-1} % Manually set the question number since we always know the theorem number 119 | \question{#2} % The actual exam class question 120 | }} 121 | \newcommand\theoremapp[3]{ 122 | \setqf{#2} 123 | \theorem{#1}{#3} 124 | \setqf{} 125 | } 126 | \newcommand\sol[1]{ 127 | \begin{solution} 128 | 129 | #1 130 | \end{solution} 131 | } 132 | 133 | % Main problem and theorem labels 134 | \def\mainprob{\textbf{Main Problem.}} 135 | \def\mainthrm{\textbf{Main Theorem.}} 136 | 137 | % Saves some typing 138 | \def\cbthrm{Cantor-Bernstein Theorem} 139 | 140 | % Book title 141 | \def\booktitle{ 142 | Introduction to Set Theory \\ 143 | Third Edition, Revised and Expanded \\ 144 | by Karel Hrbacek and Thomas Jech 145 | } 146 | -------------------------------------------------------------------------------- /notes.txt: -------------------------------------------------------------------------------- 1 | Unfinished or needs work: 2 | 1. Section about transfinite recursion (esp exercise 6.4.1) 3 | 2. Add title pages to solutions and theorems? (Errata already has this) 4 | 5 | Notes: 6 | 1. Theorems from theorems doc suffixed with T 7 | 8 | Incomplete or problematic exercises: 9 | -First chapters 10 | -Exercise 5.2.2 (Rudin) 11 | -Exercise 5.2.7 (Tedious and Rudin) 12 | -Exercise 6.5.11e (Could not solve) 13 | -------------------------------------------------------------------------------- /prelude.tex: -------------------------------------------------------------------------------- 1 | \documentclass[answers]{exam} 2 | 3 | % AMS packages 4 | \usepackage{amsmath} 5 | \usepackage{amsthm} 6 | \usepackage{amsfonts} 7 | \usepackage{amssymb} 8 | 9 | % Used so the PDF is hyperlinked for convenience, and chapter/section numbers are in the PDF bookmarks 10 | \usepackage[bookmarksnumbered,hidelinks]{hyperref} 11 | 12 | % Used in copmlex tabular/table 13 | % Needed for a table in one of the solutions 14 | \usepackage{multirow} 15 | 16 | % Make the margins not so large 17 | % Cannot use for now because the page number at the bottom goes off page 18 | %\usepackage[margin=0.75in]{geometry} 19 | 20 | % Used to indent whole paragraphs 21 | \usepackage{changepage} 22 | 23 | % Needed to be able to do arithmetic when setting counters 24 | \usepackage{calc} 25 | 26 | % Needed to make title pages without page numbers in exam class 27 | \usepackage{titling} 28 | 29 | % Needed for customization of section headers 30 | \usepackage{titlesec} 31 | 32 | % For diagrams (e.g. Venn diagrams in section 1.4) 33 | \usepackage{tikz} 34 | 35 | % TODO: Should we migrate \newcommand to this, which better enforces arguments and gives much better error messages for missing arguments 36 | % Useful for more advanced macro arguments (e.g. boolean values) 37 | \usepackage{xparse} 38 | 39 | % For Creative Commons license icons 40 | \usepackage[scale=1.5]{ccicons} 41 | -------------------------------------------------------------------------------- /python/ex_8.1.2.py: -------------------------------------------------------------------------------- 1 | X = set((0, 2, 3, 4, 7)) 2 | Y = set((0, 2, 3, 5)) 3 | #Y = set((0, 3, 5)) 4 | Z = set((0, 4, 5, 6)) 5 | 6 | le = lambda A : sorted(A)[0] 7 | 8 | def pt(l, v) : 9 | print(l + " = " + str(v)) 10 | 11 | pt("X", X) 12 | pt("Y", Y) 13 | pt("Z", Z) 14 | 15 | x = le(X ^ Y) 16 | y = le(Y ^ Z) 17 | pt("x", x) 18 | pt("y", y) 19 | 20 | if x not in X or y not in Y : 21 | print("BAD SETS!") 22 | 23 | pt("X - Z", X - Z) 24 | pt("X ^ Z", X ^ Z) 25 | 26 | a = le(X ^ Z) 27 | pt("a", a) 28 | 29 | if a not in (x, y) : 30 | print("Weird case!") 31 | -------------------------------------------------------------------------------- /python/ex_9.1.13.py: -------------------------------------------------------------------------------- 1 | import sys 2 | import argparse 3 | import itertools as it 4 | 5 | # Parse command line arguments 6 | parser = argparse.ArgumentParser(description="Form injection as in Exercise 9.1.13 for sets of given size.") 7 | parser.add_argument("kappas", metavar="KAPPA", nargs="+", type=int, help="Set sizes for each i in I") 8 | args = parser.parse_args() 9 | 10 | I = tuple(range(len(args.kappas))) 11 | Ais = [tuple(range(k)) for k in args.kappas] 12 | 13 | i0 = 1 14 | 15 | def ai(ix, x, i) : 16 | (alix, beix) = (Ais[ix][0], Ais[ix][1]) 17 | (ali, bei) = (Ais[i][0], Ais[i][1]) 18 | 19 | if i == ix : 20 | return x 21 | else : 22 | if i == i0 : 23 | return ali if x == alix else bei 24 | else : 25 | return bei if x == alix else ali 26 | 27 | ys = [] 28 | for ix in I : 29 | for x in Ais[ix] : 30 | y = tuple([ai(ix, x, i) for i in I]) 31 | ys.append(y) 32 | 33 | print("f(" + ", ".join(map(str, (ix,x))) + ") = " + str(y)) 34 | 35 | print("Injective:", len(ys) == len(set(ys))) 36 | -------------------------------------------------------------------------------- /python/good_weak.py: -------------------------------------------------------------------------------- 1 | """ 2 | Calculates weak Goodstein sequence from 3 | given base (first argument). Does N (second 4 | argument) iterations. If no N is passed 5 | that runs until termination. 6 | """ 7 | from pylab import * 8 | import sys 9 | 10 | m = int(sys.argv[1]) 11 | N = (int(sys.argv[2]) if len(sys.argv) > 2 else None) 12 | 13 | def base(n, b): 14 | """ 15 | Converts a natural number n to base b. 16 | Returns list of tuples where the first 17 | element is the exponent of b and the 18 | is the coefficient of that term (the 19 | digit less than b). Only nonzero 20 | digits are returns. 21 | """ 22 | if n == 0: 23 | return [0] 24 | ret = [] 25 | while True : 26 | e = int(log(n)/log(b)) 27 | d = int(n / b**e) 28 | ret.append((e, d)) 29 | n = n % (b**e * d) 30 | if e == 0 or n == 0: 31 | break 32 | 33 | return ret 34 | 35 | k = 0 36 | while True : 37 | ed = base(m, k+2) 38 | print(k, m, ed) 39 | 40 | if m == 0 or (N != None and k == N) : 41 | break 42 | 43 | m = sum([(k+3)**e * d for e,d in ed]) - 1 44 | k += 1 45 | 46 | -------------------------------------------------------------------------------- /python/thrm_5.1.7b.py: -------------------------------------------------------------------------------- 1 | # Visualization of K^(L*M) and (K^L)^M so as to formulate a bijection 2 | import itertools as it 3 | 4 | K = set((0,1)) 5 | L = set((0,1,2)) 6 | M = set((0,1)) 7 | 8 | # Column width 9 | cw = 15 10 | def pcol(ls) : 11 | print("".join([str(e).ljust(cw) for e in ls])) 12 | 13 | print("f in K^(L*M)") 14 | pcol(["f" + str(p) for p in it.product(L, M)]) 15 | for vs in it.product(*[K for p in it.product(L, M)]) : 16 | pcol(vs) 17 | 18 | print("") 19 | print("g in (K^L)^M") 20 | pcol(["g(" + str(v) +")" for v in M]) 21 | # Members of K^L as a sequence 22 | KL = [s for s in it.product(*[K for l in L])] 23 | for vs in it.product(*[KL for m in M]) : 24 | pcol(vs) 25 | -------------------------------------------------------------------------------- /sections/sec_1_3.tex: -------------------------------------------------------------------------------- 1 | \subsection{The Axioms} 2 | 3 | \def\axcomp{Axiom Schema of Comprehension} 4 | 5 | \exercise{1}{ 6 | Show that the set of all $x$ such that $x \in A$ and $x \notin B$ exists. 7 | } 8 | \sol{ 9 | \qproof{ 10 | $A$ is a given extant set. 11 | So let $\propP{x, B}$ be the property that $x \in A$ and $x \notin B$ so that clearly $\propP{x, B}$ implies that $x \in A$. 12 | Then, by was expounded just before Example~1.3.13, the set 13 | \gath{ 14 | \braces{x \where \propP{x, B}} = \braces{x \where x \in A \land x \notin B} 15 | } 16 | uniquely exists, which is of course the set we seek. 17 | } 18 | } 19 | 20 | \exercise{2}{ 21 | Replace the Axiom of Existence by the following weaker postulate: 22 | 23 | Weak Axiom of Existence: Some set exists. 24 | 25 | Prove the Axiom of Existence using the Weak Axiom of Existence and the Comprehension Schema. 26 | [Hint: Let $A$ be a set known to exist; consider $\braces{x \in A \where x \neq x}$.] 27 | } 28 | \sol{ 29 | \qproof{ 30 | Invoking the Weak Axiom of Existence, suppose that $A$ is a set that exists. 31 | Let $\propP(x)$ denote the proprty $x \neq x$, noting that clearly $\propP{x}$ is false no matter the $x$, since otherwise identity would be violated. 32 | Then the set $B = \braces{x \in A \where \propP{x}}$ uniquely exists by the \axcomp{} and Lemma~1.3.4. 33 | 34 | Now consider any $x$. 35 | If $x \notin A$, then clearly $x \notin B$ either. 36 | If $x \in A$, then again $x \notin B$ since the property $\propP{x}$ is false (since it is false for \emph{any} $x$). 37 | Hence, in both cases $x \notin B$, which shows that 38 | \gath{ 39 | \forall x (x \notin B) \bic \lnot \exists x (x \in A) 40 | } 41 | since $x$ was arbitrary. 42 | This of course means that $B = \es$ is the unique empty set, proving the Axiom of Existence since $B$ was shown to exist. 43 | } 44 | } 45 | 46 | \exercise{3}{ 47 | (a) Prove that a ``set of all sets'' does not exist. 48 | [Hint: if $V$ is a set of all sets, consider $\braces{x \in V \where x \notin x}$.] 49 | 50 | (b) Prove that for any set $A$ there is some $x \notin A$. 51 | } 52 | \sol{ 53 | (a) \qproof{ 54 | Suppose that a set of all sets exists, and denote such a set by $V$. 55 | Then, since the property $\propP{x}$ defined by $x \notin x$ is a perfectly valid property, the set $W = \braces{x \in V \where \propP{x}}$ uniquely exists by the \axcomp{} and Lemma~1.3.3. 56 | Of course, it must be that either $W \in W$ or $W \notin W$, noting that $W \in V$ since $W$ is a set. 57 | If $W \in W$ then it must be that $\propP{W}$ by the definition of $W$ since $W \in V$. 58 | However, $\propP{W}$ means that $W \notin W$, which is a contradiction. 59 | Hence, it must be that $W \notin W$, but in this case $\propP{W}$ is true so that $W \in W$ by definition since $W \in V$, which is again a contradiction. 60 | Since a contradiction occurs no matter what, it must be that our original supposition that a set of all sets exists is untrue. 61 | } 62 | 63 | (b) \qproof{ 64 | Suppose that $A$ is a set. 65 | To avoid duplicating effort, we jump ahead a bit and utilize the result of Exercise~1.3.6 below. 66 | There it is proved that $\pset{A}$, which uniquely exists by the Axiom of Power Set, is not a subset of $A$. 67 | Therefore, there must be some $x \in \pset{A}$ such that $x \notin A$, which proves our result. 68 | This is because, by the definition of subset, we have the following logical equivalences: 69 | \ali{ 70 | \pset{A} \not\ss A &\bic \lnot \forall x \squares{x \in \pset{A} \to x \in A} \\ 71 | &\bic \exists x \lnot \squares{x \in \pset{A} \to x \in A} \\ 72 | &\bic \exists x \lnot \squares{x \notin \pset{A} \lor x \in A} \\ 73 | &\bic \exists x \squares{x \in \pset{A} \land x \notin A}. 74 | } 75 | } 76 | } 77 | 78 | \exercise{4}{ 79 | Let $A$ and $B$ be sets. 80 | Show that there exists a unique set $C$ such that $x \in C$ if and only if either $x \in A$ and $x \notin B$ or $x \in B$ and $x \notin B$. 81 | } 82 | \sol{ 83 | \qproof{ 84 | Since $A$ and $B$ are sets, the set $A \cup B$ uniquely exists by the Axiom of Union. 85 | Now let $\propP{x}$ be the property defined by $x \in A$ and $x \notin B$ or $x \in B$ and $x \notin B$. 86 | Consider any $x$ such that $\propP{x}$ holds. 87 | If $x \in A$ and $x \notin B$, then obviously $x \in A$. 88 | On the other hand, if $x \in B$ and $x \notin A$, then obviously $x \in B$. 89 | Thus, in either case, $x \in A$ or $x \in B$ so that $x \in A \cup B$. 90 | Therefore, $\propP{x}$ implies that $x \in A \cup B$. 91 | Therefore, clearly the set we seek is $C = \braces{x \where \propP{x}} = \braces{x \in A \cup B \where \propP{x}}$, which uniquely exists by the \axcomp{}. 92 | } 93 | 94 | It is worth noting that the set $C$ here is called the \emph{symmetric difference} of $A$ and $B$ and is introduced in the next section. 95 | } 96 | 97 | \exercise{5}{ 98 | (a) Given $A$, $B$, and $C$, there is a set $P$ such that $x \in P$ if and only if $x = A$ or $x = B$ or $x = C$. 99 | 100 | (b) Generalize to four elements. 101 | } 102 | \sol{ 103 | (a) \qproof{ 104 | By the Axiom of Pair, the sets $D = \braces{A, B}$ and $E = \braces{C}$ uniquely exist. 105 | Then, by the Axiom of Union the set $P = D \cup E$ uniquely exists, which we claim is exactly the set we seek. 106 | 107 | ($\to$) Suppose that $x \in P = D \cup E$ so that $x \in D$ or $x \in E$. 108 | If $x \in D = \braces{A, B}$ then either $x = A$ or $x = B$. 109 | On the other hand, if $x \in E = \braces{C}$ then it must be that $x = C$. 110 | So in all cases it is true that $x = A$ or $x = B$ or $x = C$ as desired. 111 | 112 | 113 | ($\leftarrow$) Suppose that $x = A$ or $x = B$ or $x = C$. 114 | In the first two cases clearly $x \in D = \braces{A, B}$ so that $x \in D \cup E = P$. 115 | In the last case $x = C$ so that clearly $x \in E = \braces{C}$ so that again $x \in D \cup E = P$. 116 | Note that $x \in P$ in both cases. 117 | 118 | This proves the results, and we can denote our set $P$ by $\braces{A, B, C}$. 119 | } 120 | 121 | (b) \qproof{ 122 | Suppose that the four elements are $A$, $B$, $C$, and $D$. 123 | That is, we want to prove the existence of a set $P$ such that $x \in P$ if and only if $x = A$, $x = B$, $x = C$, or $x = D$. 124 | By part (a) just above, the set $E = \braces{A, B, C}$ exists as does $F = \braces{D}$ by the Axiom of Pair. 125 | Predictably, our set is then $P = E \cup F$ which exists by the Axiom of Union. 126 | We can denote this set $P$ in kind by $\braces{A, B, C, D}$. 127 | The proof that $P$ is the set we seek is directly analogous to the corresponding proof in part (a), so we do not repeat it here. 128 | } 129 | } 130 | 131 | \exercise{6}{ 132 | Show that $\pset{X} \ss X$ is false for any $X$. 133 | In particular, $\pset{X} \neq X$ for any $X$. 134 | This proves again that a ``set of all sets'' does not exist. 135 | [Hint: Let $Y = \braces{u \in X \where u \notin u}$; $Y \in \pset{X}$ but $Y \notin X$.] 136 | } 137 | \sol{ 138 | \qproof{ 139 | Following the hint, let $Y = \braces{u \in X \where u \notin u}$. 140 | Clearly $Y \ss X$ so that $Y \in \pset{X}$ since any element in $Y$ must also be in $X$. 141 | Now suppose that also $Y \in X$. 142 | We treat two cases. 143 | 144 | Case: $Y \in Y$. Then $Y \in X$ and $Y \notin Y$ by the definition of $Y$. 145 | This is of course a contradiction. 146 | 147 | Case: $Y \notin Y$. Then, since we supposed that $Y \in X$, the property in the definition of $Y$ is satisfied so that $Y \in Y$. 148 | This is again a contradiction. 149 | 150 | Since both exhaustive cases result in a contradiction, our supposition that $Y \in X$ cannot be true! 151 | So, since $Y \in \pset{X}$ but $Y \notin X$, this suffices to show that $\pset{X}$ is not a subset of $X$ as desired by the definition of inclusion. 152 | } 153 | 154 | This can also serve as a lemma to prove (differently than Exercise~1.3.3a) that a ``set of all sets'' cannot exist. 155 | To see this, suppose that $V$ is such a set. 156 | Clearly then $\pset{V}$ is a set that contains other sets. 157 | However, it was just shown that there is a $Y \in \pset{V}$ such that $Y \notin V$. 158 | But $Y$ is a \emph{set}, so it must be in $V$ by definition, a contradiction. 159 | } 160 | 161 | \exercise{7}{ 162 | The Axiom of Pair, the Axiom of Union, and the Axiom of Power Set can be replaced by the following weaker versions. 163 | 164 | Weak Axiom of Pair: For any $A$ and $B$, there is a set $C$ such that $A \in C$ and $B \in C$. 165 | 166 | Weak Axiom of Union: For any $S$, there exists $U$ such that if $X \in A$ and $A \in S$, then $X \in U$. 167 | 168 | Weak Axiom of Power Set: For any set $S$, there exists $P$ such that $X \ss S$ implies $X \in P$. 169 | 170 | Prove the Axiom of Pair, the Axiom of Union, and the Axiom of Power Set using these weaker versions. 171 | [Hint: Use also the Comprehension Schema.] 172 | } 173 | \sol{ 174 | All the below proofs follow the same basic structure: 175 | \begin{enumerate} 176 | \item Establish a property $\propP{x}$ that exactly matches the defining property of the set guaranteed to exist by the normal axiom. 177 | \item Instantiate the set $C$ that exists by the weak form of the axiom. 178 | \item Show that $\propP{x} \to x \in C$ for all $x$. 179 | \item This proves that the set $\braces{x \where \propP{x}}$ exists by the \axcomp{}, which is exactly the set asserted to exist by the normal axiom. 180 | \end{enumerate} 181 | 182 | The Axiom of Pair 183 | \qproof { 184 | Suppose that $A$ and $B$ are sets and define the property $\propP{x}$ by $x = A$ or $x = B$. 185 | Now let $C$ be a set guaranteed to exist by the Weak Axiom of Pair such that $A \in C$ and $B \in C$. 186 | Now suppose that $\propP{x}$ holds for $x$, hence $x = A$ or $x = B$. 187 | If $x = A$ then $x \in C$ since $A \in C$. 188 | Similarly, in the other case in which $x = B$ then also $x \in C$ since $B \in C$. 189 | Hence, in both cases $x \in C$ so that $\propP{x}$ implies that $x \in C$. 190 | Therefore, the set $P = \braces{x \where \propP{x}}$ = $\braces{x \in C \where \propP{x}}$ exists by the \axcomp{}. 191 | Of course $P$ is exactly the set guaranteed to exist by the (normal) Axiom of Pair, proving the result. 192 | } 193 | 194 | The Axiom of Union 195 | \qproof{ 196 | Let $S$ be any set and define the property $\propP{x}$ as $x \in A$ for some $A \in S$. 197 | Now let $C$ be a set guaranteed to exist by the Weak Axiom of Union so that, for every $A \in S$, $x \in A$ implies that $x \in C$ also. 198 | Suppose that $\propP{x}$ holds for $x$ so that there is a particular $A \in S$ such that $x \in A$. 199 | Then, since $A \in S$ and $x \in A$, it follows from the Weak Axiom of Union that $x \in C$. 200 | Hence, we have established that $\propP{x}$ implies that $x \in C$. 201 | Therefore, by the \axcomp{}, the set $\braces{x \where \propP{X}} = \braces{x \in C \where \propP{x}}$ exists, which is exactly the set guaranteed to exist by the (normal) Axiom of Union. 202 | } 203 | 204 | The Axiom of Power Set 205 | \qproof{ 206 | Let $S$ be any set and define the property $\propP{X}$ by $X \ss S$. 207 | Now let $C$ be a set that exists by the Weak Axiom of Power Set so that $X \ss S$ implies that $X \in C$ for all $X$. 208 | Next, suppose that $\propP{X}$ holds for $X$ so that $X \ss S$. 209 | Then, by the above, we have that $X \in C$ so that clearly $\propP{X}$ implies that $X \in C$. 210 | Then the set $P = \braces{X \where \propP{X}} = \braces{X \in C \where \propP{X}}$ exists by the \axcomp{}. 211 | Of course $P$ is exactly the set guaranteed to exist be the (normal) Axiom of Power Set, proving our result. 212 | } 213 | } 214 | -------------------------------------------------------------------------------- /sections/sec_5_1.tex: -------------------------------------------------------------------------------- 1 | \subsection{Cardinal Arithmetic} 2 | 3 | \exercise{1}{ 4 | Prove properties (a)-(n) of cardinal arithmetic stated in the text of this section. 5 | These are 6 | 7 | (a) $\k + \l = \l + \k$ 8 | 9 | (b) $\k + (\l + \mu) = (\k + \l) + \mu$ 10 | 11 | (c) $\k \leq \k + \l$ 12 | 13 | (d) If $\k_1 \leq \k_2$ and $\l_1 \leq \l_2$, then $\k_1 + \l_1 \leq \k_2 + \l_2$ 14 | 15 | (e) $\k \cdot \l = \l \cdot \k$ 16 | 17 | (f) $\k \cdot (\l \cdot \mu) = (\k \cdot \l) \cdot \mu$ 18 | 19 | (g) $\k \cdot (\l + \mu) = \k \cdot \l + \k \cdot \mu$ 20 | 21 | (h) $\k \leq \k \cdot \l$ if $\l > 0$ 22 | 23 | (i) If $\k_1 \leq \k_2$ and $\l_1 \leq \l_2$, then $\k_1 \cdot \l_1 \leq \k_2 \cdot \l_2$ 24 | 25 | (j) $\k + \k = 2 \cdot \k$ 26 | 27 | (k) $\k + \k \leq \k \cdot \k$, whenever $\k \geq 2$ 28 | 29 | (l) $\k \leq \k^\l$ if $\l > 0$ 30 | 31 | (m) $\l \leq \k^\l$ if $\k > 1$ 32 | 33 | (n) If $\k_1 \leq \k_2$ and $\l_1 \leq \l_2$, then $\k_1^{\l_1} \leq \k_2^{\l_1}$ 34 | } 35 | \sol{ 36 | For solutions (a) through (c) suppose that 37 | \ali{ 38 | \k &= |K| & 39 | \l &= |L| & 40 | \mu &= |M|, 41 | } 42 | where $K$, $L$, and $M$ are mutually disjoint sets. 43 | 44 | (a) 45 | \qproof{ 46 | It is obvious that 47 | $$ 48 | \k + \l = |K \cup L| = |L \cup K| = \l + \k 49 | $$ 50 | since $K \cup L = L \cup K$ and $K$ and $L$ are disjoint. 51 | } 52 | 53 | (b) 54 | \qproof{ 55 | First we note that clearly 56 | $$ 57 | K \cup (L \cup M) = K \cup L \cup M = (K \cup L) \cup M. 58 | $$ 59 | Now suppose that there is an $x \in K \cap (L \cup M)$ so that $x \in K$ and $x \in L \cup M$ 60 | If $x \in L$ then $x \in K \cap L$ and if $x \in M$ then $x \in K \cap M$, either of which is a contradiction since all three sets are mutually disjoint. 61 | Hence $K$ and $L \cup M$ are disjoint. 62 | A similar argument show that $K \cup L$ and $M$ are disjoint. 63 | Thus we have the following: 64 | $$ 65 | \k + (\l + \mu) = |K| + |L \cup M| = |K \cup (L \cup M)| = |(K \cup L) \cup M| = |K \cup L| + |M| = (\k + \l) + \mu 66 | $$ 67 | as desired. 68 | } 69 | 70 | (c) 71 | \qproof{ 72 | Define the function $f: K \to K \cup L$ by simply the identity $f(k) = k$ for any $k \in K$. 73 | Obviously this is an injective function so that $\k = |K| \leq |K \cup L| = \k + \l$. 74 | } 75 | 76 | (d) 77 | \qproof{ 78 | Suppose that 79 | \ali{ 80 | \k_1 &= |K_1| & 81 | \k_2 &= |K_2| & 82 | \l_1 &= |L_1| & 83 | \l_2 &= |L_2| 84 | } 85 | for sets $K_1$, $K_2$, $L_1$, and $L_2$ where $K_1 \cap L_1 = \es$ and $K_2 \cap L_2 = \es$. 86 | Also suppose that $\k_1 \leq \k_2$ and $\l_1 \leq \l_2$. 87 | Thus $|K_1| = \k_1 \leq \k_2 = |K_2|$ so that there is an injective function $f$ from $K_1$ to $K_2$. 88 | Similarly there is an injective function $g : L_1 \to L_2$ since $|L_1| = \l_1 \leq \l_2 = |L_2|$. 89 | Now define $h : K_1 \cup L_1 \to K_2 \cup L_2$ by 90 | $$ 91 | h(x) = \begin{cases} 92 | f(x) & x \in K_1 \\ 93 | g(x) & x \in L_1. 94 | \end{cases} 95 | $$ 96 | We show that $h$ is injective so consider $x$ and $y$ in $K_1 \cup L_1$ where $x \neq y$. 97 | 98 | Case: $x \in K_1$, $y \in K_1$. 99 | Then 100 | $$ 101 | h(x) = f(x) \neq f(y) = h(y) 102 | $$ 103 | since $f$ is injective and $x \neq y$. 104 | 105 | Case: $x \in L_1$, $y \in L_1$. 106 | Then 107 | $$ 108 | h(x) = g(x) \neq g(y) = h(y) 109 | $$ 110 | since $g$ is injective and $x \neq y$. 111 | 112 | Case: $x \in K_1$, $y \in L_1$. 113 | Then we have $h(x) = f(x) \in K_2$ and $h(y) = g(y) \in L_2$ so that $h(x) \neq h(y)$ since $K_2$ and $L_2$ are disjoint. 114 | Note that this is the same as the case in which $x \in L_1$ and $y \in K_1$ since we simply switch $x$ and $y$. 115 | 116 | Since these cases are exhaustive and $h(x) \neq h(y)$ in each this shows that $h$ is injective. 117 | Hence we have demonstrated that 118 | $$ 119 | \k_1 + \l_1 = |K_1 \cup L_1| \leq |K_2 \cup L_2| = \k_2 + \l_2 120 | $$ 121 | as desired. 122 | } 123 | 124 | For solutions (e) through (h) suppose that 125 | \ali{ 126 | \k &= |A| & \l &= |B| & \mu &= |C| 127 | } 128 | for sets $A$, $B$, and $C$. 129 | 130 | (e) 131 | \qproof{ 132 | First we show that $|A \times B| = |B \times A|$ by constructing a bijection $f : A \times B \to B \times A$. 133 | For $(a,b) \in A \times B$ define 134 | $$ 135 | f(a,b) = (b,a) \in B \times A, 136 | $$ 137 | which is clearly a function. 138 | Then for $(a,b) \in A \times B$ and $(c,d) \in A \times B$ where $f(a,b) = f(c,d)$ we have 139 | $$ 140 | f(a,b) = (b,a) = f(c,d) = (d,c) 141 | $$ 142 | so that $b=d$ and $a=c$. 143 | Hence $(a,b) = (c,d)$ so that $f$ is injective. 144 | Now consider any $(b,a) \in B \times A$ so that clearly $f(a,b) = (b,a)$, noting that $(a,b) \in A \times B$. 145 | Clearly this shows that $f$ is surjective. 146 | 147 | Hence $f$ is bijective so that 148 | $$ 149 | \k \cdot \l = |A \times B| = |B \times A| = \l \cdot \k 150 | $$ 151 | as required. 152 | } 153 | 154 | (f) 155 | \qproof{ 156 | Similar part (e) above, it is trivial to find a bijection from $A \times (B \times C)$ to $(A \times B) \times C$ so that 157 | $$ 158 | \k \cdot (\l \cdot \mu) = |A \times (B \times C)| = |(A \times B) \times C| = (\k \cdot \l) \cdot \mu 159 | $$ 160 | as desired. 161 | } 162 | 163 | (g) 164 | \qproof{ 165 | Here suppose additionally that $B \cap C = \es$. 166 | First we note that since $B$ and $C$ are disjoint that $A \times B$ and $A \times C$ are also disjoint. 167 | Suppose that this is not the case so that there is an $(a,b) \in A \times B$ where $(a,b) \in A \times C$ also. 168 | Then clearly $b \in B$ and $b \in C$, which is a contradiction since they are disjoint. 169 | Now, it is also trivial to show the equality 170 | $$ 171 | A \times (B \cup C) = (A \times B) \cup (A \times C). 172 | $$ 173 | Hence we have that 174 | $$ 175 | \k \cdot (\l + \mu) = \k \cdot |B \cup C| = |A \times (B \cup C)| = |(A \times B) \cup (A \times C)| 176 | = |A \times B| + |A \times C| = \k \cdot \l + \k \cdot \mu 177 | $$ 178 | as desired. 179 | } 180 | 181 | (h) 182 | \qproof{ 183 | Here suppose that $\l > 0$ so that $B \neq \es$. 184 | Here we construct a bijection $f: A \to A \times B$, from which it follows that 185 | $$ 186 | \k = |A| \leq |A \times B| = \k \cdot \l. 187 | $$ 188 | Since $B \neq \es$ there exists a $b \in B$. 189 | So for any $a \in A$ define 190 | $$ 191 | f(a) = (a, b), 192 | $$ 193 | which is clearly a function. 194 | So for $a_1, a_2 \in A$ where $a_1 \neq a_2$ we have that 195 | $$ 196 | f(a_1) = (a_1, b) \neq (a_2, b) = f(a_2) 197 | $$ 198 | so that $f$ is injective. 199 | } 200 | 201 | (i) 202 | \qproof{ 203 | Suppose that 204 | \ali{ 205 | \k_1 &= |A_1| & \k_2 &= |A_2| & \l_1 &= |B_1| & \l_2 &= |B_2| 206 | } 207 | for sets $A_1$, $A_2$, $B_1$, and $B_2$ where $\k_1 = |A_1| \leq |A_2| = \k_2$ and $\l_1 = |B_1| \leq |B_2| = \l_2$. 208 | Hence there is an injective function $f: A_1 \to A_2$ and injective function $g: B_1 \to B_2$. 209 | We shall construct an injective function $h: A_1 \times B_1 \to A_2 \times B_2$ so that it immediately follows that 210 | $$ 211 | \k_1 \cdot \l_1 = |A_1 \times B_1| \leq |A_2 \times B_2| = \k_2 \cdot \l_2 212 | $$ 213 | as required. 214 | So for $(a, b) \in A_1 \times B_1$ define 215 | $$ 216 | h(a, b) = (f(a), g(b)) 217 | $$ 218 | Suppose then $(a, b) \in A_1 \times B_1$ and $(c, d) \in A_1 \times B_1$ where $(a,b) \neq (c,d)$. 219 | If $a \neq c$ then $f(a) \neq f(c)$ since $f$ is injective so that 220 | $$ 221 | h(a,b) = (f(a), g(b)) \neq (f(c), g(d)) = h(c,d) 222 | $$ 223 | Similarly if $b \neq d$ then $g(b) \neq g(d)$ since $g$ is injective. 224 | Hence again 225 | $$ 226 | h(a,b) = (f(a), g(b)) \neq (f(c), g(d)) = h(c,d) 227 | $$ 228 | Thus in all cases $h(a,b) \neq h(c,d)$ so that $h$ is injective. 229 | } 230 | 231 | (j) This is adequately proven in the text. 232 | 233 | For solutions (k) through (m) suppose that 234 | \ali{ 235 | \k &= |A| & \l &= |B| 236 | } 237 | for sets $A$ and $B$. 238 | 239 | (k) 240 | \qproof{ 241 | Suppose here that $\k \geq 2$. 242 | Then $2 \leq \k$ and $\k \leq \k$ so that by property (i) we have 243 | $$ 244 | 2 \cdot \k \leq \k \cdot \k. 245 | $$ 246 | Then by property (j) we have 247 | $$ 248 | \k + \k = 2 \cdot \k \leq \k \cdot \k 249 | $$ 250 | as desired. 251 | } 252 | 253 | (l) 254 | \qproof{ 255 | Here suppose that $\l = |B| > 0$ so that $B \neq \es$. 256 | Hence there exists a $b \in B$. 257 | We shall construct an injective $f: A \to A^B$, from which it follows that 258 | $$ 259 | \k = |A| \leq |A^B| = \k^\l. 260 | $$ 261 | So for any $a \in A$ define $f(a) = g$ where $g : B \to A$ is a function defined by $g(b) = a$ for all $b \in B$, noting that $g \neq \es$ since $B \neq \es$. 262 | 263 | Now consider any $a_1, a_2 \in A$ where $a_1 \neq a_2$ so that for any $b \in B$ we have 264 | $$ 265 | f(a_1)(b) = a_1 \neq a_2 = f(a_2)(b). 266 | $$ 267 | From this it follows that $f(a_1) \neq f(a_2)$ so that $f$ is injective. 268 | } 269 | 270 | (m) 271 | \qproof{ 272 | Here suppose that $\k = |A| > 1$ so that there are $a_1, a_2 \in A$ where $a_1 \neq a_2$. 273 | We shall construct an injective function $f: B \to A^B$ so that 274 | $$ 275 | \l = |B| \leq |A^B| = \k^\l. 276 | $$ 277 | So for any $b \in B$ define $f(b) = g$ where $g: B \to A$ is a function defined by 278 | $$ 279 | g(c) = 280 | \begin{cases} 281 | a_1 & c = b \\ 282 | a_2 & c \neq b 283 | \end{cases} 284 | $$ 285 | for $c \in B$. 286 | Now suppose that $b_1, b_2 \in B$ where $b_1 \neq b_2$. 287 | We then have 288 | $$ 289 | f(b_1)(b_1) = a_1 \neq a_2 = f(b_2)(b_1) 290 | $$ 291 | since $b_1 \neq b_2$. 292 | From this it follows that $f(b_1) \neq f(b_2)$ so that $f$ is injective. 293 | } 294 | 295 | (n) 296 | \qproof{ 297 | Suppose that 298 | \ali{ 299 | \k_1 &= |A_1| & \k_2 &= |A_2| & \l_1 &= |B_1| & \l_2 &= |B_2| 300 | } 301 | for sets $A_1$, $A_2$, $B_1$, and $B_2$ where $\k_1 = |A_1| \leq |A_2| = \k_2$ and $\l_1 = |B_1| \leq |B_2| = \l_2$. 302 | 303 | The theorem as presented in the text is actually not true in full generality. 304 | As a counterexample suppose that $A_1 = A_2 = B_1 = \es$ so that $\k_1 = \k_2 = \l_1 = 0$ and $B_2 = 1$ so that $\l_2 = 1$. 305 | Then certainly the hypotheses above are true but we also have 306 | $$ 307 | \k_1^{\l_1} = 0^0 = 1 > 0 = 0^1 = \k_2^{\l_2} 308 | $$ 309 | where we have used the results of Exercises~5.1.2 and 5.1.3. 310 | 311 | However, if we add the restriction that $\k_2 > 0$ then it becomes true. 312 | To prove this first note that this implies that $A_2 \neq \es$ so that there is an $a_2 \in A_2$. 313 | Also there is an injective function $f: A_1 \to A_2$ and an injective function $g: B_1 \to B_2$. 314 | We shall construct an injective $F: A_1^{B_1} \to A_2^{B_2}$, from which it follows that 315 | $$ 316 | \k_1^{\l_1} = |A_1^{B_1}| \leq |A_2^{B_2}| = \k_2^{\l_2}. 317 | $$ 318 | So for any $h_1 \in A_1^{B_1}$ define $F(h_1) = h_2$ where $h_2 \in A_2^{B_2}$ is defined by 319 | $$ 320 | h_2(b) = \begin{cases} 321 | f(h_1(g^{-1}(b))) & b \in \ran(h_1) \\ 322 | a_2 & b \notin \ran(h_1) 323 | \end{cases} 324 | $$ 325 | for any $b \in B_2$, noting that $g^{-1}$ is a function on $\ran(h_1)$ since $g$ is injective. 326 | Clearly $F$ is a function but now we show that it is injective. 327 | 328 | So consider any $h_1, h_2 \in A_1^{B_1}$ where $h_1 \neq h_2$. 329 | Then there is a $b_1 \in B_1$ such that $h_1(b_1) \neq h_2(b_1)$. 330 | So let $b_2 = g(b_1)$ so that clearly $b_2 \in \ran(g)$ and $b_1 = g^{-1}(b_2)$. 331 | Hence we have 332 | $$ 333 | F(h_1)(b_2) = f(h_1(g^{-1}(b_2))) = f(h_1(b_1)) \neq f(h_2(b_1)) = f(h_2(g^{-1}(b_2))) = F(h_2)(b_2) 334 | $$ 335 | since $h_1(b_1) \neq h_2(b_1)$ and $f$ is injective. 336 | It thus follows that $F(h_1) \neq F(h_2)$ so that we have shown that $F$ is injective. 337 | } 338 | } 339 | 340 | \exercise{2}{ 341 | Show that $\k^0 = 1$ and $\k^1 = \k$ for all $\k$. 342 | } 343 | \sol{ 344 | \qproof{ 345 | Suppose that $\k = |A|$ for a set $A$. 346 | 347 | We claim that $A^\es = \braces{\es} = 1$ so that clearly 348 | $$ 349 | \k^0 = |A^\es| = |1| = 1. 350 | $$ 351 | First consider any $f \in A^\es$. 352 | Suppose that $f \neq \es$ so that there is a $(b,a) \in f \subseteq \es \times A$. 353 | But then $b \in \es$, which is a contradiction. 354 | Hence $f = \es$. 355 | So if there are any $f \in A^\es$ then $f = \es$ but are there any $f \in A^\es$? 356 | Clearly the empty set is a function from $\es$ to $A$ since it is vacuously true that for every $b \in \es$ there is a unique $a \in A$ such that $(b,a) \in \es$. 357 | Hence $\es \in A^\es$ so that $A^\es = \braces{\es} = 1$. 358 | 359 | We also claim that $|A^1| = |A|$ so that 360 | $$ 361 | \k^1 = |A^1| = |A| = \k. 362 | $$ 363 | To this end for any $f \in A^1$ define $F(f) = f(\es)$, noting that $1 = \braces{\es}$, so that clearly $F: A^1 \to A$. 364 | Now consider any $f,g \in A^1$ where $f \neq g$. 365 | Then it has to be that 366 | $$ 367 | F(f) = f(\es) \neq g(\es) = F(g) 368 | $$ 369 | so that $F$ is injective. 370 | Now consider any $a \in A$ and define $f \in A^1$ by $f(\es) = a$. 371 | Then clearly 372 | $$ 373 | F(f) = f(\es) = a 374 | $$ 375 | so that $F$ is surjective. 376 | Hence we've shown that $F$ is bijective. 377 | } 378 | } 379 | 380 | \exercise{3}{ 381 | Show that $1^\k = 1$ for all $\k$ and $0^\k = 0$ for all $\k > 0$. 382 | } 383 | \sol{ 384 | \qproof{ 385 | Suppose that $\k = |A|$ for a set $A$. 386 | 387 | Note first that if $\k = 0$ then by Exercise~5.1.2 it follows that 388 | $$ 389 | 1^\k = 1^0 = 1. 390 | $$ 391 | In the case where $\k > 0$ we claim that there is a unique $f \in 1^A$ so that clearly then 392 | $$ 393 | 1^\k = |1^A| = 1. 394 | $$ 395 | For existence define $f:A \to 1$ by $f(x) = \es$ for all $x \in A$ so that clearly $f \in 1^A$. 396 | For uniqueness consider any $f_1, f_2 \in 1^A$. 397 | Since $1 = \braces{\es}$ it has to be that $f_1(x) = f_2(x) = \es$ for all $x \in A$. 398 | Hence $f_1 = f_2$. 399 | 400 | Now suppose also that $\k > 0$ so that $A \neq \es$. 401 | Hence there is an $a \in A$. 402 | We claim that in this case that $\es^A = \es$ so that 403 | $$ 404 | 0^\k = |\es^A| = |\es| = 0. 405 | $$ 406 | So suppose that $\es^A \neq \es$ so that there \emph{is} an $f \in \es^A$. 407 | Then it's true that for every $a \in A$ there is a unique $b \in \es$ such that $f(a) = b$. 408 | But since there \emph{is} an $a \in A$ this implies that there is also a $b \in \es$, which is a contradiction. 409 | Hence there can be no $f \in \es^A$ so that $\es^A = \es$. 410 | } 411 | } 412 | 413 | \exercise{4}{ 414 | Prove that $\k^\k \leq 2^{\k \cdot \k}$. 415 | } 416 | \sol{ 417 | \qproof{ 418 | Suppose that $\k = |A|$ for a set $A$. 419 | Then we construct an injective $F : A^A \to 2^{A \times A}$ so that 420 | $$ 421 | \k^\k = |A^A| \leq |2^{A \times A}| = |2|^{|A \times A|} = 2^{\k \cdot \k}. 422 | $$ 423 | So for any $f \in A^A$ define $F(f) = g$ where $g \in 2^{A\times A}$ is defined by 424 | $$ 425 | g(a_1, a_2) = 426 | \begin{cases} 427 | 0 & f(a_1) \neq a_2 \\ 428 | 1 & f(a_1) = a_2 429 | \end{cases} 430 | $$ 431 | for $(a_1, a_2) \in A \times A$. 432 | To show that $F$ is injective consider any $f,g \in A^A$ where $f \neq g$. 433 | Then there is an $a \in A$ such that $f(a) \neq g(a)$. 434 | Now let $a_1 = f(a)$ and $a_2 = g(a)$ so that 435 | $$ 436 | f(a) = a_1 \neq a_2 = g(a). 437 | $$ 438 | Since $f(a) = a_1$ it follows by definition that $F(f)(a,a_1) = 1$. 439 | Similarly since $g(a) = a_2 \neq a_1$ it follows that $F(g)(a,a_1) = 0$. 440 | Hence we have 441 | $$ 442 | F(f)(a, a_1) = 1 \neq 0 = F(g)(a, a_1) 443 | $$ 444 | so that clearly $F(f) \neq F(g)$. 445 | Thus $F$ is injective. 446 | } 447 | } 448 | 449 | \exercise{5}{ 450 | If $\abs{A} \leq \abs{B}$ and if $A \neq \es$, then there is a mapping of $B$ onto $A$. 451 | We later show, with the help of the Axiom of Choice, that the converse is also true: If there is a mapping of $B$ onto $A$, then $\abs{A} \leq \abs{B}$. 452 | } 453 | \sol{ 454 | \qproof{ 455 | Suppose that $|A| \leq |B|$ for sets $A$ and $B$ where $A \neq \es$. 456 | Then there is an $a \in A$. 457 | There is also an injective $f : A \to B$ so that $f^{-1}$ is a function from $\ran(f) \to A$. 458 | So let $g$ be a mapping from $B$ to $A$ defined by 459 | $$ 460 | g(b) = \begin{cases} 461 | f^{-1}(b) & b \in \ran(f) \\ 462 | a & b \notin \ran(f). 463 | \end{cases} 464 | $$ 465 | To show that $g$ is onto consider any $x \in A$ and let $b = f(a)$. Thus $b \in \ran(f)$ so that 466 | $$ 467 | g(b) = f^{-1}(b) = f^{-1}(f(a)) = a. 468 | $$ 469 | Hence $g$ is onto since $a$ was arbitrary. 470 | } 471 | } 472 | 473 | \exercise{6} { 474 | If there is a mapping of $B$ onto $A$, then $2^{\abs{A}} \leq 2^{\abs{B}}$. 475 | [Hint: Given $g$ mapping $B$ onto $A$, let $f(X) = \inv{g}[X]$, for all $X \ss A$.] 476 | } 477 | \sol{ 478 | \qproof{ 479 | Suppose that $f$ is a mapping from $B$ \emph{onto} $A$. 480 | We shall construct an injective $F: 2^A \to 2^B$ so that 481 | $$ 482 | 2^{|A|} = |2|^{|A|} = |2^A| \leq |2^B| = |2|^{|B|} = 2^{|B|}. 483 | $$ 484 | So for any $g \in 2^A$ let $F(g) = h$ where $h \in 2^B$ is defined by 485 | $$ 486 | h(b) = g(f(b)) 487 | $$ 488 | for $b \in B$. 489 | To show that $F$ is injective consider any $g_1, g_2 \in 2^A$ where $g_1 \neq g_2$. 490 | Then there is an $a \in A$ such that $g_1(a) \neq g_2(a)$. 491 | Since $f : B \to A$ is onto there is a $b \in B$ such that $f(b) = a$. 492 | Thus we have 493 | $$ 494 | F(g_1)(b) = g_1(f(b)) = g_1(a) \neq g_2(a) = g_2(f(b)) = F(g_2)(b) 495 | $$ 496 | so that $F(g_1) \neq F(g_2)$. 497 | Thus $F$ is injective. 498 | } 499 | } 500 | 501 | \exercise{7}{ 502 | Use Cantor's Theorem to show that ``the set of all sets'' does not exist. 503 | } 504 | \sol{ 505 | \qproof{ 506 | Suppose that $X$ is the set of all sets. 507 | Consider any $Z \in \pset{X}$. 508 | Since clearly $Z$ is a set we have $Z \in X$. 509 | Thus since $Z$ was arbitrary it follows that $\pset{X} \subseteq X$ so that by Exercise~4.1.3 $|\pset{X}| \leq |X|$. 510 | However, this contradicts Cantor's Theorem, according to which $|\pset{X}| > |X|$. 511 | Thus $X$ cannot be the set of all sets. 512 | } 513 | } 514 | 515 | \exercise{8}{ 516 | Let $X$ be a set and let $f$ be a one-to-one mapping of $X$ into itself such that $f[X] \pss X$. 517 | Then $X$ is infinite. 518 | } 519 | \sol{ 520 | \qproof{ 521 | For a set $X$ suppose that $f : X \to X$ is injective. 522 | Also suppose that $\ran(f)$ is a proper subset of $X$. 523 | 524 | Now suppose that $X$ is finite so that there is an $n \in \nats$ such that there is a bijective $g : n \to X$. 525 | We also note that clearly $g^{-1} : X \to n$ is also a bijection. 526 | Now define a function $h : n \to n$ by 527 | $$ 528 | h(k) = (g^{-1} \circ f \circ g)(k) = g^{-1}(f(g(k)) 529 | $$ 530 | for any $k \in n$. 531 | Since $g$, $f$, and $g^{-1}$ are all injective it follows from Exercise~2.3.5 that $h$ is also injective. 532 | 533 | We now claim that $\ran(h)$ is proper subset of $n$. 534 | First we note that for any $m \in n$ we have 535 | $$ 536 | g(h(m)) = g(g^{-1}(f(g(m)))) = f(g(m)) 537 | $$ 538 | since $g$ is a function. 539 | Now, since $\ran(f) \subset X$ there is an $x \in X$ such that $x \notin \ran(f)$. 540 | So let $k = g^{-1}(x)$ so that $g(k) = x$. 541 | Now suppose that there is an $m \in n$ such that $h(m) = k$. 542 | Then per the above we have 543 | $$ 544 | f(g(m)) = g(h(m)) = g(k) = x, 545 | $$ 546 | which is impossible since $x \notin \ran(f)$. 547 | So it must be that there is no such $m$ so that $k \notin \ran(h)$ 548 | Hence since $k \in n$ it follows that $\ran(g) \subset n$. 549 | 550 | Now clearly $h$ is a surjective mapping from $n$ to $\ran(h)$. 551 | But since $h$ is also injective it is thus a bijection from $n$ to $\ran(n)$. 552 | However, according to Lemma~4.2.2 there is no bijective mapping from $n$ to $\ran(n)$ since $\ran(h) \subset n$. 553 | We have thus arrived at a contradiction so that, if the hypotheses hold, then $X$ cannot be finite. 554 | Hence by definition $X$ is infinite. 555 | } 556 | } 557 | 558 | \exercise{9}{ 559 | Every countable set is Dedekind infinite. 560 | } 561 | \sol{ 562 | \begin{lem}\label{lem:card:equidede} 563 | If sets $X$ and $Y$ are equipotent (i.e. $|X| = |Y|$) and $Y$ is Dedekind infinite then $X$ is also Dedekind infinite. 564 | \end{lem} 565 | \qproof{ 566 | Since $X$ and $Y$ are equipotent there is a bijective $f : X \to Y$ so that $\inv{f}$ is also bijective. 567 | Also since $Y$ is Dedekind infinite there is a $Z \subset Y$ such that there is a bijective $g : Y \to Z$ so that $\inv{g}$ is also bijective. 568 | So since $Z \subset Y$ there is a $y \in Y$ such that $y \notin Z$. 569 | So let $S = X - \braces{\inv{f}(y)}$. 570 | Clearly since $\inv{f}(y) \in X$ it follows that $S \subset X$ since $\inv{f}(y) \notin S$. 571 | Now define $h: X \to S$ by 572 | $$ 573 | h(x) = (\inv{f} \circ g \circ f)(x) = \inv{f}(g(f(x)))) 574 | $$ 575 | for $x \in X$. 576 | Since $f$ is a function this implies that 577 | $$ 578 | f(h(x)) = f(\inv{f}(g(f(x)))) = g(f(x)). 579 | $$ 580 | Now suppose for a moment that there is an $x \in X$ such that $h(x) = \inv{f}(y)$. 581 | Then 582 | $$ 583 | g(f(x)) = f(h(x)) = f(\inv{f}(y)) = y, 584 | $$ 585 | which is impossible since $y \notin Z$ but $\ran(g) \subseteq Z$. 586 | Hence there is no such $x$ so that $h$ really is a map from $X$ to $S$ (as opposed to $X$ to $X$). 587 | 588 | Now, since $\inv{f}$, $g$, and $f$ are all injective it follows from Exercise~2.3.5 that $h$ is injective as well. 589 | Then consider any $s \in S$ and let $x = \inv{f}(\inv{g}(f(s)))$, noting that $\inv{g}(f(s))$ exists since $s \neq \inv{f}(y)$. 590 | Then we have 591 | $$ 592 | h(x) = \inv{f}(g(f(\inv{f}(\inv{g}(f(s))))))) = \inv{f}(g(\inv{g}(f(s)))) = \inv{f}(f(s)) = s 593 | $$ 594 | so that $h$ is surjective since $s$ was arbitrary. 595 | Hence $h$ is a bijective map from $X$ to $S$, and since $S \subset X$ this means that $X$ is Dedekind infinite. 596 | } 597 | 598 | \begin{lem}\label{lem:card:natdede} 599 | $\nats$ is Dedekind infinite. 600 | \end{lem} 601 | \qproof{ 602 | Let $N = \nats - \braces{0}$ so that clearly $N$ is proper subset of $\nats$. 603 | Then we define the map $f : \nats \to N$ by 604 | $$ 605 | f(n) = n+1 606 | $$ 607 | for $n \in \nats$. 608 | Consider any $n,m \in \nats$ where $n \neq m$. 609 | Then clearly 610 | \gath{ 611 | n \neq m \\ 612 | n+1 \neq m+1 \\ 613 | f(n) \neq f(m) 614 | } 615 | so that $f$ is injective. 616 | Now consider any $n \in N$ so that clearly $f(n-1) = (n-1) + 1 = n$, noting that since $n \neq 0$ we have $n \geq 1$ so that $n-1 \geq 0$. 617 | Hence $n-1 \in \nats$. 618 | This shows that $f$ is surjective. 619 | Hence $f$ is a bijection from $\nats$ onto a proper subset $N$ so that by definition $\nats$ is Dedekind infinite. 620 | } 621 | 622 | \mainprob 623 | \qproof{ 624 | Suppose that $X$ is a countable set. 625 | Then by definition $X$ is equipotent to $\nats$. 626 | Hence since $\nats$ is Dedekind infinite (Lemma~\ref{lem:card:natdede}) it follows that $X$ is as well by Lemma~\ref{lem:card:equidede}. 627 | } 628 | } 629 | 630 | \exercise{10}{ 631 | If $X$ contains a countable subset, then $X$ is Dedekind infinite. 632 | } 633 | \sol{ 634 | \qproof{ 635 | Suppose that $X$ is a set with a countable subset $Y$. 636 | Then by Exercise~5.1.9 $Y$ is Dedekind infinite so that there is a $Z \subset Y \subseteq X$ such that there is a bijective $f : Y \to Z$. 637 | So define the following $g: X \to X$ by 638 | $$ 639 | g(x) = \begin{cases} 640 | f(x) & x \in Y \\ 641 | x & x \notin Y 642 | \end{cases} 643 | $$ 644 | for any $x \in X$. 645 | Now since $Z \subset Y$ there is a $y \in Y$ such that $y \notin Z$, noting that since $Y \subseteq X$, $y \in X$. 646 | 647 | First we claim that $y \notin \ran(g)$. 648 | So suppose that it is so that there is an $x \in X$ such that $g(x) = y$. 649 | If $x \in Y$ then by definition $g(x) = f(x) = y$, but this is a contradiction since $f : Y \to Z$ but $y \notin Z$. 650 | On the other hand if $x \notin Y$ then we have $g(x) = x = y$, which is also a contradiction since $y = x \in Y$. 651 | Since a contradiction follows in either case it must be that there is no such $x$ so that $y \notin \ran(g)$. 652 | Hence $\ran(g) \subset X$. 653 | 654 | Clearly $g$ is a surjective map from $X$ to $\ran(g)$ so we now show that it is injective. 655 | So consider any $x_1, x_2 \in X$ where $x_1 \neq x_2$. 656 | 657 | Case: $x_1 \in Y$ and $x_2 \in Y$. 658 | Then 659 | $$ 660 | g(x_1) = f(x_1) \neq f(x_2) = g(x_2) 661 | $$ 662 | since $f$ is injective. 663 | 664 | Case: $x_1 \notin Y$ and $x_2 \notin Y$. 665 | Then 666 | $$ 667 | g(x_1) = x_1 \neq x_2 = g(x_2). 668 | $$ 669 | 670 | Case: $x_1 \in Y$ and $x_2 \notin Y$. 671 | Then $g(x_1) = f(x_1) \in Z$ but $g(x_2) = x_2 \notin Y$ so that $x_2 \notin Z$ either since $Z \subset Y$. 672 | Hence $f(x_1) \neq x_2$ so that 673 | $$ 674 | g(x_1) = f(x_1) \neq x_2 = g(x_2). 675 | $$ 676 | Thus in all cases $g(x_1) \neq g(x_2)$ so that $g$ is injective since $x_1$ and $x_2$ were arbitrary. 677 | 678 | Therefore we have shown that $g$ is a bijective map from $X$ to $\ran(g) \subset X$ so that $X$ is Dedekind infinite by definition. 679 | } 680 | } 681 | 682 | \exercise{11}{ 683 | If $X$ is Dedekind infinite, then it contains a countable subset. 684 | [Hint: Let $x \in X - f[X]$; define $x_0 = x$, $x_1 = f(x_0)$, \ldots, $x_{n+1} = f(x_n)$, \ldots. 685 | The set $\braces{x_n \where n \in \nats}$ is countable.] 686 | } 687 | \sol{ 688 | \qproof{ 689 | Suppose that $X$ is a Dedekind infinite set. 690 | Then there is a $Y \subset X$ such that there is a bijective $f : X \to Y$. 691 | Since $Y \subset X$ there is an $x \in X$ such that $x \notin Y$. 692 | So first define $x_0 = x$ and then for $n \in \nats$ define $x_{n+1} = f(x_n)$. 693 | 694 | We claim that $x_n \neq x_m$ for any $n,m \in \nats$ where $n \neq m$, from which it clearly follows that 695 | $$ 696 | Z = \braces{x_n \where n \in \nats} 697 | $$ 698 | is a countable set. 699 | So consider any $n,m \in \nats$ where $n \neq m$ 700 | Without loss of generality we can assume that $n < m$. 701 | Suppose that $x_n = x_m$. 702 | We now show by induction that $x_{n-k} = x_{m-k}$ for all $n \geq k \geq 0$. 703 | If $k=0$ then we clearly have 704 | $$ 705 | x_{n-k} = x_{n-0} = x_n = x_m = x_{m-0} = x_{m-k}. 706 | $$ 707 | Now suppose that $x_{n-k} = x_{m-k}$. 708 | We then have 709 | $$ 710 | f(x_{n-(k+1)}) = f(x_{n-k-1}) = x_{n-k} = x_{m-k} = f(x_{m-k-1}) = f(x_{m-(k+1)}) 711 | $$ 712 | Since $f$ is injective this implies that $x_{n-(k+1)} = x_{m-(k+1)}$ so that inductive proof is complete. 713 | So since this holds for $k=n$ we have that 714 | $$ 715 | x_0 = x_{n-n} = x_{m-n} = f(x_{m-n-1}), 716 | $$ 717 | Noting that $m-n-1 \geq 0$ since $m \geq n +1$. 718 | But $x_0 = x \notin Y$ and $f(x_{m-n-1}) \in Y$ since $f:X \to Y$ so that we have a contradiction. 719 | So it must be that $x_n \neq x_m$. 720 | Hence $Z$ is countable. 721 | Since also clearly $Z \subseteq X$ the proof is complete. 722 | } 723 | } 724 | 725 | \exercise{12}{ 726 | If $A$ and $B$ are Dedekind infinite, the $A \cup B$ is Dedekind infinite. 727 | [Hint: Use Exercise~1.11.] 728 | } 729 | \sol{ 730 | \qproof{ 731 | Suppose that sets $A$ and $B$ are both Dedekind infinite. 732 | Then $A$ contains a countable subset $C$ by Exercise~5.1.11. 733 | Clearly $C \subseteq A \cup B$ so that $C$ is a countable subset of $A \cup B$. 734 | Hence by Exercise~5.1.10 $A \cup B$ is Dedekind infinite. 735 | } 736 | } 737 | 738 | \exercise{13}{ 739 | If $A$ and $B$ are Dedekind infinite, then $A \times B$ is Dedekind infinite. 740 | [Hint: Use Exercise~1.11.] 741 | } 742 | \sol{ 743 | \qproof{ 744 | Suppose that $A$ and $B$ are both Dedekind infinite. 745 | Then $A$ contains a countable subset $C$ by Exercise~5.1.11. 746 | Also since $B$ is Dedekind infinite it is not finite by Exercise~5.1.8. 747 | Hence $B \neq \es$ so that there is a $b \in B$. 748 | Clearly then the set 749 | $$ 750 | D = \braces{(a,b) \where a \in C} 751 | $$ 752 | is a countable subset of $A \times B$ so that $A \times B$ is Dedekind infinite by Exercise~5.1.10. 753 | } 754 | } 755 | 756 | \exercise{14}{ 757 | If $A$ is infinite, then $\pset{\pset{A}}$ is Dedekind infinite. 758 | [Hint: For each $n \in \nats$, let $S_n = \braces{X \pss A \where \abs{X} = n}$. 759 | The set $\braces{S_n \where n \in \nats}$ is a countable subset of $\pset{\pset{A}}$.] 760 | } 761 | \sol{ 762 | \begin{lem}\label{lem:card:infss} 763 | If $A$ is an infinite set then for any $n \in \nats$ there is a $B \subseteq A$ such that $|B| = n$. 764 | \end{lem} 765 | \qproof{ 766 | Suppose that $A$ is an infinite set and consider any $n \in \nats$. 767 | Then $\cnats \leq |A|$ so that there is an injective $f : \nats \to A$. 768 | Now $n \in \nats$ but also $n \subseteq \nats$. 769 | So define the set 770 | $$ 771 | B = \braces{f(k) \where k \in n}. 772 | $$ 773 | Clearly $B \subseteq A$ and we show that $|B| = n$ by defining a mapping $g : n \to B$ by 774 | $$ 775 | g(k) = f(k) 776 | $$ 777 | for $k \in n$. 778 | Since $f$ is injective clearly $g$ is. 779 | Now consider any $b \in B$. 780 | By definition then there is a $k \in n$ such that $f(k) = b$. 781 | Hence $g(k) = f(k) = b$ so that $g$ is surjective. 782 | Hence since $g$ is bijective $|B| = n$ as desired. 783 | } 784 | 785 | \mainprob 786 | \qproof{ 787 | Suppose that $A$ is infinite. 788 | Then for any $n \in \nats$ define 789 | $$ 790 | S_n = \braces{X \in \pset{A} \where |X| = n}, 791 | $$ 792 | noting that $S_n \neq \es$ by Lemma~\ref{lem:card:infss}. 793 | We also note that for $n,m \in \nats$ where $n \neq m$ we have $S_n \neq S_m$ since for any $X \in S_n$ and $Y \in S_m$ we have 794 | $$ 795 | |X| = n \neq m = |Y| 796 | $$ 797 | so that $X \neq Y$. 798 | From this it follows that 799 | $$ 800 | S = \braces{S_n \where n \in \nats} 801 | $$ 802 | is a countable set. 803 | Also, for an $S_n \in S$ we have that each $X \in S_n$ is in $\pset{A}$ so that $S_n \subseteq \pset{A}$. 804 | Hence each $S_n \in \pset{\pset{A}}$. 805 | Thus $S \subseteq \pset{\pset{A}}$. 806 | Since $S$ is countable it then follows that $\pset{\pset{A}}$ is Dedekind infinite by Exercise~5.1.10. 807 | } 808 | } 809 | -------------------------------------------------------------------------------- /sections/sec_5_2.tex: -------------------------------------------------------------------------------- 1 | \subsection{The Cardinality of the Continuum} 2 | 3 | \exercise{1}{ 4 | Prove that the set of all finite sets of reals has cardinality $2^{\al_0}$. 5 | We remark here that the set of all countable sets of reals also has cardinality $2^{\al_0}$, but the proof of this requires the Axiom of Choice. 6 | } 7 | \sol{ 8 | \qproof{ 9 | Let $F$ denote the set of all finite sets of reals. 10 | First we construct an injective $f: F \to \reals^\nats$ 11 | So consider any $A \in F$. 12 | Then $|A|=n$ for an $n \in \nats$ so that there is a finite sequence $\angles{a_k \where k \in n}$ where $\ran(a) = A$. 13 | Now we define an infinite sequence of reals $\hat{a} \in \reals^\nats$ by 14 | $$ 15 | \hat{a}_k = \begin{cases} 16 | a_k & k \in n\, \text{(i.e. $0 \leq k < n$)} \\ 17 | a_0 & k \notin n\, \text{(i.e. $k \geq n$)} 18 | \end{cases} 19 | $$ 20 | so that clearly we have $\ran(\hat{a}) = A$ as well. 21 | Note that this only works if $A \neq \es$ since otherwise there is no $a_0$. 22 | In the case where $A = \es$ we set $\hat{a}_k = k$ for $k \in \nats$ so that $\ran(\hat{a}) = \nats$. 23 | In any case we set $f(A) = \hat{a}$. 24 | 25 | Now we claim that $f$ is injective. 26 | So consider any $A,B \in F$ where $A \neq B$. 27 | If one of them is the empty set, say $A$, then since $B$ is finite $m = \max(\ceil{\max(B)} + 1, 0)$ exists so that clearly $m \notin B$. 28 | Hence $m \notin \ran(f(B)) = B$. 29 | However $m \in \ran(f(A)) = \nats$ since $m \in \nats$. 30 | It thus follows that $\ran(f(B)) \neq \ran(f(A))$ so that $f(A) \neq f(B)$. 31 | On the other hand if neither $A$ nor $B$ is the empty set (but still $A \neq B$) then there is an $a \in A$ where $a \notin B$ or vice versa. 32 | Without loss of generality we need only consider the first case. 33 | Clearly then $a \in \ran(f(A)) = A$ but $a \notin \ran(f(B)) = B$ so that again $f(A) \neq f(B)$. 34 | Hence in all cases we've shown that $f$ is injective. 35 | 36 | Thus we have that 37 | $$ 38 | |F| \leq |\reals^\nats| = \ccont, 39 | $$ 40 | where the last equality was shown in Theorem~5.2.3d. 41 | Now define 42 | $$ 43 | E = \braces{\braces{x} \where x \in \reals} 44 | $$ 45 | so that clearly $E \subseteq F$ and $|E| = |\reals|$. 46 | Hence we have 47 | $$ 48 | \ccont = |\reals| = |E| \leq |F| 49 | $$ 50 | by Exercise 4.1.3. 51 | Thus by the \cbthrm{} $|F| = \ccont$ as required. 52 | } 53 | } 54 | 55 | \exercise{2}{ 56 | A real number $x$ is \emph{algebraic} if it is a solution of some equation 57 | $$ 58 | a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0, 59 | $$ 60 | where $a_0, \ldots, a_n$ are integers. 61 | If $x$ is not algebraic, it is called \emph{transcendental}. 62 | Show that the set of algebraic numbers is countable and hence the set of all transcendental numbers has cardinality $2^{\al_0}$. 63 | } 64 | \sol{ 65 | I did not prove this here as I have already done so when studying Rudin's Principles of Mathematical Analysis, Exercise~2.2. 66 | } 67 | 68 | \exercise{3}{ 69 | If a linearly ordered set $P$ has a countable dense subset, then $\abs{P} \leq 2^{\al_0}$. 70 | } 71 | \sol{ 72 | Note that the countable dense subset is dense in $P$ and not just in itself as explained in the errata list. 73 | \qproof{ 74 | Suppose that $(P,<)$ is our linearly ordered set and $R$ is the countable dense subset of $P$. 75 | We construct an $f : P \to \pset{R}$ by defining 76 | $$ 77 | f(x) = \braces{y \in R \where y < x} 78 | $$ 79 | for any $x \in P$. 80 | Clearly for such $x$ we have that $f(x) \ss R$ so that $f(x) \in \pset{R}$. 81 | 82 | Now we claim that $f$ is injective. 83 | So consider any $x,y \in P$ such that $x \neq y$. 84 | Without loss of generality we can assume that $x < y$. 85 | Since $R$ is dense in $P$ there is a $z \in R$ where $x < z < y$. 86 | From this it follows that $z \in f(y)$ but that $z \notin f(x)$ since it is not true that $z < x$. 87 | Hence clearly $f(x) \neq f(y)$ so that we have shown that $f$ is injective. 88 | 89 | Thus we have 90 | $$ 91 | |P| \leq |\pset{R}| = 2^{|R|} = \ccont, 92 | $$ 93 | where we have used Theorem~5.1.9. 94 | } 95 | } 96 | 97 | \exercise{4}{ 98 | The set of all closed subsets of reals has cardinality $2^{\al_0}$. 99 | } 100 | \sol{ 101 | \qproof{ 102 | Let $C$ denote all the closed subsets of $\reals$ and $O$ the open sets. 103 | We form a mapping $f: C \to O$ defined by 104 | $$ 105 | f(A) = \reals - A 106 | $$ 107 | for $A \in C$. 108 | Clearly by definition $f(A)$ is open for every $A \in C$ since $A$ is closed. 109 | 110 | Now consider any $A,B \in C$ where $A \neq B$. 111 | Then there is an $a \in A$ such that $a \notin B$ or vice versa. 112 | Without loss of generality we can assume the former. 113 | Then since $a \in A$ it follows that $a \notin \reals - A$. 114 | But also since $a \notin B$ (but $a \in \reals$) we have that $a \in \reals - B$. 115 | Thus 116 | $$ 117 | f(A) = \reals - A \neq \reals - B = f(B) 118 | $$ 119 | so that $f$ is injective. 120 | 121 | Now consider any $B \in O$ and let $A = \reals - B$. 122 | $$ 123 | f(A) = \reals - A = \reals - (\reals - B) = B 124 | $$ 125 | so that $f$ is also surjective. 126 | Hence we have that 127 | $$ 128 | |C| = |O| = \ccont 129 | $$ 130 | by Theorem~2.6b. 131 | } 132 | } 133 | 134 | \exercise{5}{ 135 | Show that, for $n > 0$, $n \cdot 2^\ccont = \cnats \cdot 2^\ccont = \ccont \cdot 2^\ccont = 2^\ccont \cdot 2^\ccont = \parens{2^\ccont}^n = \parens{2^\ccont}^\cnats = \parens{2^\ccont}^\ccont = 2^\ccont$. 136 | } 137 | \sol{ 138 | \begin{lem}\label{lem:card:multone} 139 | For any cardinal number $\k$ 140 | $$ 141 | 1 \cdot \k = \k. 142 | $$ 143 | \end{lem} 144 | \qproof{ 145 | Suppose that $\k = |A|$ for a set $A$. 146 | We define $f : A \to 1 \times A$ by 147 | $$ 148 | f(a) = (0, a) 149 | $$ 150 | for $a \in A$, noting that $1 = \braces{0}$. 151 | Clearly by simple inspection this is bijective so that 152 | $$ 153 | 1 \cdot \k = |1 \times A| = |A| = \k 154 | $$ 155 | as desired. 156 | } 157 | 158 | \mainprob 159 | \qproof{ 160 | First we note that clearly since $\cnats \leq \ccont$ we have 161 | $$ 162 | \ccont \leq 2^\ccont 163 | $$ 164 | by property (n) in section~5.1. 165 | So consider any cardinal $n \in \nats$ where $n > 0$ so that $1 \leq n$. 166 | We then have 167 | \ali{ 168 | 2^\ccont &= 1 \cdot 2^\ccont & \text{(by Lemma~\ref{lem:card:multone})} \\ 169 | &\leq n \cdot 2^\ccont \leq \cnats \cdot 2^\ccont \leq \ccont \cdot 2^\ccont \leq 2^\ccont \cdot 2^\ccont & \text{(repeated property (i) of 5.1)} \\ 170 | &= \parens{2^\ccont}^2 & \text{(by property (o) of 5.1)}\\ 171 | &= 2^{2 \cdot \ccont} & \text{(by Theorem 5.1.7b)} \\ 172 | &= 2^\ccont. & \text{(by Theorem 5.2.2b)} 173 | } 174 | We also have 175 | \ali{ 176 | 2^\ccont &= \parens{2^\ccont}^1 & \text{(by Exercise 5.1.2)} \\ 177 | &\leq \parens{2^\ccont}^n \leq \parens{2^\ccont}^\cnats \leq \parens{2^\ccont}^\ccont & \text{(repeated property (n) of 5.1)} \\ 178 | &= 2^{\ccont \cdot \ccont} & \text{(by Theorem 5.1.7b)} \\ 179 | &= 2^\ccont & \text{(by Theorem 5.2.2b)} 180 | } 181 | Clearly these together with the \cbthrm{} shows the desired result. 182 | } 183 | } 184 | 185 | \exercise{6}{ 186 | The cardinality of the set of all discontinuous functions is $2^\ccont$. 187 | [Hint: Using Exercise~2.5, show that $\abs{\reals^\reals - C} = 2^\ccont$ whenever $\abs{C} \leq \ccont$.] 188 | } 189 | \sol{ 190 | \begin{lem}\label{lem:card:contminus} 191 | If $B$ is a set with $|B| = 2^\ccont$ and $A$ is a subset of $B$ with $|A| \leq \ccont$ then $|B - A| = 2^\ccont$. 192 | \end{lem} 193 | \qproof{ 194 | The proof is analogous to that of Theorem~5.2.4. 195 | So suppose that $C$ is a set with $|C| = 2^\ccont$. 196 | Let $B = C \times C$ so that by Exercise~5.2.5 we have 197 | $$ 198 | |B| = |C \times C| = 2^\ccont \cdot 2^\ccont = 2^\ccont. 199 | $$ 200 | Also suppose that $A \subseteq B$ where $|A| = \ccont$. 201 | Now define a set 202 | $$ 203 | P = \braces{x \in C \where \exists y \in C ((x,y) \in A)}. 204 | $$ 205 | Clearly then $|P| \leq |A| = \ccont$. 206 | Since also $|C| = 2^\ccont$ but $P \subseteq C$ it follows that there is an $x_0 \in C$ where $x_0 \notin P$. 207 | If we let $X = \braces{x_0} \times C$ then any $(x,y) \in X$ is not in $A$ so that $(x,y) \in C\times C - A = B - A$. 208 | Hence $X \subseteq B - A$ but also since there is an obvious bijection between $X$ and $C$ we have 209 | $$ 210 | 2^\ccont = |C| = |X| \leq |B-A|. 211 | $$ 212 | Since also clearly $B - A \subseteq B$ we also have that 213 | $$ 214 | |B-A| \leq |B| = 2^\ccont. 215 | $$ 216 | Hence by the \cbthrm{} $|B-A| = 2^\ccont$ as desired. 217 | } 218 | 219 | \mainprob 220 | \qproof{ 221 | By Lemma~5.2.7 $|\reals^\reals| = 2^\ccont$. 222 | Also by Theorem~5.2.6a the set $C$ of all continuous $f: \reals \to \reals$ has cardinality of $\ccont$. 223 | Thus clearly the set of all discontinuous functions from $\reals \to \reals$ is simply 224 | $$ 225 | D = \reals^\reals - C. 226 | $$ 227 | But then by Lemma~\ref{lem:card:contminus} above we have that 228 | $$ 229 | |D| = |\reals^\reals| = 2^\ccont 230 | $$ 231 | as desired. 232 | } 233 | } 234 | 235 | \exercise{7}{ 236 | Construct a one-to-one mapping of $\reals \times \reals$ onto $\reals$. 237 | [Hint: If $a,b \in [0, 1]$ have decimal expansions $0.a_1 a_2 a_3 \cdots$ and $0.b_1 b_2 b_3 \cdots$, map the ordered pair $(a,b)$ onto $0.a_1 b_1 a_2 b_2 a_3 b_3 \cdots \in [0,1]$. 238 | Make adjustments to avoid sequences where the digit 9 appears from some place onward.] 239 | } 240 | \sol{ 241 | Skipping this problem due to the obviousness of it in principle but the fact that the details are tedious, and I have proved similar problems when studying Rudin's Principles of Mathematical Analysis. 242 | } 243 | -------------------------------------------------------------------------------- /sections/sec_6_1.tex: -------------------------------------------------------------------------------- 1 | \subsection{Well-Ordered Sets} 2 | 3 | \exercise{1}{ 4 | Give an example of a linearly ordered set $(L,<)$ and an initial segment $S$ of $L$ which is not of the form $\braces{x \where x < a}$, for any $a \in L$. 5 | } 6 | \sol{ 7 | We claim that $L = \reals$ and $S = \braces{x \in L \where x \leq 0}$ with the usual order meet the criteria. 8 | \qproof{ 9 | First, clearly $L = \reals$ is a linearly ordered set. 10 | So consider any $a \in S$ and any $x < a$ so that we have 11 | $$ 12 | x < a \leq 0. 13 | $$ 14 | Hence $x \in S$ also so that by definition $S$ is an initial segment of $\reals$. 15 | Now suppose that $S$ does have the form 16 | $$ 17 | S = \braces{x \in L \where x < a} 18 | $$ 19 | for some $a \in L$. 20 | Since $0 \leq 0$ clearly $0 \in S$ by the original definition so that by the above $0 < a$. 21 | But now consider $a/2$, which is clearly in $L = \reals$. 22 | By the above $a/2 < a$ since $a > 0$ so $a/2 \in S$ but we also have $0 < a/2$ (hence it is not true that $a/2 \leq 0$) since $0 < a$ so that by the original definition $a/2 \notin S$. 23 | Since we have a contradiction it must be that $S$ cannot be expressed in such a form. 24 | } 25 | } 26 | 27 | \exercise{2}{ 28 | $\w + 1$ is not isomorphic to $\w$ (in the well-ordering by $\in$). 29 | } 30 | \sol{ 31 | \qproof{ 32 | Since $\w = \nats$ and $\w + 1 = \w \cup \braces{\w}$ clearly $\w$ is a proper subset of $\w + 1$ (since $\w \notin \w$ but $\w \in \w + 1$). 33 | Now consider any $a \in \w = \nats$ and any $x < a$. 34 | Then clearly also $x \in \nats$ so $x \in \w$. 35 | Thus $\w$ is an initial segment of $\w + 1$. 36 | Then, since it has already been shown that both $\w$ and $\w + 1$ are well-ordered sets, it follows from Corollary~6.1.5a that they cannot be isomorphic. 37 | } 38 | } 39 | 40 | \exercise{3}{ 41 | There exist $\ccont$ well-orderings of the set of natural numbers. 42 | } 43 | \sol{ 44 | \begin{lem}\label{lem:ord:isnats} 45 | Suppose that $A$ is a subset of $\nats$ (including $A = \nats$). 46 | Then every initial segment of $A$ with the standard ordering is finite. 47 | \end{lem} 48 | \qproof{ 49 | Consider any initial segment $S$ of $(A, <)$. 50 | Then by Lemma~6.1.2 there is an $n \in A \ss \nats$ such that $S = \braces{k \in A \where k < n}$. 51 | So consider any $k \in S$ so that $k \in A \ss \nats$ and $k < n$. 52 | Then by the definition of $<$ we have that $k \in n$. 53 | Since $k$ was arbitrary this shows that $S \ss n$ so that $|S| \leq n$. 54 | From this it clearly follows that $S$ is finite since $n$ is. 55 | } 56 | 57 | \mainprob 58 | \qproof{ 59 | Throughout the following let $<$ denote the standard well-ordering on $\nats$ and let $R$ be the set of all well-orderings defined on $\nats$. 60 | 61 | First we construct an injective $F: R \to \nats^\nats$. 62 | So for any $\prec \in R$ we have that $(\nats, <)$ and $(\nats, \prec)$ are two well-orderings of $\nats$. 63 | Consider then Theorem~6.1.3. 64 | We show that (c) cannot be the case, i.e. that an initial segment of $(\nats, <)$ cannot be isomorphic to $(\nats, \prec)$. 65 | So suppose that this is the case so that $f$ is an isomorphism from an initial segment $S$ of $(\nats, <)$ to $(\nats, \prec)$. 66 | Then by Lemma~\ref{lem:ord:isnats}, $S$ is finite whereas $\nats$ is infinite, but since $f$ is a bijection this is impossible since it would imply that $|S| = |\nats| = \cnats$. 67 | Hence it must be that (a) $(\nats, <)$ is isomorphic to $(\nats, \prec)$ or (b) the former is isomorphic to an initial segment of the latter. 68 | In either case such an isomorphism $f$ is unique by Corollary~6.1.5c. 69 | So define $F(\prec) = f$, noting that clearly $f \in \nats^\nats$. 70 | 71 | Now we show that $F$ is injective by considering two $\prec_1, \prec_2 \in R$ where $\prec_1 \neq \prec_2$. 72 | Without loss of generality we can the assume that there is an $(n,m) \in \prec_1$ where $(n,m) \notin \prec_2$. 73 | Thus $n \prec_1 m$ but since $\prec_2$ is a linear, strict ordering and $\lnot (n \prec_2 m)$ it has to be that $m \prec_2 n$ since $n \neq m$. 74 | Now let $f_1 = F(\prec_1)$ and $f_2 = F(\prec_2)$. 75 | Since both $f_1$ and $f_2$ are bijective there are $k_1,l_1,k_2,l_2 \in \nats$ such that 76 | \ali{ 77 | f_1(k_1) &= n & f_2(k_2) &= n \\ 78 | f_1(l_1) &= m & f_2(l_2) &= m 79 | } 80 | Since $f_1$ is an isomorphism and $f_1(k_1) = n \prec_1 m = f_1(l_1)$ it follows that 81 | $$ 82 | k_1 < l_1 83 | $$ 84 | and similarly since $f_2$ is an isomorphism and $f_2(l_2) = m \prec_2 n = f_2(k_2)$ it follows that 85 | $$ 86 | l_2 < k_2. 87 | $$ 88 | Now we claim that either $f_1(k_1) \neq f_2(k_1)$ or $f_1(l_1) \neq f_2(l_1)$. 89 | Either case shows that $F(\prec_1) = f_1 \neq f_2 = F(\prec_2)$ so that $F$ is injective. 90 | To this end suppose that $f_1(k_1) = f_2(k_1) = n = f_2(k_2)$. 91 | Then since $f_2$ is injective it follows that $k_1 = k_2$. 92 | Hence with the above we have 93 | $$ 94 | l_2 < k_2 = k_1 < l_1 95 | $$ 96 | so that $m = f_2(l_2) \prec_2 f_2(l_1)$ and hence $m \neq f_2(l_1)$. 97 | Thus we have $f_1(l_1) = m \neq f_2(l_1)$ so that the disjunction is shown (since $\lnot P \to Q \equiv P \lor Q$) and $F$ is injective. 98 | 99 | Hence since $F: R \to \nats^\nats$ is injective we have that 100 | $$ 101 | |R| \leq |\nats^\nats| = \cnats^\cnats = \ccont 102 | $$ 103 | by Theorem~5.2.2c. 104 | 105 | Now suppose that $B$ is the set of all bijections from $\nats$ to $\nats$. 106 | We then construct an injective $G: 2^\nats \to B$. 107 | So for any infinite sequence $a \in 2^\nats$ we define an $f \in \nats^\nats$ by 108 | \ali{ 109 | f(2n) &= \begin{cases} 110 | 2n & a_n = 0 \\ 111 | 2n+1 & a_n = 1 112 | \end{cases} 113 | & 114 | f(2n+1) &= \begin{cases} 115 | 2n+1 & a_n = 0 \\ 116 | 2n & a_n = 1. 117 | \end{cases} 118 | } 119 | for $n \in \nats$, i.e we swap $2n$ and $2n+1$ if $a_n = 1$ and leave them alone if $a_n = 0$. 120 | We then assign $G(a) = f$. 121 | It is trivial but tedious to show that $f$ is bijective so that indeed $f \in B$. 122 | 123 | Now consider any $a, b \in 2^\nats$ where $a \neq b$ and let $f = G(a)$ and $g = G(b)$. 124 | Since $a \neq b $ there is an $n \in \nats$ where $a_n \neq b_n$. 125 | Without loss of generality we can assume that $a_n = 0 \neq 1 = b_n$. 126 | Then we have 127 | $$ 128 | f(2n) = 2n \neq 2n+1 = g(2n) 129 | $$ 130 | since $a_1 = 0$ but $b_n = 1$. 131 | Hence $f \neq g$ so that $G$ is injective, from which it follows that $|2^\nats| \leq |B|$. 132 | 133 | Lastly we construct an injective $H : B \to R$. 134 | So for an $f \in B$ define 135 | $$ 136 | \prec = \braces{(f(n), f(m)) \where (n,m) \in \nats \times \nats \land n < m} 137 | $$ 138 | and set $H(f) = \prec$. 139 | Clearly by definition since $f$ is bijective it is an isomorphism from $(\nats, <)$ to $(\nats, \prec)$. 140 | This means that $(\nats, \prec)$ is isomorphic to $(\nats, <)$ so that clearly $\prec$ is a well-ordering since $<$ is. 141 | Hence indeed $H(f) = \prec \in R$. 142 | 143 | Now we show that $H$ is injective. 144 | So consider $f_1, f_2 \in B$ where $\prec_1 = H(f_1) = H(f_2) = \prec_2$. 145 | Then $\inv{f_1} \circ f_2$ is an isomorphism from $(\nats, \prec_1)$ to $(\nats, \prec_2)$ 146 | But since $\prec_1 = \prec_2$ these are the same well-ordered set so that it follows from Corollary~6.1.5b that the only isomorphism between them is the identity $i_\nats$. 147 | Hence $\inv{f_1} \circ f_2 = i_\nats$, from which it follows that $f_1 = f_2$. 148 | Therefore $H$ is injective so that $|B| \leq |R|$. 149 | 150 | Putting this together results in 151 | $$ 152 | \ccont = |2^\nats| \leq |B| \leq |R|. 153 | $$ 154 | It then follows from the \cbthrm{} that $|R| = \ccont$ as desired. 155 | } 156 | } 157 | 158 | \exercise{4}{ 159 | For every infinite subset $A$ of $\nats$, $(A,<)$ is isomorphic to $(\nats,<)$. 160 | } 161 | \sol{ 162 | \qproof{ 163 | Let $A$ be an infinite subset of $\nats$. 164 | Then $(A, <)$ (where $<$ is the standard well-ordering of $\nats$) is a well-ordering since any $B \ss A$ is also a subset of $\nats$ and therefore has a least element. 165 | Hence by Theorem~6.1.3 either: 166 | \begin{enumerate} 167 | \item $(A,<)$ and $(\nats,<)$ are isomorphic, 168 | \item An initial segment of $(A,<)$ is isomorphic to $(\nats,<)$, or 169 | \item $(A,<)$ is isomorphic to an initial segment of $(\nats,<)$ 170 | \end{enumerate} 171 | We show that they must be isomorphic (1) by showing that (2) and (3) lead to contradictions. 172 | 173 | Suppose (2), i.e. that an initial segment $S$ of $(A,<)$ is isomorphic to $(\nats,<)$. 174 | Then since $A \ss \nats$ it follows from Lemma~\ref{lem:ord:isnats} that $S$ is finite. 175 | But since this is isomorphic to $\nats$ it means that $|S| = |\nats| = \cnats$, which is a contradiction! 176 | 177 | Now suppose (3) so that $(A,<)$ is isomorphic to an initial segment $S$ of $(\nats, <)$. 178 | Again Lemma~\ref{lem:ord:isnats} tells us that $S$ is finite whereas $A$ is infinite. 179 | But since they are isomorphic this implies that $|S| = |A| = \cnats$, which is again a contradiction! 180 | 181 | Hence it has to be that $(A,<)$ and $(\nats,<)$ are isomorphic. 182 | } 183 | } 184 | 185 | \exercise{5}{ 186 | Let $(W_1, <_1)$ and $(W_2,<_2)$ be disjoint well-ordered sets, each isomorphic to $(\nats,<)$. 187 | Show that the sum of the two linearly ordered sets (as defined in Lemma~4.5 in Chapter~4) is a well-ordering, and is isomorphic to the ordinal number $\w + \w = \braces{0,1,2,\ldots,\w,\w+1,\w+2,\ldots}$. 188 | } 189 | \sol{ 190 | \qproof{ 191 | Suppose that $(W, \prec)$ is the sum and associated order as defined in Lemma~4.4.5. 192 | By that lemma $\prec$ is a linear ordering but we must show that it is also a well-ordering. 193 | 194 | First we note that clearly $W_1$ and $W_2$ are both well-orderings since they are both isomorphic to $(\nats, <)$. 195 | So consider any non-empty subset of $A$ of $W = W_1 \cup W_2$. 196 | Let $A_1 = A \cap W_1$ and $A_2 = A \cap W_2$ so that clearly they are disjoint since $W_1$ and $W_2$ are and $A_1 \ss W_1$ and $A_2 \ss W_2$. 197 | Also since $A$ is not empty either $A_1$ or $A_2$ (or both) are also not empty. 198 | If $A_1$ is not empty then since $A_1 \ss W_1$ and $(W_1, <_1)$ is a well-ordering there is a least element $a \in A_1$. 199 | Otherwise if $A_1$ is empty then $A_2$ is not and it has a least element $a$ since it is a non-empty subset of the well-ordered $(W_2, <_2)$. 200 | Now consider any $b \in A$ so that also $b \in W$. 201 | If $b \in W_1$ then $b \in A_1$ so that $A_1$ is not empty. 202 | In this case since $a$ is the least element of $A_1$ we have $a \leq_1 b$ so that by definition $a \prece b$. 203 | On the other hand if $b \in W_2$ then $b \in A_2$. 204 | If $A_1$ was empty then $a$ is the least element of $A_2$ and $b \in A_2$ so that again $a \leq_2 b$, hence by definition $a \prece b$. 205 | If $A_1$ is not empty then $a \in A_1 \ss W_1$ so that by the definition of the sum $(W, \prec)$ we have that $a \prec b$ since $b \in W_2$. 206 | Hence also $a \prece b$. 207 | Thus in all cases $a \prece b$ so that $a$ is the least element of $A$ since $b$ was arbitrary. 208 | 209 | Now we show that $(W, \prec)$ is isomorphic to $(\w + \w, <)$. 210 | First, since $(W_1, <_1)$ and $(W_2, <_2)$ are both isomorphic to $(\nats, <)$ let $f_1 : W_1 \to \nats$ and $f_2 : W_2 \to \nats$ be isomorphisms. 211 | Now we define $g : W \to \w + \w$ by 212 | $$ 213 | g(w) = \begin{cases} 214 | f_1(w) & w \in W_1 \\ 215 | \w + f_2(w) & w \in W_2 216 | \end{cases} 217 | $$ 218 | for $w \in W = W_1 \cup W_2$, noting that $g$ is well defined since $W_1$ and $W_2$ are disjoint. 219 | Clearly since $\ran(f_1) = \ran(f_2) = \nats$ we have that $g(w) \in \w + \w$ for all $w \in W$. 220 | 221 | Consider any $k \in \w + \w$ so that $k \in \nats$ or $k = \w + n$ for some $n \in \nats$. 222 | In the former case let $w = \inv{f_1}(k)$, which exists since $f_1$ is bijective. 223 | Thus $w \in W_1$ so that by definition $g(w) = f_1(w) = k$. 224 | In the latter case let $w = \inv{f_2}(n)$, which exists since $f_2$ is bijective. 225 | Thus $w \in W_2$ so that by definition $g(w) = \w + f_2(w) = \w + n = k$. 226 | This shows that $g$ is surjective. 227 | 228 | Now we show that $g$ is an increasing function. 229 | So consider any $w_1, w_2 \in W$ where $w_1 \prec w_2$. 230 | 231 | Case: $w_1, w_2 \in W_1$. 232 | Then since $w_1 \prec w_2$ we have that $w_1 <_1 w_2$. 233 | It then follows that $g(w_1) = f(w_1) < f_(w_2) = g(w_2)$ since $f_1$ is an isomorphism. 234 | 235 | Case: $w_1, w_2 \in W_2$. 236 | Then since $w_1 \prec w_2$ we have that $w_1 <_2 w_2$. 237 | It then follows that $f_2(w_1) < f_2(w_2)$ since $f_2$ is an isomorphism. 238 | Hence we clearly then have $g(w_1) = \w + f_2(w_1) < \w + f_2(w_2) = g(w_2)$. 239 | 240 | Case: $w_1 \in W_1$ and $w_2 \in W_2$. 241 | Then we have that $g(w_1) = f_1(w_1) \in \nats$ and $g(w_2) = \w + f_2(w_2)$ so that clearly $g(w_1) < \w \leq g(w_2)$ since $f_2(w_2) \in \nats$. 242 | 243 | Case: $w_2 \in W_1$ and $w_1 \in W_2$. 244 | If this were the case then by the definition of $\prec$ we would have that $w_2 \prec w_1$, which contradicts the established hypothesis that $w_1 \prec w_2$. 245 | Hence this case is impossible. 246 | 247 | Hence in all cases $g(w_1) < g(w_2)$ so that $g$ is increasing. 248 | Therefore it is also injective and an isomorphism (since we've shown that it is surjective as well). 249 | Thus we've shown that $W$ is isomorphic to $\w + \w$ as desired. 250 | } 251 | } 252 | 253 | \exercise{6}{ 254 | Show that the lexicographic product $(\nats \times \nats, <)$ (see Lemma~4.6 in Chapter~4) is isomorphic to $\w \cdot \w$. 255 | } 256 | \sol{ 257 | \qproof{ 258 | Suppose that $\prec$ is the lexicographic ordering of $\nats \times \nats$. 259 | Now we define $f: \nats \times \nats \to \w \cdot \w$ by 260 | $$ 261 | f(n,m) = \w \cdot n + m 262 | $$ 263 | for any $(n,m) \in \nats \times \nats$. 264 | Clearly $f(n,m) \in \w \times \w$. 265 | 266 | First we show that $f$ is surjective. 267 | So consider any $k \in \w \cdot \w$ so that there are $n,m \in \nats$ where $k = \w \cdot n + m$. 268 | Then we clearly have that $f(n,m) = \w \cdot n + m = k$. 269 | Since clearly $(n,m) \in \nats \times \nats$ it follows that $f$ is surjective. 270 | 271 | Now we show that $f$ is an increasing function. 272 | To this end consider any $(n_1, m_1), (n_2, m2) \in \nats \times \nats$ where $(n_1, m_1) \prec (n_2, m_2)$. 273 | 274 | Case: $n_1 = n_2$. 275 | Then since $(n_1, m_1) \prec (n_2, m_2)$ it must be that $m_1 < m_2$. 276 | Hence we have that $f(n_1, m_1) = \w \cdot n_1 + m_1 = \w \cdot n_2 + m_1 < \w \cdot n_2 + m_2 = f(n_2, m_2)$. 277 | 278 | Case: $n_1 \neq n_2$. 279 | Then since $(n_1, m_1) \prec (n_2, m_2)$ it must be that $n_1 < n_2$. 280 | Hence we have that $f(n_1, m_1) = \w \cdot n_1 + m_1 < \w \cdot n_2 \leq \w \cdot n_2 + m_2 = f(n_2, m_2)$. 281 | 282 | Thus in all cases $f(n_1, m_1) < f(n_2, m_2)$ so that $f$ is increasing. 283 | It then follows that $f$ is injective and isomorphic. 284 | Hence $(\nats \times \nats, \prec)$ is isomorphic to $\w \cdot \w$. 285 | } 286 | } 287 | 288 | \exercise{7}{ 289 | Let $(W,<)$ be a well-ordered set, and let $a \notin W$. 290 | Extend $<$ to $W' = W \cup \braces{a}$ by making $a$ greater than all $x \in W$. 291 | Then $W$ has a smaller order type than $W'$. 292 | } 293 | \sol{ 294 | This problem is looking ahead to future sections where order types and how to compare them are defined. 295 | \qproof{ 296 | Suppose that $\a$ is the order type of $W$ and that $f : W \to \a$ is the isomorphism. 297 | Now let $\b = S(\a) = \a \cup \braces{\a}$. 298 | We then claim that $W'$ is isomorphic to $\b$. 299 | So define a $g : W' \to \b$ by 300 | $$ 301 | g(w) = \begin{cases} 302 | f(w) & w \in W \\ 303 | \a & w \notin W 304 | \end{cases} 305 | $$ 306 | for $w \in W'$. 307 | Clearly we have that $g(w) \in \b$ for any $w \in W'$. 308 | 309 | Now consider any $x \in \b$. 310 | If $x = \a$ then set $w = a \notin W$ so that $g(w) = \a = x$. 311 | If $x \neq \a$ then $x \in \a$ so set $w = \inv{f}(x)$ so that then $w \in W$. 312 | We then have that $g(w) = f(w) = f(\inv{f}(x)) = x$. 313 | Therefore $g$ is surjective. 314 | 315 | Now consider any $w_1, w_2 \in W'$ where $w_1 < w_2$ 316 | 317 | Case: $w_1, w_2 \in W$. 318 | Then since $f$ is an isomorphism and $w_1 < w_2$ we have that $g(w_1) = f(w_1) < f_(w_2) = g(w_2)$. 319 | 320 | Case: $w_1 \in W$ and $w_2 = a$. 321 | Then $g(w_1) = f(w_1) \in \a$ and $w_2 \notin W$ so that $g(w_2) = \a$. 322 | Hence $g(w_1) \in g(w_2)$ so that by the definition of $<$ we have that $g(w_1) < g(w_2)$. 323 | 324 | Note that these cases are exhaustive since it can't be that $w_1 = w_2 = \a$ since $w_1 < w_2$ (and therefore $w_1 \neq w_2$). 325 | It also cannot be that $w_2 \in W$ but $w_1 = a$ since then it would be that $w_2 \leq w_1$ since $a$ is the greatest element of $W'$, which contradicts $w_1 < w_2$. 326 | Thus in all cases $g(w_1) < g(w_2)$ so that $g$ is increasing, and therefore injective and an isomorphism. 327 | 328 | Hence $\b$ is the order type of $W'$, $\a$ is the order type of $W$, and $\a < \b$ since $\a \in \b$. 329 | } 330 | } 331 | 332 | \exercise{8}{ 333 | The sets $W = \nats \times \braces{0,1}$ and $W' = \braces{0,1} \times \nats$, ordered lexicographically, are nonisomorphic well-ordered sets. 334 | (See the remark following Theorem~4.7 in Chapter~4.) 335 | } 336 | \sol{ 337 | \qproof{ 338 | Let $\prec$ be the lexicographic ordering of $W = \nats \times \braces{0,1}$ and $\prec'$ be the lexicographic ordering of $W' = \braces{0,1} \times \nats$. 339 | 340 | First we define $f : W \to \w$ by 341 | $$ 342 | f(n,m) = 2n + m 343 | $$ 344 | for $(n, m) \in W$. 345 | Clearly each $f(n,m) \in \nats = \w$. 346 | 347 | Now consider any $k \in \w = \nats$. 348 | If $k$ is even then $k = 2n$ for some $n \in \nats$ so set $w = (n, 0) \in W$. 349 | Then clearly $f(w) = f(n,0) = 2n = k$. 350 | On the other hand if $k$ is even then $k = 2n + 1$ for some $n \in \nats$ so set $w = (n,1) \in W$. 351 | Then clearly $f(w) = f(n,1) = 2n+1 = k$. 352 | This shows that $f$ is surjective. 353 | 354 | Now consider any $w_1 = (n_1,m_1)$ and $w_2 = (n_2, m_2)$ in $W$ where $w_1 \prec w_2$. 355 | 356 | Case: $n_1 = n_2$. 357 | Then since $w_1 \prec w_2$ it has to be that $m_1 < m_2$, and since $m_1,m_2 \in \braces{0,1}$ it has to be that $m_1=0$ and $m_2 = 1$. 358 | From this it follows that 359 | $$ 360 | f(w_1) = f(n_1,m_1) = f(n_1,0) = 2n_1 < 2n_1 + 1 = 2n_2 + 1 = f(n_2,1) = f(n_2,m_2) = f(w_2). 361 | $$ 362 | 363 | Case: $n_1 \neq n_2$. 364 | Then since $w_1 \prec w_2$ it has to be that $n_1 < n_2$. 365 | Then $n_1 + 1 \leq n_2$ and since also $m_1 < 2$ we have 366 | $$ 367 | f(w_1) = f(n_1, m_1) = 2n_1 + m_1 < 2n_1 + 2 = 2(n_1 + 1) \leq 2n_2 \leq 2n_2 + m_2 = f(n_1, m_2) = f(w_2). 368 | $$ 369 | Hence in all cases $f(w_1) < f(w_2)$ so that $f$ is increasing and therefore injective and isomorphic. 370 | Therefore $W$ is isomorphic to $\w$. 371 | 372 | Now we define $g: W' \to \w+\w$ by 373 | $$ 374 | g(n,m) = \begin{cases} 375 | m & n = 0 \\ 376 | \w + m & n = 1 377 | \end{cases} 378 | $$ 379 | for $(n,m) \in W'$. 380 | Clearly since $m \in \nats$ we have that $g(n,m) \in \w+\w$ for all $(n,m) \in W'$. 381 | 382 | Now consider any $\a \in \w + \w$. 383 | If $\a \in \w = \nats$ then $(0, \a) \in W'$ and $g(0,\a) = \a$. 384 | On the other hand if $\a = \w + m$ for some $m \in \nats$ then $(1, m) \in W'$ and $g(1,m) = \w + m = \a$. 385 | Therefore $g$ is surjective. 386 | 387 | Now consider any $w_1 = (n_1,m_1)$ and $w_2 = (n_2, m_2)$ in $W'$ where $w_1 \prec' w_2$. 388 | 389 | Case: $n_1 = n_2$. 390 | Then since $w_1 \prec' w_2$ it has to be that $m_1 < m_2$. 391 | If $n_1 = n_2 = 0$ then 392 | $$ 393 | g(w_1) = g(n_1,m_1) = g(0,m_1) = m_1 < m_2 = g(0, m_2) = g(n_2,m_2) = g(w_2). 394 | $$ 395 | On the other hand if $n_1 = n_2 = 1$ then 396 | $$ 397 | g(w_1) = g(n_1,m_1) = g(1,m_1) = \w + m_1 < \w + m_2 = g(1, m_2) = g(n_2,m_2) = g(w_2). 398 | $$ 399 | 400 | Case: $n_1 \neq n_2$. 401 | Then since $w_1 \prec' w_2$ it has to be that $n_1 < n_2$. 402 | Moreover since $n_1,n_2 \in \braces{0,1}$ it has to be that $n_1 = 0$ and $n_2 = 1$ so that 403 | $$ 404 | g(w_1) = g(n_1,m_1) = g(0,m_1) = m_1 < \w + m_2 = g(1,m_2) = g(n_2,m_2) = g(w_2). 405 | $$ 406 | 407 | Hence in all cases $g(w_1) < g(w_2)$ so that $g$ is increasing and therefore injective and an isomorphism. 408 | Therefore $W'$ is isomorphic to $\w + \w$. 409 | 410 | Now, since $w \in \w+\w$ we have that $\w < \w+\w$ and so are distinct ordinals. 411 | Therefore by the remarks following Theorem~6.2.10 $\w$ and $\w+\w$ are not isomorphic. 412 | If $W$ and $W'$ were isomorphic with $h$ as the isomorphism then $g \circ h \circ \inv{f}$ would be an isomorphism from $\w$ to $\w + \w$, which is impossible. 413 | So it must be that $W$ and $W'$ are not isomorphic. 414 | } 415 | } 416 | -------------------------------------------------------------------------------- /sections/sec_6_2.tex: -------------------------------------------------------------------------------- 1 | \subsection{Ordinal Numbers} 2 | 3 | \exercise{1}{ 4 | A set $X$ is transitive if and only if $X \ss \pset{X}$. 5 | } 6 | \sol{ 7 | \qproof{ 8 | ($\to$) Suppose that $X$ is a transitive set and consider any $x \in X$. 9 | Then $x \ss X$ since $X$ is transitive. 10 | Thus $x \in \pset{X}$ so that, since $x$ was arbitrary, $X \ss \pset{X}$. 11 | 12 | ($\leftarrow$) Now suppose that $X \ss \pset{X}$ and consider any $x \in X$. 13 | Then also $x \in \pset{X}$ so that $x \ss X$. 14 | Hence, since $x$ was arbitrary, $X$ is transitive by definition. 15 | } 16 | } 17 | 18 | \exercise{2}{ 19 | A set $X$ is transitive if and only if $\bigcup X \ss X$. 20 | } 21 | \sol{ 22 | \qproof{ 23 | ($\to$) Suppose that $X$ is transitive and consider any $y \in \bigcup X$. 24 | Then there is an $x \in X$ such that $y \in x$. 25 | Since $X$ is transitive and $x \in X$ we have that $x \ss X$ so that $y \in X$ as well. 26 | Since $y$ was arbitrary this shows that $\bigcup X \ss X$. 27 | 28 | ($\leftarrow$) Now suppose that $\bigcup X \ss X$ and consider any $x \in X$. 29 | If $x = \es$ then clearly $x \ss X$. 30 | So suppose that $x \neq \es$ and consider any $y \in x$. 31 | Then since $x \in X$ it follows that $y \in \bigcup X$ so that also $y \in X$. 32 | So since $y$ was arbitrary it follows that $x \ss X$. 33 | Since $x$ was arbitrary by definition $X$ is transitive. 34 | } 35 | } 36 | 37 | \exercise{3}{ 38 | Are the following sets transitive? 39 | 40 | (a) $\braces{\es, \braces{\es}, \braces{\braces{\es}}}$, 41 | 42 | (b) $\braces{\es, \braces{\es}, \braces{\braces{\es}}, \braces{\es, \braces{\es}}}$, 43 | 44 | (c) $\braces{\es, \braces{\braces{\es}}}$. 45 | } 46 | \sol{ 47 | 48 | (a) We claim that $X = \braces{\es, \braces{\es}, \braces{\braces{\es}}}$ is transitive. 49 | \qproof{ 50 | Suppose $x \in X$. 51 | If $x = \es$ then obviously $x \ss X$. 52 | If $x = \braces{\es}$ then $x \ss X$ since $\es \in X$. 53 | If $x = \braces{\braces{\es}}$ then $x \ss X$ since $\braces{\es} \in X$. 54 | Thus since the cases are exhaustive we've shown that $x \ss X$ so that $X$ is transitive by definition. 55 | } 56 | 57 | (b) We claim that $X = \braces{\es, \braces{\es}, \braces{\braces{\es}}, \braces{\es, \braces{\es}}}$ is transitive. 58 | \qproof{ 59 | For $x \in X$ the three cases in part (a) above have the same results and, if $x = \braces{\es, \braces{\es}}$, then $x \ss X$ since $\es \in X$ and $\braces{\es} \in X$. 60 | Hence again $X$ is transitive by definition. 61 | } 62 | 63 | (c) We claim that $X = \braces{\es, \braces{\braces{\es}}}$ is \emph{not} transitive. 64 | \qproof{ 65 | If $x = \braces{\braces{\es}}$ we have that $x$ is not a subset of $X$ since $\braces{\es} \in x$ but $\braces{\es} \notin X$. 66 | Hence $X$ is not transitive. 67 | } 68 | } 69 | 70 | \exercise{4}{ 71 | Which of the following statements are true? 72 | 73 | (a) If $X$ and $Y$ are transitive, then $X\cup Y$ is transitive. 74 | 75 | (b) If $X$ and $Y$ are transitive, then $X \cap Y$ is transitive. 76 | 77 | (c) If $X \in Y$ and $Y$ is transitive, then $X$ is transitive. 78 | 79 | (d) If $X \ss Y$ and $Y$ is transitive, then $X$ is transitive. 80 | 81 | (e) If $Y$ is transitive and $S \ss \pset{Y}$, then $Y \cup S$ is transitive. 82 | } 83 | \sol{ 84 | 85 | (a) We claim that this is true. 86 | \qproof{ 87 | Consider any $x \in X \cup Y$. 88 | If $x \in X$ then $x \ss X$ since $X$ is transitive. 89 | Since also $X \ss X \cup Y$ we clearly have that $x \ss X \ss X \cup Y$. 90 | We can make the same argument if it is the case that $x \in Y$. 91 | Hence since $x$ was arbitrary this shows that $X \cup Y$ is indeed transitive. 92 | } 93 | 94 | (b) We claim that this is true. 95 | \qproof{ 96 | Consider any $x \in X \cap Y$. 97 | Then $x \in X$ and $x \in Y$. 98 | Since $X$ and $Y$ are transitive this means that $x \ss X$ and $x \ss Y$. 99 | So consider any $y \in x$ then $y \in X$ and $y \in Y$ so that $y \in X \cap Y$. 100 | Hence since $y$ was arbitrary it follows that $x \ss X \cap Y$ so that $X \cap Y$ is transitive since $x$ was arbitrary. 101 | } 102 | 103 | (c) We claim that this is \emph{not} true. 104 | \qproof{ 105 | It was shown in Exercise~6.2.3 part (a) that $Y = \braces{\es, \braces{\es}, \braces{\braces{\es}}}$ is transitive. 106 | So let $X = \braces{\braces{\es}}$ so that clearly $X \in Y$. 107 | If then $x = \braces{\es}$ then $x \in X$ but $x$ is not a subset of $X$ since $\es \notin X$. 108 | Hence the original hypothesis is not true. 109 | } 110 | 111 | (d) We claim that this is \emph{not} true. 112 | \qproof{ 113 | Again $Y = \braces{\es, \braces{\es}, \braces{\braces{\es}}}$ is transitive so let $X = \braces{\braces{\es}, \braces{\braces{\es}}}$ so that clearly $X \ss Y$. 114 | Then if $x = \braces{\es}$ then $x \in X$ but $x$ is not a subset of $X$ because $\es \notin X$. 115 | Thus the original hypothesis is false. 116 | } 117 | 118 | (e) We claim that this is true. 119 | \qproof{ 120 | Consider any $x \in Y \cup S$. 121 | If $x \in Y$ then since $Y$ is transitive $x \ss Y$. 122 | Hence $x \ss Y \ss Y \cup S$. 123 | On the other hand if $x \in S$ then $x \in \pset{Y}$ since $S \ss \pset{Y}$ 124 | Hence $x \ss Y \ss Y \cup S$. 125 | Since in all cases $x \ss Y \cup S$ and $x$ was arbitrary this shows that $Y \cup S$ is transitive by definition. 126 | } 127 | } 128 | 129 | \exercise{5}{ 130 | If every $X \in S$ is transitive, then $\bigcup S$ is transitive. 131 | } 132 | \sol{ 133 | \qproof{ 134 | Consider any $x \in \bigcup S$. 135 | Then there is an $X \in S$ where $x \in X$. 136 | Since $X$ is transitive it follows that $x \ss X$. 137 | So consider any $y \in x$ so that also $y \in X$. 138 | Thus also $y \in \bigcup S$ since $X \in S$. 139 | Since $y$ was arbitrary this shows that $x \ss \bigcup S$. 140 | Since $x$ was arbitrary this shows by definition that $\bigcup S$ is transitive. 141 | } 142 | } 143 | 144 | \exercise{6}{ 145 | An ordinal $\a$ is a natural number if and only if every nonempty subset of $\a$ has a greatest element. 146 | } 147 | \sol{ 148 | \qproof{ 149 | ($\to$) Suppose that $n$ is a natural number and consider any nonempty subset $A$ of $n$. 150 | Since $A \ss n$ it follows that $|A| \leq |n| = n$ so that $A$ is finite. 151 | Thus $A$ is a finite set of natural numbers and so has a greatest element. 152 | This can be proven by a trivial inductive argument. 153 | 154 | ($\leftarrow$) We show this by contrapositive. 155 | Suppose that $\a$ is an ordinal such that $\a \notin \nats$. 156 | Then $\a \notin \w = \nats$ so that $\a \nless \w$, from which it follows that $\a \geq \w$ 157 | Hence $\a = \w$ or $\a > \w$, in which case $\w \in \a$ so that $\w \ss \a$ since $\a$ is transitive. 158 | Thus in either case $\nats = \w \ss \a$. 159 | Clearly $\nats$ has no greatest element (since if $n$ were such a greatest element then $n+1 \in \nats$ but $n < n+1$). 160 | Thus there is a nonempty subset $A$ of $\a$ such that $A$ has no greatest element. 161 | } 162 | } 163 | 164 | \exercise{7}{ 165 | If a set of ordinals $X$ does not have a greatest element, then $\sup{X}$ is a limit ordinal. 166 | } 167 | \sol{ 168 | 169 | \begin{lem}\label{lem:ord:ordleq} 170 | If $\a$ and $\b$ are ordinals and $\a < \b$ then $\a+1 \leq \b$. 171 | \end{lem} 172 | \qproof{ 173 | To the contrary, suppose that $\a+1 > \b$. 174 | Then by the definition of $<$ we have that $\b \in \a+1 = \a \cup \braces{\a}$ and since $\b \neq \a$ it has to be that $\b \in \a$. 175 | But then $\b < \a$, which is a contradiction. 176 | } 177 | 178 | \begin{lem}\label{lem:ord:ordleq2} 179 | If $\a$ and $\b$ are ordinals and $\a < \b+1$ then $\a \leq \b$. 180 | \end{lem} 181 | \qproof{ 182 | Since $\a < \b+1$ we have that $\a \in \b+1 = \b \cup \braces{\b}$. 183 | Hence $\a \in \b$ or $\a = \b$. 184 | Thus $\a < \b$ or $\a = \b$, i.e. $\a \leq \b$. 185 | } 186 | 187 | \mainprob 188 | \qproof{ 189 | Suppose that $X$ is a set of ordinals with no greatest element. 190 | Let $\b = \sup{X} = \bigcup X$. 191 | Then by the remarks following the proof of Theorem~6.2.6 $\b \notin X$ since $X$ has no greatest element. 192 | Now also suppose that $\b$ is a successor so that there is an ordinal $\a$ such that $\b = \a+1$. 193 | 194 | If $\a \in X$ then since $X$ has no greatest element there is a $\g \in X$ such that $\a < \g$. 195 | Then by Lemma~\ref{lem:ord:ordleq} $\b = \a + 1 \leq \g$. 196 | It cannot be that $\g = \b$ since $\g \in X$ but $\b \notin X$ so it must be that $\b < \g$. 197 | But then since $\b$ is an upper bound of $X$ it follows that $\g$ is also. 198 | However, since $\g \in X$ this would make $\g$ the greatest element of $X$, which is a contradiction. 199 | 200 | On the other hand if $\a \notin X$ then consider any $\g \in X$. 201 | Then $\g < \b = \a+1$ so that by Lemma~\ref{lem:ord:ordleq2} $\g \leq \a$. 202 | Since $\g$ was arbitrary this shows that $\a$ is an upper bound of $X$. 203 | However, since $\a < \b$ this contradicts the definition of $\b$ as being the least upper bound of $X$, according to which $\a \geq \b$. 204 | 205 | Since all cases lead to a contradiction it cannot be that $\b = \sup{X}$ is a successor and therefore by definition is a limit ordinal. 206 | } 207 | } 208 | 209 | \exercise{8}{ 210 | If $X$ is a nonempty set of ordinals, then $\bigcap X$ is an ordinal. 211 | Moreover $\bigcap X$ is the least element of $X$. 212 | } 213 | \sol{ 214 | \qproof{ 215 | Suppose that $X$ is a set of ordinals. 216 | Then by Theorem~6.2.6d $X$ has a least element $\a$. 217 | We shall show that $\a = \bigcap X$, which simultaneously shows that $\bigcap X$ is an ordinal and the least element of $X$. 218 | 219 | Consider any $\b \in X$. 220 | Since $\a$ is the least element $\a \leq \b$ so that $\a = \b$ or $\a < \b$. 221 | Clearly $\a \ss \a = \b$ in the former case. 222 | In the latter case we have $\a \in \b$ so that $\a \ss \b$ as well since $\b$ is transitive (since it is an ordinal). 223 | Since $\b$ was arbitrary any $x \in \a$ is also in every $\b \in X$ so that $x \in \bigcap X$ so that $\a \ss \bigcap X$ since $x$ was arbitrary. 224 | 225 | Now consider any $x \in \bigcap X$. 226 | Then clearly $x \in \a$ since $\a \in X$ so that $\bigcap X \ss \a$ since $x$ was arbitrary. 227 | 228 | Thus we have shown that $\a = \bigcap X$ as desired. 229 | } 230 | } 231 | -------------------------------------------------------------------------------- /sections/sec_6_3.tex: -------------------------------------------------------------------------------- 1 | \subsection{The Axiom of Replacement} 2 | 3 | \exercise{1}{ 4 | Let $\prop{P}(x,y)$ be a property such that for every $x$ there is at most one $y$ for which $\prop{P}(x,y)$ holds. 5 | Then for every set $A$ there is a set $B$ such that, for all $x \in A$, if $\prop{P}(x,y)$ holds for some $y$, then $\prop{P}(x,y)$ holds for some $y \in B$. 6 | } 7 | \sol{ 8 | \qproof{ 9 | Define a property $\prop{R}(x,y)$ such that $\prop{R}(x,y)$ holds if and only if 10 | \begin{enumerate} 11 | \item $\prop{P}(x,y)$ holds, or 12 | \item $y = \es$ and there is not a $z$ such that $\prop{P}(x,z)$ holds. 13 | \end{enumerate} 14 | Clearly this property is such that for every $x$ there is a unique $y$ for which $\prop{R}(x,y)$ holds. 15 | 16 | Now consider any set $A$. 17 | Then by the Axiom Schema of Replacement there is a set $B$ such that, for every $x \in A$, there is a $y \in B$ for which $\prop{R}(x,y)$ holds. 18 | Consider any $x \in A$. 19 | Then by the above there is a $y \in B$ such that $\prop{R}(x,y)$ holds. 20 | Now suppose that $\prop{P}(x,z)$ holds for some $z$. 21 | Then option 2 above cannot be the case so that, $\prop{P}(x,y)$ holds (option 1) since $\prop{R}(x,y)$ does. 22 | Thus $\prop{P}(x,y)$ holds for some $y \in B$ as we were required to show. 23 | } 24 | } 25 | 26 | \exercise{2}{ 27 | Use Theorem~6.3.6 to prove the existence of 28 | 29 | (a) The set $\braces{\es, \braces{\es}, \braces{\braces{\es}}, \braces{\braces{\braces{\es}}}, \ldots}$. 30 | 31 | (b) The set $\braces{\nats, \pset{\nats}, \pset{\pset{\nats}}, \ldots}$. 32 | 33 | (c) The set $\w + \w = \w \cup \braces{\w, \w+1, (\w+1)+1, \ldots}$. 34 | } 35 | \sol{ 36 | (a) 37 | \qproof{ 38 | Define the operation $\prop{G}(x,n)$ for set a $x$ and $n \in \nats$ by 39 | $$ 40 | \prop{G}(x,n) = \braces{x}. 41 | $$ 42 | Then by Theorem~6.3.6 there is a unique sequence $\angles{a_n \where n \in \nats}$ where 43 | \ali{ 44 | a_0 &= \es \\ 45 | a_{n+1} &= \prop{G}(a_n, n) = \braces{a_n} 46 | } 47 | for all $n \in \nats$. 48 | Clearly the range of $\angles{a_n}$ is the set we seek. 49 | } 50 | 51 | (b) 52 | \qproof{ 53 | Similarly define the operation $\prop{G}(x,n)$ for a set $x$ and $n \in \nats$ by 54 | $$ 55 | \prop{G}(x,n) = \pset{x}, 56 | $$ 57 | noting that this set exists by the Axiom of Power Set. 58 | Then by Theorem~6.3.6 there is a sequence $\angles{a_n \where n \in \nats}$ defined by 59 | \ali{ 60 | a_0 &= \nats \\ 61 | a_{n+1} &= \prop{G}(a_n, n) = \pset{a_n}, 62 | } 63 | noting that $a_0 = \nats$ exists by the Axiom of Infinity. 64 | Clearly then the range of $\angles{a_n}$ is the set we seek. 65 | } 66 | 67 | (c) 68 | \qproof{ 69 | Define the operation $\prop{G}(x,n)$ for a set $x$ and $n \in \nats$ by 70 | $$ 71 | \prop{G}(x,n) = S(x) = x \cup \braces{x}, 72 | $$ 73 | Then by Theorem~6.3.6 there is a sequence $\angles{a_n \where n \in \nats}$ defined by 74 | \ali{ 75 | a_0 &= \w \\ 76 | a_{n+1} &= \prop{G}(a_n, n) = S(a_n) = a_n+1, 77 | } 78 | noting that $a_0 = \w = \nats$ exists by the Axiom of Infinity. 79 | Clearly then the range of $\angles{a_n}$ is $A = \braces{\w, \w+1, \w+2, \ldots}$. 80 | It then follows that $\w + \w = \w \cup A$ is the set we seek. 81 | } 82 | } 83 | 84 | \exercise{3}{ 85 | Use Theorem~6.3.6 to define 86 | \ali{ 87 | V_0 &= \es; \\ 88 | V_{n+1} &= \pset{V_n} \hspace{0.5cm} (n \in \w); \\ 89 | V_\w &= \bigcup_{n \in \w} V_n. 90 | } 91 | } 92 | \sol{ 93 | \qproof{ 94 | Define the operation $\prop{G}(x,n)$ for a set $x$ and $n \in \nats$ by 95 | $$ 96 | \prop{G}(x,n) = \pset{x}, 97 | $$ 98 | noting that this set exists by the Axiom of Power Set. 99 | Then by Theorem~6.3.6 there is a sequence $\angles{V_n \where n \in \nats}$ defined by 100 | \ali{ 101 | V_0 &= \es \\ 102 | V_{n+1} &= \prop{G}(V_n, n) = \pset{V_n}, 103 | } 104 | noting that $V_0 = \es$ exists by the Axiom of Existence. 105 | Then we let 106 | $$ 107 | V_\w = \bigcup_{n \in \w} V_n, 108 | $$ 109 | noting that $\w = \nats$. 110 | This set exists by the Axiom of Union. 111 | } 112 | } 113 | 114 | \exercise{4}{ 115 | (a) Every $x \in V_\w$ is finite. 116 | 117 | (b) $V_\w$ is transitive. 118 | 119 | (c) $V_\w$ is an inductive set. 120 | } 121 | \sol{ 122 | 123 | (a) 124 | \qproof{ 125 | First we show by induction that every $V_n$ is finite (for $n \in \nats$). 126 | For $n=0$ we have $V_n = V_0 = \es$, which is clearly finite. 127 | Now suppose that $V_n$ is finite then we have $V_{n+1} = \pset{V_n}$, which is finite by Theorem~4.2.8. 128 | 129 | Now consider any $x \in V_\w = \bigcup_{n \in \w} V_n$ so that there is an $n \in \w$ such that $x \in V_n$. 130 | We note that $n \neq 0$ since $V_0 = \es$ so it cannot be that $x \in V_0 = \es$. 131 | Hence $V_{n-1}$ is a set and moreover $V_n = \pset{V_{n-1}}$. 132 | So since $x \in V_n$ it follows that $x \in \pset{V_{n-1}}$ so that $x \ss V_{n-1}$. 133 | Thus it follows that $|x| \leq |V_{n-1}|$ so that clearly $x$ is finite since $V_{n-1}$ is (shown above). 134 | } 135 | 136 | (b) 137 | \qproof{ 138 | Consider any $x \in V_w$. 139 | Then by the same argument as in part (a) above it follows that $x \in V_n$ where $n \neq 0$. 140 | Hence again $V_{n-1}$ is a set and $V_n = \pset{V_{n-1}}$ so that $x \ss V_{n-1}$. 141 | Then for any $y \in x$ we have that $y \in V_{n-1}$, from which it follows that clearly $y \in \bigcup_{k \in \w} V_k = V_\w$. 142 | Hence since $y$ was arbitrary $x \ss V_\w$, and since $x$ was arbitrary this shows that $V_\w$ is transitive by definition. 143 | } 144 | 145 | (c) 146 | \qproof{ 147 | First we show by induction that each $V_n$ (where $n \in \w$) is transitive. 148 | For $n=0$ we have $V_n = V_0 = \es$, which is clearly vacuously transitive. 149 | Now suppose that $V_n$ is transitive and consider any $x \in V_{n+1} = \pset{V_n}$ so that $x \ss V_n$. 150 | Now consider any $y \in x$ so that also $y \in V_n$. 151 | But since $V_n$ is transitive $y \ss V_n$ so that $y \in \pset{V_n} = V_{n+1}$. 152 | Hence since $y$ was arbitrary this shows that $x \ss V_{n+1}$ and since $x$ was arbitrary this shows by definition that $V_{n+1}$ is transitive, thereby completing the inductive proof. 153 | 154 | Now we show that $V_\w$ is inductive. 155 | So first note that $V_1 = \pset{V_0} = \pset{\es} = \braces{\es}$ so that $0 = \es \in V_1$. 156 | From this is clearly follows that $0 \in \bigcup_{n \in \w} V_n = V_\w$. 157 | 158 | Now suppose that $n \in V_\w = \bigcup_{k \in \w} V_k$ so that there is an $m \in \w$ such that $n \in V_m$. 159 | Since it was shown above that $V_m$ is transitive we have that $n \ss V_m$ as well. 160 | So consider any $x \in n+1 = n \cup \braces{n}$. 161 | If $x \in n$ then also $x \in V_m$ since $n \ss V_m$. 162 | On the other hand if $x \in \braces{n}$ then $x = n \in V_m$. 163 | Since $x$ was arbitrary this shows that $n+1 \ss V_m$ so that $n+1 \in \pset{V_m} = V_{m+1}$. 164 | From this it clearly follows that $n+1 \in \bigcup_{k \in \w} V_k = V_\w$. 165 | This shows that $V_\w$ is inductive by definition. 166 | } 167 | } 168 | 169 | \exercise{5}{ 170 | (a) If $x \in V_\w$ and $y \in V_\w$, then $\braces{x,y} \in V_\w$. 171 | 172 | (b) If $X \in V_\w$, then $\bigcup X \in V_\w$ and $\pset{X} \in V_\w$. 173 | 174 | (c) If $A \in V_\w$ and $f$ is a function on $A$ such that $f(x) \in V_\w$ for each $x \in A$, then $f[X] \in V_\w$. 175 | 176 | (d) If $X$ is a finite subset of $V_\w$, then $X \in V_\w$. 177 | 178 | Note that part (c) differs slightly from the book; see the Errata List. 179 | } 180 | \sol{ 181 | 182 | (a) 183 | \qproof{ 184 | First we show that if $x \in V_n$ for some $n \in \w$ then $x \in V_m$ for all $m \geq n$. 185 | We show this by induction on $m$. 186 | So for $m=n$ clearly $x \in V_n = V_m$. 187 | Now suppose that $x \in V_m$. 188 | Then it was shown in Exercise~6.3.4 part (c) that $V_m$ is transitive so that $x \ss V_m$. 189 | Hence $x \in \pset{V_m} = V_{m+1}$, thereby completing the inductive proof. 190 | 191 | Now suppose that $x,y \in V_\w$. 192 | Then there are $n,m \in \w$ such that $x \in V_n$ and $y \in V_m$. 193 | Without loss of generality we can assume that $n \leq m$ (since if this is not the case then we simply reverse the roles of $x$ and $y$). 194 | So since $m \geq n$ it follows from what was shown above that $x \in V_m$ as well. 195 | Hence we have that clearly $\braces{x,y} \ss V_m$ since both $x \in V_m$ and $y \in V_m$. 196 | Then $\braces{x,y} \in \pset{V_m} = V_{m+1}$ from which it clearly follows that $\braces{x,y} \in \bigcup_{k \in \w} V_k = V_\w$. 197 | } 198 | 199 | (b) 200 | \qproof{ 201 | Suppose that $X \in V_\w = \bigcup_{k \in \w} V_k$. 202 | Then there is an $n \in \w$ such that $X \in V_n$. 203 | It was shown in Exercise 6.3.4 part (c) that $V_n$ is transitive so that $X \ss V_n$. 204 | 205 | First we show that $\bigcup X \in V_\w$. 206 | So consider any $x \in \bigcup X$ so that there is a $Y \in X$ such that $x \in Y$. 207 | Then since $X \ss V_n$ we have that $Y \in V_n$. 208 | Since again $V_n$ is transitive we have that $Y \ss V_n$ so that $x \in V_n$ since $x \in Y$. 209 | Since $x$ was arbitrary it follows that $\bigcup X \ss V_n$ so that $\bigcup X \in \pset{V_n} = V_{n+1}$. 210 | From this it clearly follows that $\bigcup X \in \bigcup_{k \in \w} V_k = V_\w$. 211 | 212 | Next we show that $\pset{X} \in V_\w$. 213 | So consider any $Y \in \pset{X}$ so that $Y \ss X$. 214 | Now consider any $y \in Y$ so that also $y \in X$. 215 | Since $X \ss V_n$ we have that $y \in V_n$. 216 | But since $y \in Y$ was arbitrary it follows that $Y \ss V_n$ so that $Y \in \pset{V_n} = V_{n+1}$. 217 | Then since $Y \in \pset{X}$ was arbitrary it follows that $\pset{X} \ss V_{n+1}$ so that $\pset{X} \in \pset{V_{n+1}} = V_{n+2}$. 218 | From this it clearly follows that $\pset{X} \in \bigcup_{k \in \w} V_k = V_\w$. 219 | } 220 | 221 | (c) 222 | \qproof{ 223 | Note the issue with this part in the errata list. 224 | Since $A \in V_\w$ we have by Exercise~6.3.4 part (a) that $A$ is finite. 225 | Then by Theorem~2.2.5 it follows that $f[A]$ is finite. 226 | Also clearly $f[A]$ is a subset of $V_\w$ and hence is a finite subset. 227 | Therefore by part (d) below $f[A] \in V_\w$. 228 | } 229 | 230 | (d) 231 | \qproof{ 232 | Consider any finite $X \ss V_\w$. 233 | Suppose then that $|X| = n$ for some $n \in \nats$. 234 | Then for each $x_k \in X$, where $k \in n$, we have that $x_k \in V_\w = \bigcup_{m \in \w} V_m$ so that there is an $m_k \in \w$ where $x_k \in V_{m_k}$. 235 | Now let $m = \max_{k \in n} m_k$, which exists since $n$ is finite. 236 | Then, for any $k \in n$, by what was shown in Exercise~6.3.5 part (a) we have $x_k \in V_m$ since $x_k \in V_{m_k}$ and $m \geq m_k$. 237 | Hence it follows that $X \ss V_m$ so that $X \in \pset{V_m} = V_{m+1}$. 238 | Clearly then $X \in \bigcup_{k \in \w} V_k = V_\w$. 239 | } 240 | } 241 | -------------------------------------------------------------------------------- /sections/sec_6_4.tex: -------------------------------------------------------------------------------- 1 | \subsection{Transfinite Induction and Recursion} 2 | 3 | \def\P{\prop{P}} 4 | \def\G{\prop{G}} 5 | \def\R{\prop{R}} 6 | \def\F{\prop{F}} 7 | \exercise{1}{ 8 | Prove a more general Transfinite Recursion Theorem (Double Recursion Theorem): 9 | Let $\G$ be an operation in two variables. 10 | Then there is an operation $\F$ such that $\F(\a,\b) = \G(\F \rest (\b \times \a))$ for all ordinals $\b$ and $\a$. 11 | [Hint: Computations are functions on $(\b+1) \times (\a+1)$.] 12 | } 13 | \sol{ 14 | \qproof{ 15 | Since in these Recursion Theorems the arguments of interest of the given operation $\G$ are typically functions, we assume that $\G$ still takes a single variable despite what the text says. 16 | Hence for each $x$ there is a unique $y$ such that $y = \G(x)$. 17 | It just happens that in this case the variables of interest are functions of two variables. 18 | 19 | For ordinals $\a$ and $\b$ we let $f_{\a,\b}$ be the isomorphism from the ordinal $(\b+1)\cdot(\a+1)$ to the lexicographic ordering of $(\a+1) \times (\b+1)$, which exists by Theorem~6.5.8. 20 | We also note that according to Exercise~6.5.2 we have 21 | $$ 22 | (\b+1)\cdot(\a+1) = (\b+1)\cdot\a + (\b+1)\cdot 1 = (\b+1)\cdot\a + (\b+1) = \squares{(\b+1)\cdot\a + \b} + 1 23 | $$ 24 | so that $(\b+1)\cdot(\a+1)$ is a successor ordinal. 25 | Then for $\g < (\b+1)\cdot(\a+1)$ define 26 | $$ 27 | X_{\a,\b,\g} : \braces{(\d,\e) \in (\a+1) \times (\b+1) \where (\d,\e) \prec f_{\a,\b}(\g)} 28 | $$ 29 | 30 | Then for $\g < (\b+1) \cdot (\a+1)$ we say that $t$ is a computation of length $\g$ if $t$ is a transfinite sequence whose domain is $\g+1$ and such that $t(\d) = \G(t \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d}))$ for all $\d \leq \g$. 31 | We note that since $(\b+1)\cdot(\a+1)$ is a successor that $\g+1 \leq (\b+1)\cdot(\a+1)$. 32 | 33 | Now we define the property $\P(x,y,z)$ such that $\P(x,y,z)$ holds if and only if 34 | \begin{enumerate} 35 | \item $x$ and $y$ are ordinal numbers and $z = t(\inv{f_{x,y}}(x,y))$ for some computation $t$ of length $(y+1)\cdot x + y$ (with respect to $\G$), or 36 | \item $x$ or $y$ is \emph{not} an ordinal and $z = \es$. 37 | \end{enumerate} 38 | 39 | We prove that $\P$ defines an operation. 40 | Hence we have to show that for any $x$ and $y$ there is a unique $z$ such that $\P(x,y,z)$ holds. 41 | So consider any sets $\a$ and $\b$. 42 | If $\a$ or $\b$ is not an ordinal than clearly $\P(\a,\b,\es)$ holds and $\es$ is unique. 43 | So suppose that both $\a$ and $\b$ are ordinals. 44 | Then it suffices to show that there is a unique computation of length $(y+1)\cdot x + y$ (with respect to $\G$) since this will make $z = t(\inv{f_{\a,\b}}(\a,\b))$ unique. 45 | We show this via transfinite induction. 46 | 47 | So consider any ordinal $\g < (\b+1) \cdot (\a+1)$ so that $\g \leq (y+1)\cdot x + y$ and assume that for all $\d < \g$ that there is a unique computation of length $\d$ and we show that there exists a unique computation of length $\g$, which completes the proof that $\P$ defines an operation. 48 | 49 | \emph{Existence.} First define a property $\R(x,y)$ such that $\R(x,y)$ holds if and only if 50 | \begin{enumerate} 51 | \item $x$ is an ordinal where $x < \g$ and $y$ is a computation of length $x$ (with respect to $\G$), or 52 | \item $x$ is is an ordinal and $x \geq \g$ and $y = \es$, or 53 | \item $x$ is not an ordinal and $y = \es$. 54 | \end{enumerate} 55 | Clearly by the induction hypothesis this property has a unique $y$ for every $x$. 56 | Hence we can apply the Axiom Schema of Replacement, according to which there is a set $T$ such that for every $\d \in \g$ (so that $\d < \g$) there is a $t$ in $T$ such that $\R(\d, t)$ holds. 57 | That is 58 | $$ 59 | T = \braces{t \where t \text{ is the unique computation of length $\d$ for all $\d < \g$}} 60 | $$ 61 | Now, $T$ is a system of transfinite sequences (which are functions) so define $\bar{t} = \bigcup T$ and let $\t = \bar{t} \cup \braces{(\g, \G(\bar{t} \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\g}))}$. 62 | 63 | \emph{Claim 1:} $\dom(\t) = \g+1$. 64 | So consider any $\e \in \dom(\t)$. 65 | Clearly if $\e = \g$ then $\e \in \g+1$. 66 | On the other hand if $\e \in \dom(\bar{t})$ then there is a $t \in T$ such that $\e \in \dom(t)$. 67 | But since $t$ is a computation of length $\d$ and $\d < \g$ it follows that $\e \leq \d < \g < \g+1$ so that $\e \in \g+1$. 68 | Hence since $\e$ was arbitrary $\dom(\t) \ss \g+1$. 69 | 70 | Now consider any $\e \in \g+1$ so that $\e \leq \g$. 71 | If $\e = \g$ then clearly by definition $\e \in \dom(\t)$. 72 | On the other hand if $\e \neq \g$ then $\e < \g$. 73 | So consider the $t \in T$ where $t$ is the unique computation of length $\e$ (which exists since $\e < \g$). 74 | Then clearly $\e \in \dom(t)$ so that $\e \in \dom(\bar{t})$. 75 | From this it follows that clearly $\e \in \dom(\t)$ so that $\g+1 \ss \dom(\t)$ since $\e$ was arbitrary. 76 | This proves the claim. 77 | 78 | \emph{Claim 2:} $\t$ is a function. 79 | Consider any $\e \in \dom(\t) = \g+1$ so that again $\e \leq \g$. 80 | If $\e = \g$ then clearly $\t(\e) = \t(\g) = \G(\bar{t} \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\g}))$ is unique since $\G$ is an operation. 81 | On the other hand if $\e < \g$ then $\t$ is a function so long as $\bar{t}$ is, and this is the case so long as $T$ is a compatible system of functions since $\bar{t} = \bigcup T$. 82 | We show this presently. 83 | 84 | So consider any arbitrary $t_1, t_2 \in T$ where $t_1$ is the computation of length $\e_1$ and $t_2$ is the computation of length $\e_2$. 85 | Without loss of generality we can assume that $\e_1 \leq \e_2$. 86 | We must show that $t_1(\d) = t_2(\d)$ for all $\d \leq \e_1$. 87 | This we show by transfinite induction. 88 | So suppose that $t_1(\k) = t_2(\k)$ for all $\k < \d \leq \e_1$. 89 | Then clearly $t_1 \rest \d = t_2 \rest \d$, from which it follows that $t_1 \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d}) = t_2 \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d})$ and since $\G$ is an operation we have $t_1(\d) = \G(t_1 \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d})) = \G(t_2 \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d})) = t_2(\d)$. 90 | This completes the proof of the claim. 91 | 92 | \emph{Claim 3:} $\t(\d) = \G(\t \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d}))$ for all $\d \leq \g$. 93 | So consider any such $\d$. 94 | If $\d = \g$ then since $\t \rest \g = \bar{t}$ we clearly have $\t(\d) = \t(\g) = \G(\bar{t} \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\g})) = \G(\t \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\g})) = \G(\t \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d}))$\,. 95 | On the other hand if $\d < \g$ then let $t \in T$ be the computation of length $\d$ (which exists since $\d < \g$). 96 | Then $\t(\d) = t(\d) = \G(t \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d})) = \G(\t \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d}))$ since $t$ is a computation (with respect to $\G$) and clearly $t \ss \t$. 97 | 98 | Claims 1 through 3 show that $\t$ is a computation of length $\g$ and hence that such a computation exists. 99 | 100 | \emph{Uniqueness.} Now let $\s$ be another computation of length $\g$. 101 | We show that $\s = \t$, which proves uniqueness. 102 | Since both $\s$ and $\t$ are functions with $\dom(\s) = \g+1 = \dom(\t)$ it suffices to show that $\s(\d) = \t(\d)$ for all $\d \leq \g$. 103 | We show this once again by using transfinite induction. 104 | So suppose that $\s(\e) = \t(\e)$ for all $\e < \d \leq \g$. 105 | It then follows that $\s \rest \d = \t \rest \d$ so that $\s \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d}) = \t \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d})$. 106 | Then since $\s$ and $\t$ are computations we have that $\s(\d) = \G(\s \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d})) = \G(\t \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,\d})) = \t(\d)$, thereby completing the uniqueness proof. 107 | 108 | This completes the proof that $\P$ defines an operation. 109 | 110 | So let $\F$ be the operation defined by $\P$. 111 | The last thing we need to show to complete the proof of the entire theorem is that $\F(\a, \b) = \G(\F \rest (\a \times \b))$ for all ordinals $\a$ and $\b$, noting that we are treating $\F$ as a function even though it is an operation. 112 | Thus, for any set $X$, $\F \rest X$ denotes the set $\braces{(x, \F(x)) \where x \in X}$, which forms a function with domain $X$. 113 | The range of this function is a set whose existence is guaranteed by the Axiom Schema of Replacement since $\F$ is an operation. 114 | 115 | So consider any ordinals $\a$ and $\b$ and the unique computation $t$ of length $(\b+1)\cdot\a + \b$. 116 | Then clearly for any $\g \leq (\b+1)\cdot\a + \b$ we have that $t_\g = t \rest (\g+1)$ is a computation of length $\g$. 117 | Since this computation is the unique computation of length $\g$, by the definition of $\F$ as it relates to $\P$ we have $\F(f(\g)) = t_\g(\g) = t(\g)$. 118 | Since $\g$ was arbitrary this shows that $\F \rest (\a \times \b) = t \rest (\b+1)\cdot\a + \b$. 119 | Then clearly we have $\F(\a, \b) = t((\b+1)\cdot\a + \b) = \G(t \circ (\inv{f_{\a,\b}} \rest X_{\a,\b,(\b+1)\cdot\a + \b})) = \G(t \rest (\b+1)\cdot\a + \b) = \G(\F \rest (\a \times \b))$ by what was just shown above. 120 | } 121 | 122 | NOTE: This is a very nasty exercise and there may be a simpler way to do this. 123 | } 124 | 125 | \def\G{\prop{G}} 126 | \def\F{\prop{F}} 127 | \exercise{2}{ 128 | Using the Recursion Theorem~6.4.9 show that there is a binary operation $\F$ such that 129 | 130 | (a) $\F(x,1) = 0$ for all $x$. 131 | 132 | (b) $\F(x, n+1) = 0$ if and only if there exist $y$ and $z$ such that $x = (y,z)$ and $\F(y,n) = 0$. 133 | 134 | We say that $x$ is an \emph{$n$-tuple} (where $n \in \w$, $n > 0$) if $\F(x,n) = 0$. 135 | Prove that this definition of $n$-tuples coincides with the one given in Exercise~5.17 in Chapter~3. 136 | } 137 | \sol{ 138 | \qproof{ 139 | Note that there is no such operation that can exactly satisfy both conditions as they actually contradict each other. 140 | To see this suppose there is such an operation $\F$. 141 | Then define the set $x = \es$ and $n = 0$ 142 | Then by (a) we have that $\F(x,n+1) = \F(x, 1) = 0$. 143 | It then follows from (b) that there are a $y$ and $z$ such that $x = (y,z)$, but clearly this is not the case for $x=\es$. 144 | Hence a contradiction. 145 | 146 | To remedy this we simply add a condition to (b), which when restated becomes 147 | 148 | (b) $n > 0$ and $\F(x,n+1) = 0$ if and only if there exists $y$ and $z$ such that $x = (y,z)$ and $\F(y,n) = 0$. 149 | 150 | Now, define an operation $\G$ by $z = \G(x,y_u)$ if and only if either 151 | \begin{enumerate} 152 | \item $y_u$ is a function with parameter $u$, $\dom(y_u) = 1$, and $z = 0$, or 153 | \item $y_u$ is a function with parameter $u$, $\dom(y_u) = \a+1$ for some ordinal $\a$, there are $p$ and $q$ where $x = (p,q)$, $y_p(\a) = 0$, and $z = 0$ or 154 | \item None of the above hold and $z = 1$. 155 | \end{enumerate} 156 | 157 | Then by Theorem~6.4.9 there is an operation $\F$ such that $\F(x,\a) = \G(x, \F_x \rest \a)$ for all ordinals $\a$ and sets $x$. 158 | 159 | Then for any set $x$ we have that clearly $\F \rest 1$ is a function with domain $1$ (and parameter $x$) so that by definition 160 | $$ 161 | \F(x,1) = \G(x, \F_x \rest 1) = 0. 162 | $$ 163 | This shows (a). 164 | 165 | To show (b) consider any ordinal $n$ and set $x$. 166 | 167 | ($\to$) Suppose that $n > 0$ and $\F(x,n+1) = 0$ so that clearly $(3)$ above cannot be the case. 168 | Also $(1)$ cannot be the case since $\F(x,n+1) = \G(x, \F_x \rest n+1)$ and $\dom(\F_x \rest n+1) = n + 1 > 1$. 169 | Hence (2) is the case so that there are $y$ and $z$ such that $x = (y,z)$ and $(\F_y \rest n+1)(n) = 0$. 170 | Hence it follows that $F(y,n) = 0$. 171 | 172 | ($\leftarrow$) Now suppose that there are $y$ and $z$ such that $x = (y,z)$ and $F(y,n) = 0$. 173 | Then $F(y,n) = G(y, \F_y \rest n) = 0$. 174 | 175 | So if $n=0$ then $\dom(\F_y \rest n) = \dom(\F_y \rest 0) = 0 \neq 1$ so (1) cannot be the case. 176 | Also since $0$ is not a successor ordinal (2) cannot be the case either (since $\dom(\F_y \rest 0) \neq \a+1$ for any ordinal $\a$). 177 | Hence (3) must be the case, but this implies that $F(y,n) = 1$, which is a contradiction. 178 | So we must have that $n \neq 0$ so $n > 0$. 179 | 180 | Since $\F(y, n) = 0$ clearly we have that $(\F_y \rest n+1)(n) = 0$. 181 | Since also $\dom(\F_x \rest n+1) = n+1$ we find that (2) holds for $\G(x, \F_x \rest n+1)$. 182 | From this it follows that $\F(x,n+1) = \G(x, \F_x \rest n+1) = 0$. 183 | 184 | This completes the proof. 185 | } 186 | } 187 | -------------------------------------------------------------------------------- /sections/sec_6_6.tex: -------------------------------------------------------------------------------- 1 | \subsection{The Normal Form} 2 | 3 | \exercise{1}{ 4 | Show that $\w^\e = \e$. 5 | } 6 | \sol{ 7 | This was shown in Exercise~6.5.15c. 8 | } 9 | 10 | \exercise{2}{ 11 | Find the first few terms of the Goodstein sequence starting at $m = 28$. 12 | } 13 | \sol{ 14 | We show the first 5 terms of the Goodstein sequences for $m = 28$: 15 | \ali{ 16 | m_0 &= m = 28 = 2^{2^2} + 2^{2 + 1} + 2^2 \\ 17 | m_1 &= 3^{3^3} + 3^{3 + 1} + 3^3 - 1 = 3^{3^3} + 3^{3 + 1} + 3^2 \cdot 2 + 3 \cdot 2 + 2 \approx 7.626 \times 10^{12} \\ 18 | m_2 &= 4^{4^4} + 4^{4 + 1} + 4^2 \cdot 2 + 4 \cdot 2 + 1 \approx 1.341 \times 10^{154} \\ 19 | m_3 &= 5^{5^5} + 5^{5 + 1} + 5^2 \cdot 2 + 5 \cdot 2 \approx 1.911 \times 10^{2182} \\ 20 | m_4 &= 6^{6^6} + 6^{6 + 1} + 6^2 \cdot 2 + 6 \cdot 2 - 1 = 6^{6^6} + 6^{6 + 1} + 6^2 \cdot 2 + 6 + 5 \approx 2.659 \times 10^{36305}. 21 | } 22 | } 23 | -------------------------------------------------------------------------------- /sections/sec_7_1.tex: -------------------------------------------------------------------------------- 1 | \subsection{Initial Ordinals} 2 | 3 | \exercise{1}{ 4 | If $X$ is an infinite well-orderable set, then $X$ has nonisomorphic well-orderings. 5 | } 6 | \sol{ 7 | \input{shared/lem_aleph_infwo_succ} 8 | 9 | \begin{lem}\label{lem:aleph:infwo:reord} 10 | If an infinite set $A$ with order $\prec$ is isomorphic to an ordinal $\a$ then it can also be re-ordered to be isomorphic to $\a + 1$. 11 | \end{lem} 12 | \qproof{ 13 | Since $A$ is infinite and the isomorphism from $A$ to $\a$ is a bijective function, it follows that they are equipotent so that $\a$ is also infinite. 14 | Then by Lemma~\ref{lem:aleph:infwo:succ} $\a$ is equipotent to $\a+1$ so that $A$ is also equipotent to $\a+1$. 15 | Hence there is an $f:A \to \a+1$ that is bijective. 16 | We then simply re-order $A$ according to $\a+1$, i.e. we create the following order on $A$: 17 | $$ 18 | R = \braces{(a,b) \in A \times A \where f(a) < f(b)} 19 | $$ 20 | so that clearly $(A, R)$ is isomorphic to $(\a+1, <)$. 21 | } 22 | 23 | \mainprob 24 | \qproof{ 25 | For an infinite well-orderable set $X$ we show that $X$ has an infinite number of non-isomorphic well-orderings. 26 | So let $\prec$ be a well-ordering of $X$ so that by Theorem~6.3.1 $(X, \prec)$ is isomorphic to some ordinal $\a$. 27 | We then show by induction that, for any natural number $n$, there is an ordering $R_n$ of $X$ such that it is isomorphic to $\a + n$. 28 | For $n=0$ we have that, for $R_0 = \prec$, clearly $(X, R_0)$ is isomorphic to $(\a, <)$ by what has already been established. 29 | Now suppose that there is an ordering $R_n$ of $X$ such that $(X, R_n)$ is isomorphic to $(\a + n, <)$. 30 | Then since $X$ is an infinite set it follows from Lemma~\ref{lem:aleph:infwo:reord} that there is an ordering $R_{n+1}$ such that $X$ is isomorphic to $(\a + n) + 1 = \a + (n+1)$. 31 | This completes the inductive proof. 32 | We note that clearly each of these re-orderings are mutually non-isomorphic since different ordinals are not isomorphic to each other. 33 | } 34 | } 35 | 36 | \exercise{2}{ 37 | If $\a$ and $\b$ are at most countable ordinals then $\a + \b$, $\a \cdot \b$, and $\a^\b$ are at most countable. 38 | [Hint: Use the representation of ordinal operations from Theorems~5.3 and 5.8 and Exercise~5.16 in Chapter~6. 39 | Another possibility is a proof by transfinite induction.] 40 | } 41 | \sol{ 42 | First we show that $\a + \b$ is at most countable. 43 | \qproof{ 44 | First we define two sets: 45 | \ali{ 46 | W_1 &= \braces{(0,\g) \where \g \in \a} & 47 | W_2 &= \braces{(1,\g) \where \g \in \b}. 48 | } 49 | We also define the order $<_1$ on $W_1$ so that $(0,\g) <_1 (0,\d)$ if and only if $\g < \d$ for $(0, \g)$ and $(0, \d)$ in $W_1$ (so that $\g$ and $\d$ are in $\a$). 50 | Similarly we define the order $<_2$ on $W_2$ so that $(1,\g) <_2 (1,\d)$ if and only if $\g < \d$ for $(1, \g)$ and $(1, \d)$ in $W_2$ (so that $\g$ and $\d$ are in $\b$). 51 | 52 | Clearly $W_1$ and $W_2$ are disjoint, $(W_1, <_1)$ is isomorphic to $\a$, and $(W_2, <_2)$ is isomorphic to $\b$. 53 | It then follows from Theorem~6.5.3 that the sum $(W, <)$ is isomorphic to $\a + \b$. 54 | 55 | Now, since they are isomorphic, clearly $W_1$ is equipotent to $\a$ and therefore is at most countable. 56 | Similarly $W_2$ is at most countable by virtue of being isomorphic to $\b$. 57 | It then follows from Theorem~4.2.6 and Theorem~4.3.5 that $W = W_1 \cup W_2$ is at most countable. 58 | Then, since $(W, <)$ is isomorphic to $\a + \b$, $W$ and $\a + \b$ are equipotent so that $\a + \b$ must be at most countable too. 59 | } 60 | 61 | Next we show that $\a \cdot \b$ is at most countable. 62 | \qproof{ 63 | Since $\a$ and $\b$ are at most countable it follows from Exercise~4.2.2 and Theorem~4.3.7 that $\a \times \b$ is at most countable. 64 | Then, since $\a \cdot \b$ is isomorphic to the antilexicographic ordering of $\a \times \b$ by Theorem~6.5.8, it follows that $\a \cdot \b$ is equipotent to $\a \times \b$ and there for at most countable. 65 | } 66 | 67 | Lastly we show that $\a^\b$ is at most countable. 68 | \qproof{ 69 | First if we have that $\a = 0$ then then either $\a^\b = 0^\b = 1$ (if $\b = 0$) or $\a^\b = 0^\b = 0$ (if $\b > 0$). 70 | Clearly both $0$ and $1$ are both at most countable so in the following we assume that $\a \neq 0$. 71 | 72 | \def\seqab{\Seq(\a \cdot \b)} 73 | \def\sba{S(\b, \a)} 74 | Now, let $\seqab$ be the set of all finite sequences of elements of $\a \cdot \b$, and $\sba$ be the set as defined in Exercise~6.5.16 such that $\sba$ with the order defined there is isomorphic (and therefore equipotent) to $\a^\b$. 75 | We shall construct a function $g: \sba \to \seqab$. 76 | 77 | So consider any $f \in \sba$ so that $f:\b \to \a$ and $s(f) = \braces{\x < \b \where f(\x) \neq 0}$ (as defined in the exercise) is finite. 78 | Hence there is a natural number $n$ such that $s(f)$ can be expressed as an increasing sequence $h: n \to s(f)$. 79 | We now define another sequence $t: n \to \a \cdot \b$ by 80 | $$ 81 | t(k) = \a \cdot h(k) + f(h(k)) 82 | $$ 83 | for $k \in n$. 84 | We then set $g(f) = t$. 85 | 86 | The first thing we show is that $g(f)$ is really a sequence whose elements are in $\a \cdot \b$ for any $f \in \sba$. 87 | Hence for any such $f$ again let $h$ be the increasing finite sequence whose range is $s(f)$ and let $t = g(f)$. 88 | Then for any $k \in n$ we we note that $h(k) < \b$ since $h(k) \in s(f)$ so that $h(k) + 1 \leq \b$. 89 | We also note that $f(h(k)) \in \a$ since $f: \b \to \a$ so that $f(h(k)) < \a$. 90 | Thus we have by definition that 91 | \ali{ 92 | t(k) &= \a \cdot h(k) + f(h(k)) \\ 93 | &< \a \cdot h(k) + \a & \text{(by Lemma~6.5.4)} \\ 94 | &= \a \cdot (h(k) + 1) & \text{(by Definition~6.5.6b)} \\ 95 | &\leq \a \cdot \b. & \text{(by Exercise~6.5.7 since $\a \neq 0$)} 96 | } 97 | Hence $t(k) < \a \cdot \b$ so that $t(k) \in \a \cdot \b$. 98 | Thus $t: n \to \a \cdot \b$ and since clearly this sequence is finite it follows that $t \in \seqab$. 99 | 100 | Now we show that $g$ is injective. 101 | So consider $f_1$ and $f_2$ in $\sba$ such that $f_1 \neq f_2$. 102 | Let $h_1$, $n_1$ and $h_2$, $n_2$ be the increasing sequences and natural numbers for $f_1$ and $f_2$, respectively, as defined above. 103 | Also let $t_1 = g(f_1)$ and $t_2 = g(f_2)$. 104 | 105 | Case: $n_1 \neq n_2$. 106 | In this case the sequences are different sizes so that clearly $t_1 \neq t_2$ since $t_1$ and $t_2$ are sequences of sizes $n_1$ and $n_2$, respectively. 107 | 108 | Case: $n_1 = n_2$. 109 | Since $f_1 \neq f_2$ there must be a $\g \in \b$ such that $f_1(\g) \neq f_2(\g)$. 110 | Without loss of generality we can assume that $f_1(\g) < f_2(\g)$. 111 | 112 | If $f_1(\g) = 0$ then by definition $\g \notin s(f_1)$ but $\g \in s(f_2)$. 113 | Hence there is a $k \in n_2$ such that $h_2(k) = \g$. 114 | Also since $n_1 = n_2$ clearly $k \in n_1$. 115 | However, it must be that $h_1(k) \neq \g = h_2(k)$ since otherwise it would be that $\g \in s(f_1)$. 116 | Suppose that $h_1(k) < h_2(k)$ so that $h_1(k) + 1 \leq h_2(k)$ and we have 117 | \ali{ 118 | t_1(k) &= \a \cdot h_1(k) + f_1(h_1(k)) \\ 119 | &< \a \cdot h_1(k) + \a & \text{(by Lemma~6.5.4)} \\ 120 | &= \a \cdot (h_1(k) + 1) & \text{(by Definition~6.5.6b)} \\ 121 | &\leq \a \cdot h_2(k) & \text{(by Exercise~6.5.7 since $\a \neq 0$)} \\ 122 | &\leq \a \cdot h_2(k) + f_2(h_2(k)) \\ 123 | &= t_2(k) 124 | } 125 | so that $t_1(k) \neq t_2(k)$ and therefore $t_1 \neq t_2$. 126 | The case in which $h_1(k) > h_2(k)$ is analogous. 127 | 128 | On the other hand if $f_1(\g) \neq 0$ then $0 < f_1(\g) < f_2(\g)$ so that $\g \in s(f_1)$ and $\g \in s(f_2)$. 129 | Thus there are $k_1$ and $k_2$ in $n_1 = n_2$ such that $h_1(k_1) = h_2(k_2) = \g$. 130 | If $k_1 = k_2$ then we have 131 | \ali{ 132 | t_1(k_1) &= \a \cdot h_1(k_1) + f_1(h_1(k_1)) \\ 133 | &= \a \cdot \g + f_1(\g) \\ 134 | &\neq \a \cdot \g + f_2(\g) & \text{(by Lemma~6.5.4b since $f_1(\g) \neq f_2(\g)$)} \\ 135 | &= \a \cdot h_2(k_2) + f_2(h_2(k_2)) \\ 136 | &= t_2(k_2) = t_2(k_1) 137 | } 138 | so that $t_1 \neq t_2$. 139 | If $k_1 < k_2$ then since $h_1$ is increasing we have $h_2(k_2) = h_1(k_1) < h_1(k_2)$ so that $h_2(k_2) + 1 \leq h_1(k_2)$ and so 140 | \ali{ 141 | t_2(k_2) &= \a \cdot h_2(k_2) + f_2(h_2(k_2)) \\ 142 | &< \a \cdot h_2(k_2) + \a & \text{(by Lemma~6.5.4)} \\ 143 | &= \a \cdot (h_2(k_2) + 1) & \text{(by Definition~6.5.6b)} \\ 144 | &\leq \a \cdot h_1(k_2) & \text{(by Exercise~6.5.7 since $\a \neq 0$)} \\ 145 | &\leq \a \cdot h_1(k_2) + f_1(h_1(k_2)) \\ 146 | &= t_1(k_2) 147 | } 148 | and $t_1 \neq t_2$. 149 | The final sub-case in which $k_1 > k_2$ is analogous. 150 | 151 | Hence in all cases and sub-cases $g(f_1) = t_1 \neq t_2 = g(f_2)$, which shows that $g$ is injective. 152 | From this it follows from Definition~4.1.4 that $\abs{\sba} \leq \abs{\seqab}$. 153 | However, from what was shown above it follows that $\a \cdot \b$ is at most countable since $\a$ and $\b$ are. 154 | Thus $\seqab$ is also at most countable by Exercise~4.2.4 and Theorem~4.3.10 so that $\sba$ must be at most countable since it was just shown that $\abs{\sba} \leq \abs{\seqab}$. 155 | Lastly, since $\sba$ is equipotent to $\a^\b$ it follows that $\a^\b$ is at most countable as well. 156 | } 157 | } 158 | 159 | \exercise{3}{ 160 | For any set $A$, there is a mapping of $\pset{A \times A}$ onto $h(A)$. 161 | [Hint: Define $f(R) =$ the ordinal isomorphic to $R$, if $R \ss A \times A$ is a well-ordering of its field; $f(R) = 0$ otherwise.] 162 | } 163 | \sol{ 164 | \begin{lem}\label{lem:aleph:pset:sslt} 165 | For a set $A$ and ordinal $\a$, if $\a$ is equipotent to a subset of $A$ then $\a < h(A)$. 166 | \end{lem} 167 | \qproof{ 168 | Suppose that $\a$ is equipotent to $X \ss A$ but that $\a \geq h(A)$. 169 | Clearly if $\a = h(A)$ then $h(A)$ is equipotent to $X \ss A$ (since $\a$ is), which contradicts the definition of the Hartogs number. 170 | On the other hand if $\a > h(A)$ then let $f$ be a bijection from $\a$ to $X$. 171 | Then, since $h(A) < \a$ we have that $h(A) \in \a$ and $h(A) \ss \a$ since ordinals are transitive. 172 | It then follows that $f \rest h(A)$ is a bijection from $h(A)$ to $f[h(A)] \ss X \ss A$. 173 | Hence again $h(A)$ is equipotent to a subset of $A$, contradicting the definition of the Hartogs number. 174 | So it has to be that $\a < h(A)$ as desired. 175 | } 176 | 177 | \mainprob 178 | \qproof{ 179 | We define a function $f : \pset{A \times A} \to h(A)$. 180 | So for any $R \in \pset{A \times A}$ clearly $R \ss A \times A$ so that $R$ is a relation on $A$. 181 | If $R$ is a well-ordering of some $X \ss A$ then by Theorem~6.3.1 there is a unique ordinal $\a$ such that $(X, R)$ is isomorphic to $\a$. 182 | We then set 183 | $$ 184 | f(R) = \begin{cases} 185 | \a & \text{$R$ is a well-ordering of some $X \ss A$} \\ 186 | 0 & \text{$R$ is not a well-ordering of any $X \ss A$} \\ 187 | \end{cases}. 188 | $$ 189 | 190 | First we show that $f(R)$ really is in $h(A)$ for any $R \in \pset{A \times A}$. 191 | So for any such $R$, if $R$ is not a well-ordering of some $X \ss A$ then clearly $f(R) = 0$. 192 | Then, since clearly $\es \ss A$ and $\es$ is equipotent to 0, it follows that $0 < h(A)$ by Lemma~\ref{lem:aleph:pset:sslt}. 193 | Hence $f(R) = 0 \in h(A)$.\ 194 | On the other hand if $R$ is a well-ordering of some $X \ss A$ then let $\a$ be the ordinal isomorphic to $(X,R)$ so that $f(R) = \a$. 195 | Since this means that $\a$ is equipotent to $X \ss A$, it again follows from Lemma~\ref{lem:aleph:pset:sslt} that $\a < h(A)$ so that $f(R) = \a \in h(A)$. 196 | 197 | To show that $f$ is surjective consider any $\a \in h(A)$ so that $\a < h(A)$. 198 | Since by definition $h(A)$ is the \emph{least} ordinal that is not equipotent to a subset of $A$ it follows that $\a$ has to be equipotent to an $X \ss A$. 199 | Then let $R$ be the well-ordering of $X$ such that $(X,R)$ is isomorphic to $\a$. 200 | Clearly $R$ is a relation on $X$ and therefore also a relation on $A$ since $X \ss A$. 201 | Thus $R \ss A \times A$ so that $R \in \pset{A \times A}$. 202 | Clearly also $f(R) = \a$ and since $\a$ was arbitrary this shows that $f$ is surjective. 203 | } 204 | 205 | Note that this does not mean that $h(A) \leq \abs{\pset{A \times A}}$ unless the Axiom of Choice is employed. 206 | } 207 | 208 | \exercise{4}{ 209 | $\abs{A} < \abs{A} + h(A)$ for all $A$. 210 | } 211 | \sol{ 212 | \begin{lem}\label{lem:aleph:lthart:lthart} 213 | $\abs{h(A)} \nleq \abs{A}$ for any set $A$. 214 | \end{lem} 215 | \qproof{ 216 | Suppose to the contrary that $\abs{h(A)} \leq \abs{A}$ so that there is an injective $f$ from $h(A)$ to $A$. 217 | Then let $X = f[h(A)]$ so that clearly $X \ss A$. 218 | But then $f$ considered as function from $h(A)$ to $X$ is a bijection so that $h(A)$ is equipotent to a subset of $A$, which contradicts the definition of the Hartogs number. 219 | Hence it must be that $\abs{h(A)} \nleq \abs{A}$ as desired. 220 | } 221 | 222 | Note that this does not imply that $h(a) > \abs{A}$ without using the Axiom of Choice. 223 | 224 | \mainprob 225 | \qproof{ 226 | Since we are only interested in the cardinalities of $A$ and $h(A)$ and their cardinal sum we can assume that they are disjoint. 227 | Clearly $f: A \to A \cup h(A)$ defined by $f(x) = x$ for $x \in A$ is an injective function so that $\abs{A} \leq \abs{A} + h(A)$ by the definition of cardinal addition. 228 | So suppose that $\abs{A} = \abs{A} + h(A)$. 229 | Then there is a bijective function $f$ from $A \cup h(A)$ into $A$ so that $f \rest h(A)$ is an injective function from $h(A)$ to $A$. 230 | By definition this means that $h(A) \leq \abs{A}$, but this contradicts Lemma~\ref{lem:aleph:lthart:lthart}. 231 | So it must be that $\abs{A} < \abs{A} + h(A)$ as desired. 232 | } 233 | } 234 | 235 | \exercise{5}{ 236 | $\abs{h(A)} < \abs{\pset{\pset{A \times A}}}$ for all $A$. 237 | [Hint: Prove that $\abs{\pset{h(A)}} \leq \abs{\pset{\pset{A \times A}}}$ by assigning to each $X \in \pset{h(A)}$ the set of all well orderings $R \ss A \times A$ for which the ordinal isomorphic to $R$ belongs to $X$.] 238 | } 239 | \sol{ 240 | \def\pha{\pset{h(A)}} 241 | \def\paa{\pset{\pset{A \times A}}} 242 | \qproof{ 243 | First we show that $\abs{\pha} \leq \abs{\paa}$ by constructing an injective $f : \pha \to \paa$. 244 | So consider any $X \in \pha$ so that $X \ss h(A)$. 245 | Then let $Y$ be the set of well-orderings $R \ss A \times A$ (so that $R \in \pset{A \times A}$) of subsets $B \ss A$ such that $(B,R)$ is isomorphic to some $\a \in X$. 246 | We then set $f(X) = Y$, noting that clearly $f(X) = Y \in \paa$ since for any $R \in Y$ we have that $R \in \pset{A \times A}$ so that $Y \ss \pset{A \times A}$ hence $Y \in \paa$. 247 | Note also that $Y \neq \es$ because every $\a \in h(A)$ is equipotent to some subset $B \ss A$ (by the definition of the Hartogs number) so that the well-ordering of $B$ according to $\a$ will be in $Y$. 248 | 249 | We claim that $f$ is injective. 250 | So consider $X_1$ and $X_2$ in $\pha$ (so that $X_1 \ss h(A)$ and $X_2 \ss h(A)$) such that $f(X_1) = f(X_2)$. 251 | Then consider any $\a \in X_1$. 252 | Then since $f(X_1) \neq \es$ there is a well-ordering $R \in f(X_1)$ of a subset of $A$ that is isomorphic to $\a$. 253 | Then since $f(X_1) = f(X_2)$ we have $R \in f(X_2)$ as well. 254 | It follows from this that $\a \in X_2$. 255 | Thus $X_1 \ss X_2$ since $\a$ was arbitrary. 256 | A similar argument shows that $X_2 \ss X_1$ so that we conclude that $X_1 = X_2$. 257 | This shows that $f$ is injective. 258 | 259 | Hence we have shown that $\abs{\pha} \leq \abs{\paa}$. 260 | We also have by Cantor's Theorem (Theorem~5.1.8 in the text) that $\abs{h(A)} < \abs{\pha}$. 261 | It therefore follows from Exercise~4.1.2a that $\abs{h(A)} < \abs{\paa}$ as desired. 262 | } 263 | } 264 | 265 | \def\hs{h^*} 266 | \exercise{6}{ 267 | Let $\hs(A)$ be the least ordinal $\a$ such that there exists no function with domain $A$ and range $\a$. 268 | Prove: 269 | 270 | (a) If $\a \geq \hs(A)$, then there is no function with domain $A$ and range $\a$. 271 | 272 | (b) $\hs(A)$ is an initial ordinal. 273 | 274 | (c) $h(A) \leq \hs(A)$. 275 | 276 | (d) If $A$ is well-orderable, then $h(A) = \hs(A)$. 277 | 278 | (e) $\hs(A)$ exists for all $A$. 279 | 280 | [Hint for part (e): Show that $\a \in \hs(A)$ if and only if $\a = 0$ or $\a = $ the ordinal isomorphic to $R$, where $R$ is a well-ordering of some partition of $A$ into equivalence classes.] 281 | } 282 | \sol{ 283 | \begin{lem}\label{lem:aleph:hs:onto} 284 | If $A$ is a well-orderable set and $B$ is any other set, there is a function from $A$ onto $B$ if and only if $\abs{B} \leq \abs{A}$. 285 | \end{lem} 286 | \qproof{ 287 | Suppose that $R$ is a well-ordering of $A$. 288 | 289 | \bicproof{ 290 | Suppose that $f$ is a function from $A$ onto $B$. 291 | If $A$ is empty then clearly $B$ must be as well or else $f$ could not be onto. 292 | Thus we have $\abs{B} = \abs{\es} = 0 \leq 0 = \abs{\es} = \abs{A}$. 293 | So we can assume that $A$ is nonempty so that if $B$ is empty then $\abs{B} = \abs{\es} = 0 < \abs{A}$. 294 | Hence we can assume that $B$ is nonempty as well. 295 | 296 | Then, for each $b \in B$, define the set $A_b = \braces{a \in A \where f(a) = b}$, which clearly not empty since $f$ is onto. 297 | Then, since $R$ is a well-ordering of $A$ and $A_b \ss A$, there is a unique least element of $a_b$ of $A_b$ according to $R$. 298 | We then define $g: B \to A$ by simply setting $g(b) = a_b$ for any $b \in B$. 299 | 300 | We then claim that $g$ is injective. 301 | So consider $b_1$ and $b_2$ in $B$ where $b_1 \neq b_2$. 302 | Then since $a_{b_1} \in A_{b_1}$ clearly $f(a_{b_1}) = b_1$. 303 | Similarly $f(a_{b_2}) = b_2$ so that clearly $a_{b_1} \neq a_{b_2}$ since $f$ is a function and $b_1 \neq b_2$. 304 | Hence $g(b_1) = a_{b_1} \neq a_{b_2} = g(b_2)$, which shows that $g$ is injective since $b_1$ and $b_2$ were arbitrary. 305 | Then by definition $\abs{B} \leq \abs{A}$ as desired. 306 | }{ 307 | Now suppose that $\abs{B} \leq \abs{A}$ so that there is an injective $f: B \to A$. 308 | If $B$ is empty then clearly it has to be that $\abs{B} = \abs{\es} = 0 \leq \abs{A}$ regardless of $A$. 309 | So we can assume that $B$ is nonempty so that there is a $b \in B$. 310 | Since $f$ is injective, the inverse $\inv{f}$ is a function from $\ran(f)$ onto $B$. 311 | Now we construct a function $g: A \to B$ by setting 312 | $$ 313 | g(x) = \begin{cases} 314 | \inv{f}(x) & x \in \ran(f) \\ 315 | b & x \notin \ran(f) 316 | \end{cases} 317 | $$ 318 | for any $x \in A$. 319 | It should be clear that $g$ maps $A$ onto $B$ since, for any $b \in B$, we have $f(b) \in \ran(f)$ (hence also $f(b) \in A$) so that $g(f(b)) = \inv{f}(f(b)) = b$. 320 | } 321 | } 322 | 323 | \mainprob 324 | 325 | First we note that presumably $\hs(A)$ for a set $A$ is the least \emph{nonzero} ordinal $\a$ such that there is no function from $A$ onto $\a$. 326 | The fact that $\hs(A)$ is nonzero is important since, for a non-empty set $A$, there is no function from $A$ onto $\es = 0$ so that $0$ is actually the least ordinal such that such a function does not exist! 327 | We also make note of the fact that, even for $A=\es$, the empty function $f = \es$ is a vacuously a function from $A$ onto $\es = 0$ so that $\hs(A) = 1$ anyway since there is no function from $A = \es$ onto $1 = \braces{0}$. 328 | 329 | (a) 330 | \qproof{ 331 | Suppose to the contrary that there \emph{is} a function $f$ from $A$ onto $\a$. 332 | Then since $0 < \hs(A) \leq \a$ clearly $0 \in \hs(A)$ and $\hs(A) \ss \a$. 333 | So we define a function $g : A \to \hs(A)$ by 334 | $$ 335 | g(a) = \begin{cases} 336 | f(a) & f(a) \in \hs(A) \\ 337 | 0 & f(a) \notin \hs(A) 338 | \end{cases}. 339 | $$ 340 | for any $a \in A$. 341 | Clearly each $g(a) \in \hs(A)$ but we also claim that $g$ is onto. 342 | To this end consider any $\b \in \hs(A)$. 343 | Since $\hs(A) \ss \a$ we have that $\b \in \a$ also. 344 | Then, since $f$ is onto $\a$ there is an $a \in A$ such that $f(a) = \b$. 345 | Since $f(a) = \b \in \hs(A)$ it follows by definition that $g(a) = f(a) = \b$. 346 | Since $\b$ was arbitrary this shows that $g$ is onto. 347 | However, the existence of $g$ contradicts the definition of $\hs(A)$ so that it must be that there is no such function from $A$ onto $\a$. 348 | } 349 | 350 | (b) 351 | \qproof{ 352 | Suppose to the contrary that $\hs(A)$ is \emph{not} an initial ordinal so that there is an $\a < \hs(A)$ such that $\abs{\a} = \abs{\hs(A)}$. 353 | Let $f$ then be a bijection from $\a$ onto $\hs(A)$. 354 | Also since $\a < \hs(A)$ it follows from the definition of $\hs(A)$ that there is a function $g$ from $A$ onto $\a$ (since otherwise $\hs(A)$ would not be the \emph{least} such ordinal for which such a function does not exist). 355 | But then $f \circ g$ is a function from $A$ onto $\hs(A)$, which contradicts the definition of $\hs(A)$. 356 | Hence it has to be that $\hs(A)$ is in fact an initial ordinal. 357 | } 358 | 359 | (c) 360 | \qproof{ 361 | Suppose to the contrary that $h(A) > \hs(A)$. 362 | Then by the definition of $h(A)$ there is a subset $X \ss A$ such that $\hs(A)$ is equipotent to $X$. 363 | So let $f$ be a bijection from $\hs(A)$ to $X$. 364 | We then define a function $g : A \to \hs(A)$ by 365 | $$ 366 | g(a) = \begin{cases} 367 | \inv{f}(a) & a \in X \\ 368 | 0 & a \notin X 369 | \end{cases} 370 | $$ 371 | for any $a \in A$, noting that $0 < \hs(A)$ so that $0 \in \hs(A)$. 372 | Clearly $g$ is into $\hs(A)$ but we also claim that it is onto. 373 | So consider any $\a \in \hs(A)$ and let $a = f(\a)$ so that $a \in X$ and therefore $a \in A$ since $X \ss A$. 374 | We then have $g(a) = \inv{f}(a) = \inv{f}(f(\a)) = \a$ since $a \in X$. 375 | Since $\a$ was arbitrary this shows that $g$ is onto. 376 | However, the existence of $g$ contradicts the definition of $\hs(A)$ so that it must be that in fact $h(A) \leq \hs(A)$ as desired. 377 | } 378 | 379 | (d) 380 | \qproof{ 381 | We show that $\hs(A) \leq h(A)$, from which the result clearly follows since also $\hs(A) \geq h(A)$ by part (c). 382 | So suppose to the contrary that $\hs(A) > h(A)$. 383 | Then by the definition of $\hs(A)$ it follows that there is a function from $A$ onto $h(A)$. 384 | It then follows from Lemma~\ref{lem:aleph:hs:onto} that $\abs{h(A)} \leq \abs{A}$ since $A$ is well-orderable. 385 | However, this contradicts Lemma~\ref{lem:aleph:lthart:lthart} so that it must be that $\hs(A) \leq h(A)$ so that the result follows. 386 | } 387 | 388 | (e) 389 | \qproof{ 390 | Consider any set $A$ and let $S$ denote the set of well-orderings of some partition of $A$ into equivalence classes, noting that it could be that $S = \es$. 391 | Since each $R \in S$ is isomorphic to a unique ordinal, let $H$ be the set of ordinals that are isomorphic to some $R \in S$, which exists by the Axiom Schema of Replacement. 392 | Then let $\a = \braces{0} \cup H$ and we claim that $\a = \hs(A)$. 393 | 394 | First we show that $\a$ is indeed an ordinal number. 395 | Since $\a$ is a set of ordinals clearly it is well-ordered by Theorem~6.2.6d. 396 | We also must show that $\a$ is transitive, so consider any $\b \in \a$. 397 | Then either $\b = 0$ or $\b \in H$. 398 | If $\b = 0 = \es$ then clearly $\b \ss \a$. 399 | On the other hand if $\b \in H$ then there is a partition $P$ of $A$ and a well-ordering $R$ of $P$ such that $(P, R)$ is isomorphic to $(\b, <)$. 400 | Now consider any $\g \in \b$ so that $\g < \b$. 401 | It then follows that $\g$ is isomorphic to an initial segment of $\b$ and therefore also to an initial segment $P'$ of $P$ ordered by $R$. 402 | Let $L$ be the least element of $P$ (which is also the least element of $P'$), which exists since $R$ is a well-ordering. 403 | Then let 404 | $$ 405 | L' = L \cup \parens{A - \bigcup P'}, 406 | $$ 407 | i.e. $L'$ is the set containing the elements of $L$ and any elements of $A$ that are not covered in the initial segment $P'$. 408 | Then let 409 | $$ 410 | P'' = \braces{L'} \cup \parens{P' - \braces{L}}, 411 | $$ 412 | i.e. $P''$ is $P'$ but with $L$ replaced with $L'$. 413 | It is easy to show that $P''$ is a partition of $A$ and that it is isomorphic to $\g$ with the same ordering as $R$ except with $L$ replaced by $L'$. 414 | Hence by definition we have that $\g \in \a$. 415 | Since $\g \in \b$ was arbitrary this shows that $\b \ss \a$, and since $\b \in \a$ was arbitrary this shows that $\a$ is transitive and hence an ordinal number. 416 | 417 | Now we show that there is no function from $A$ onto $\a$. 418 | So suppose to the contrary that there is such a function $f$. 419 | We then define the set 420 | $$ 421 | E = \braces{(a,b) \in A \times A \where f(a) = f(b)}. 422 | $$ 423 | It is trivial to show that this is an equivalence relation on $A$ so that $A/E$ is a partition of $A$ by Theorem~4.4.7. 424 | Moreover let $g$ be the mapping from $A/E$ to $\a$ defined as follows: for any $B \in A/E$ let $g(B)$ be the least element of $\braces{f(x) \where x \in B}$, noting that $\braces{f(x) \where x \in B}$ contains only a single element since $B$ is an equivalence class where $f(x) = f(y)$ for any $x$ and $y$ in $B$. 425 | It is trivial to show that $g$ is a bijective function so that we can well-order $A/E$ according to the ordinal $\a$ since $\a$ is the range of $g$. 426 | However, it then follows by definition that $\a \in \a$, which contradicts Lemma~6.2.7. 427 | Hence it must be that there is no function from $A$ onto $\a$. 428 | 429 | Lastly we show that there is a function from $A$ onto $\b$ for every nonzero $\b < \a$. 430 | So consider any such $\b$ so that $\b \in \a$. 431 | Then either $\b = 0$ or $\b \in H$, but since $\b$ is nonzero it must be that $\b \in H$. 432 | Then by definition there is is a partition $P$ of $A$ and well-ordering $R$ of $P$ such that $(P,R)$ is isomorphic to $(\b,<)$. 433 | Let $f$ then be the isomorphism from $P$ to $\b$. 434 | We then define the mapping $g : A \to \b$ as follows: for any $a \in A$ there is a unique $B \in P$ such that $a \in B$ since $P$ is a partition of $A$. 435 | We then set $g(a) = f(B)$. 436 | It is easy to show that $g$ is onto. 437 | 438 | It follows from what has been shown that indeed $\a = \hs(A)$. 439 | } 440 | } 441 | -------------------------------------------------------------------------------- /sections/sec_8_2.tex: -------------------------------------------------------------------------------- 1 | \subsection{The Axiom of Choice in Mathematics} 2 | 3 | \exercise{1}{ 4 | Without using the Axiom of Choice, prove that the two definitions of closure points are equivalent if $A$ is an open set. 5 | [Hint: $X_n$ is open, so $X_n \cap \rats \neq \es$, and $\rats$ can be well-ordered.] 6 | } 7 | \sol{ 8 | \begin{lem}\label{lem:aoc:intopen} 9 | If $A$ and $B$ are open subsets of $\reals$ then $A \cap B$ is open. 10 | \end{lem} 11 | \qproof{ 12 | First note that, if $A \cap B = \es$, then this is vacuously open. 13 | Otherwise, consider any $x \in A \cap B$. 14 | Then, since $A$ is open and $x \in A$, there is a real $\d_1 > 0$ such that $\abs{y - x} < \d_1$ implies that $y \in A$ for all $y \in \reals$. 15 | Similarly, $x \in B$ and $B$ is open so that there is a $\d_2 > 0$ where $\abs{y - x} < \d_2$ implies $y \in B$ for all $y \in \reals$ 16 | Let $\d = \min\braces{\d_1, \d_2}$ so that $\d \leq \d_1$ and $\d \leq \d_2$. 17 | Then consider any $y \in \reals$ where $\abs{y - x} < \d$. 18 | Then we have $\abs{y - x} < \d \leq \d_1$ so that $y \in A$. 19 | Similarly, $\abs{y - x} < \d \leq \d_2$ so that $y \in B$ as well. 20 | Hence $y \in A \cap B$. 21 | Since $y$ was arbitrary and $\d > 0$, this shows that $A \cap B$ is open. 22 | } 23 | 24 | \begin{lem}\label{lem:aoc:openrat} 25 | If $A$ is a nonempty open subset of $\reals$ then $A \cap \rats \neq \es$. 26 | \end{lem} 27 | \qproof{ 28 | Suppose that $A \ss \reals$ is nonempty and open. 29 | Then there is an $x \in A$ and, since $A$ is open, there is a real $\d > 0$ such that $(x-\d, x+\d) \ss A$. 30 | Now, since $\d > 0$, we clearly have 31 | \gath{ 32 | -\d < 0 < \d \\ 33 | x-\d < x+\d. 34 | } 35 | Since $\rats$ is dense in $\reals$ with respect to order, there is a $q \in \rats$ such that $x-\d < q < x+\d$. 36 | Then we have $q \in (x-\d, x+\d)$ so that also $q \in A$ since $(x-\d, x+\d) \ss A$. 37 | Thus $q \in \rats$ and $q \in A$ so that $q \in A \cap \rats$ as desired. 38 | } 39 | 40 | \mainprob 41 | \qproof{ 42 | A proof of this equivalence is presented in the text as Theorem~8.2.1. 43 | Recall that, in the proof that (b) implies (a), $a \in \reals$ is the closure point of $A \ss \reals$ and $X_n = \braces{x \in A \where \abs{x-a} < 1/n}$, and we know from (b) that each $X_n$ is nonempty. 44 | Note that we assume that $X_0 = A \cap (-\infty, \infty) = A \cap \reals = A$. 45 | Per the remarks after this proof, it suffices to show that the system of nonempty sets $\braces{X_n}_{n \in \nats}$ has a choice function when $A$ is open. 46 | 47 | First we show that each $X_n$ is an open set. 48 | Clearly $X_0 = A$ is open, so consider any natural $n > 0$ and let $I_n = (a-1/n, a+1/n)$. 49 | We claim that $X_n = A \cap I_n$. 50 | To this end we have 51 | \ali{ 52 | x \in X_n &\bic x \in A \land \abs{x-a} < 1/n \\ 53 | &\bic x \in A \land -1/n < x - a < 1/n \\ 54 | &\bic x \in A \land a-1/n < x < a+1/n \\ 55 | &\bic x \in A \land x \in I_n \\ 56 | &\bic x \in A \cap I_n. 57 | } 58 | for any real $x$, which of course shows that $X_n = A \cap I_n$. 59 | Then, since $A$ is open and $I_n$ is clearly an open interval, it follows from Lemma~\ref{lem:aoc:intopen} that $X_n$ is open as well. 60 | 61 | Now, since each $X_n$ is open and nonempty, it follows that $X_n \cap \rats \neq \es$ from Lemma~\ref{lem:aoc:openrat}. 62 | Then, since $\rats$ is countable, it can clearly be well-ordered. 63 | So choose a well-ordering of $\rats$. 64 | Since $X_n \cap \rats$ is clearly a nonempty subset of $\rats$, it then has a least element $x_n$ according to our well-ordering. 65 | We then define a function $g$ on $\braces{X_n}_{n \in \nats}$ by $g(X_n) = x_n$, which is clearly a choice function. 66 | } 67 | } 68 | 69 | \exercise{2}{ 70 | Prove that every continuous additive function $f$ is equal to $f_a$ for some $a \in \reals$. 71 | } 72 | \sol{ 73 | \qproof{ 74 | Consider any arbitrary continuous additive function $f : \reals \to \reals$. 75 | Then, by what was shown in the text, there is a real $a$ such that $f(q) = f_a(q) = a \cdot q$ for all $q \in \rats$; in particular $a = f(1)$. 76 | 77 | Now suppose to the contrary that $f \neq f_a$ so that there is an $x \in \reals$ where $f(x) \neq f_c(x)$. 78 | So let $\e = \abs{f(x) - f_a(x)}/2$, noting that clearly $\e > 0$ since $f(x) \neq f_a(x)$. 79 | Since $f$ is continuous there is a real $\d_1 > 0$ such that $\abs{f(y) - f(x)} < \e$ for all $y \in \reals$ where $\abs{y - x} < \d_1$. 80 | Also clearly $f_a$ is also continuous so that there is a real $\d_2 > 0$ where $\abs{f_a(y) - f_a(x)} < \e$ for all $y \in \reals$ where $\abs{y - x} < \d_2$. 81 | So let $\d = \min\braces{\d_1, \d_2}$. 82 | Then, since $\d > 0$ it follows that $x - \d < x + \d$ so that there is a $q \in \rats$ where $x - \d < q < x + \d$ since $\rats$ is order dense in $\reals$. 83 | It then clearly follows that $\abs{q - x} < \d$ so that $\abs{q - x} < \d \leq \d_1$ and $\abs{q - x} < \d \leq \d_2$. 84 | Therefore $\abs{f(q) - f(x)} < \e$ and $\abs{f_a(q) - f_a(x)} < \e$. 85 | 86 | We then have 87 | \ali{ 88 | \abs{f(x) - f_a(x)} &\leq \abs{f(x) - f(q)} + \abs{f(q) - f_a(x)} \\ 89 | &\leq \abs{f(x) - f(q)} + \abs{f(q) - f_a(q)} + \abs{f_a(q) - f_a(x)} \\ 90 | &< \e + 0 + \e = 2\e = \abs{f(x) - f_a(x)}, 91 | } 92 | which is a contradiction, noting that $\abs{f(q) - f_a(q)} = 0$ since $f(q) = f_a(q)$ since $q \in \rats$. 93 | So it must be that in fact $f = f_a$ as desired. 94 | } 95 | 96 | This proof is similar to that of Theorem~10.3.11 later in the text. 97 | That theorem is certainly more general, and this can be easily proved from it. 98 | In particular it was shown in the text that, for an arbitrary additive and continuous $f: \reals \to \reals$, $f(q) = f_a(q)$ for all $q \in \rats$ for some $a \in \reals$ so that $f \rest \rats = f_a \rest \rats$. 99 | Since $f$ and $f_a$ are both continuous and $\rats$ is order dense in $\reals$, it follows from Theorem~10.3.11 that $f = f_a$. 100 | } 101 | 102 | \exercise{3}{ 103 | Assume that $\mu$ has properties 0)-ii). 104 | Prove properties iv) and v). 105 | Also prove: 106 | \begin{enumerate} 107 | \item[\ilab{vi}] $\mu(A \cup B) = \mu(A) + \mu(B) - \mu(A \cap B)$. 108 | \item[\ilab{vii}] $\mu\parens{\bigcup_{n=0}^\infty A_n} \leq \sum_{n=0}^\infty \mu(A_n)$. 109 | \end{enumerate} 110 | } 111 | \sol{ 112 | First, for reference, we assume the following properties of $\mu$: 113 | \begin{enumerate} 114 | \item[\ilab{0}] $\mu([a,b]) = b - a$ for any $a$ and $b$ in $\reals$ where $a < b$. 115 | \item[\ilab{i}] $\mu(\es) = 0$ and $\mu(\reals) = \infty$. 116 | \item[\ilab{ii}] If $\braces{A_n}_{n=0}^\infty$ is a collection of mutually disjoint subsets of $\reals$, then 117 | \gath{ 118 | \mu\parens{\bigcup_{n=0}^\infty A_n} = \sum_{n=0}^\infty \mu(A_n). 119 | } 120 | \end{enumerate} 121 | First we show 122 | \begin{enumerate} 123 | \item[\ilab{iv}] If $A \cap B = \es$ then $\mu(A \cup B) = \mu(A) + \mu(B)$. 124 | \end{enumerate} 125 | \qproof{ 126 | Assume that $A \cap B = \es$ and define $A_0 = A$, $A_1 = B$, and $A_n = \es$ for all natural $n > 1$. 127 | Then clearly each of the sets in $\braces{A_n}_{n=0}^\infty$ are mutually disjoint. 128 | It is also trivial to show that $\bigcup_{n=0}^\infty A_n = A \cup B$. 129 | We then have by property ii) that 130 | \ali{ 131 | \mu(A \cup B) &= \mu\parens{\bigcup_{n=0}^\infty A_n} = \sum_{n=0}^\infty \mu(A_n) = \mu(A_0) + \mu(A_1) + \sum_{n=2}^\infty \mu(A_n) \\ 132 | &= \mu(A) + \mu(B) + \sum_{n=2}^\infty \mu(\es) = \mu(A) + \mu(B) + \sum_{n=2}^\infty 0 \\ 133 | &= \mu(A) + \mu(B), 134 | } 135 | noting that we have also used property i) according to which $\mu(\es) = 0$. 136 | This shows the desired result. 137 | } 138 | 139 | \begin{lem}\label{lem:aoc:measmin} 140 | For a measure $\mu$, if $A \ss B \ss \reals$ then $\mu(B-A) = \mu(B) - \mu(A)$. 141 | \end{lem} 142 | \qproof{ 143 | Clearly $A$ and $B-A$ are disjoint sets such that $A \cup (B-A) = B$ so that $\mu(A) + \mu(B-A) = \mu(B)$ by property iv). 144 | The result then clearly follows by subtracting $\mu(A)$ from both sides. 145 | } 146 | 147 | Next we show 148 | \begin{enumerate} 149 | \item[\ilab{v}] If $A \ss B$ then $\mu(A) \leq \mu(B)$. 150 | \end{enumerate} 151 | \qproof{ 152 | Suppose that $A \ss B$ so that $\mu(A) + \mu(B-A) = \mu(B)$ by Lemma~\ref{lem:aoc:measmin}. 153 | Since $\mu$ is a function into $\clop{0, \infty} \cup \braces{\infty}$, it follows that 154 | \gath{ 155 | 0 \leq \mu(B-A) \\ 156 | \mu(A) \leq \mu(B-A) + \mu(A) = \mu(A) + \mu(B-A) \\ 157 | \mu(A) \leq \mu(B) 158 | } 159 | as desired. 160 | } 161 | 162 | Now we show 163 | \begin{enumerate} 164 | \item[\ilab{vi}] $\mu(A \cup B) = \mu(A) + \mu(B) - \mu(A \cap B)$. 165 | \end{enumerate} 166 | \qproof{ 167 | Let $C = A \cap B$, $A' = A - C$, and $B' = B - C$. 168 | It is then trivial to show that $A'$, $B'$, and $C$ are mutually disjoint sets such that $A' \cup B' \cup C = A \cup B$. 169 | We then have by a straightforward extension of property iv) that 170 | \gath{ 171 | \mu(A') + \mu(B') + \mu(C) = \mu(A \cup B) \\ 172 | \mu(A - C) + \mu(B - C) + \mu(C) = \mu(A \cup B) \\ 173 | \mu(A) - \mu(C) + \mu(B) - \mu(C) + \mu(C) = \mu(A \cup B) \\ 174 | \mu(A) + \mu(B) - \mu(C) = \mu(A \cup B) \\ 175 | \mu(A) + \mu(B) - \mu(A \cap B) = \mu(A \cup B) 176 | } 177 | as desired. 178 | Note that we have also used Lemma~\ref{lem:aoc:measmin} since clearly $C \ss A$ and $C \ss B$ so that $\mu(A-C) = \mu(A) - \mu(C)$ and $\mu(B-C) = \mu(B) - \mu(C)$. 179 | } 180 | 181 | Lastly we show 182 | \begin{enumerate} 183 | \item[\ilab{vii}] $\mu\parens{\bigcup_{n=0}^\infty A_n} \leq \sum_{n=0}^\infty \mu(A_n)$. 184 | \end{enumerate} 185 | \qproof{ 186 | Supposing that we have a system of sets $\braces{A_n}_{n=0}^\infty$, first we define a sequence of corresponding sets recursively: 187 | \ali{ 188 | A_0' &= A_0 \\ 189 | A_{n+1}' &= A_{n+1} - \bigcup_{k=0}^n A_k' 190 | } 191 | Note that it is clear that $A_n' \ss A_n$ for any $n \in \nats$ so that $\mu(A_n') \leq \mu(A_n)$ by property v). 192 | 193 | We now show that each of these sets are mutually disjoint. 194 | So consider any natural $m$ and $n$ where $m \neq n$. 195 | Without loss of generality, we can then assume that $m < n$. 196 | Suppose that $A_m'$ and $A_n'$ are \emph{not} disjoint so that there is an $x \in A_m' \cap A_n'$. 197 | Thus $x \in A_n' = A_n - \bigcup_{k=0}^{n-1} A_k'$ so that $x \notin \bigcup_{k=0}^{n-1} A_k'$. 198 | However since also $x \in A_m'$ and $0 \leq m \leq n-1$, we also can conclude that $x \in \bigcup_{k=0}^{n-1} A_k'$. 199 | Since this is a contradiction, it must be that $A_m'$ and $A_n'$ are in fact disjoint, which shows mutual disjointedness since $m$ and $n$ were arbitrary. 200 | 201 | Next we show that $\bigcup_{n=0}^\infty A_n' = \bigcup_{n=0}^\infty A_n$. 202 | The $\ss$ direction is clear since, for any $x \in \bigcup_{n=0}^\infty A_n'$, there a natural $n$ where $x \in A_n'$. 203 | Since $A_n' \ss A_n$ it follows that $x \in A_n$ so that clearly $x \in \bigcup_{n=0}^\infty A_n$. 204 | Now consider any $x \in \bigcup_{n=0}^\infty A_n$ so that there is a natural $n$ where $x \in A_n$. 205 | Clearly if $n=0$ then $x \in A_0 = A_0'$. 206 | So assume that $n > 0$ so that $\bigcup_{k=0}^{n-1} A_k'$ is defined. 207 | If $x \in \bigcup_{k=0}^{n-1} A_k'$ then there is a $0 \leq k \leq n-1$ where $x \in A_k'$. 208 | On the other hand, if $x \notin \bigcup_{k=0}^{n-1} A_k'$ then clearly $x \in A_n - \bigcup_{k=0}^{n-1} A_k' = A_n'$. 209 | Thus in all cases there is a natural $k$ such that $x \in A_k'$ so that $x \in \bigcup_{n=0}^\infty A_n'$ as desired. 210 | 211 | We therefore have 212 | \ali{ 213 | \mu\parens{\bigcup_{n=0}^\infty A_n} &= \mu\parens{\bigcup_{n=0}^\infty A_n'} \\ 214 | &= \sum_{n=0}^\infty \mu(A_n') & \text{(by property ii) since the sets $\braces{A_n'}_{n=0}^\infty$ are mutually disjoint)} \\ 215 | &\leq \sum_{n=0}^\infty \mu(A_n) & \text{(since $0 \leq \mu(A_n') \leq \mu(A_n)$ for all natural $n$)} 216 | } 217 | as desired. 218 | } 219 | } 220 | 221 | \def\defsa{ 222 | \begin{enumerate} 223 | \item[(a)] $\es \in \fs$ and $S \in \fs$. 224 | \item[(b)] If $X \in \fs$ then $S - X \in \fs$. 225 | \item[(c)] If $X_n \in \fs$ for all $n$, then $\bigcup_{n=0}^\infty X_n \in \fs$ and $\bigcap_{n=0}^\infty X_n \in \fs$. 226 | \end{enumerate} 227 | } 228 | \def\fs{\mathfrak{S}} 229 | \exerciseapp{4}{*}{ 230 | Let $\fs = \braces{X \ss S \where \abs{X} \leq \al_0 \text{ or } \abs{S-X} \leq \al_0}$. 231 | Prove that $\fs$ is a $\s$-algebra. 232 | } 233 | \sol{ 234 | \begin{lem}\label{lem:aoc:setdm} 235 | If $A$, $B$, and $C$ are sets then $A - (B - C) = (A - B) \cup (A \cap C)$. 236 | \end{lem} 237 | \qproof{ 238 | For any $x$ we have 239 | \ali{ 240 | x \in A - (B - C) &\bic x \in A \land x \notin B - C \\ 241 | &\bic x \in A \land \lnot (x \in B \land x \notin C) \\ 242 | &\bic x \in A \land (x \notin B \lor x \in C) \\ 243 | &\bic (x \in A \land x \notin B) \lor (x \in A \land x \in C) \\ 244 | &\bic X \in A - B \lor x \in A \cap C \\ 245 | &\bic x \in (A - B) \cup (A \cap C). 246 | } 247 | } 248 | 249 | \begin{lem}\label{lem:aoc:smss} 250 | If $A \ss B$ and $S$ is another set, then $S - B \ss S - A$. 251 | \end{lem} 252 | \qproof{ 253 | Consider any $x \in S - B$ so that $x \in S$ and $x \notin B$. 254 | Then it has to be that also $x \notin A$ since otherwise it would not be that $A \ss B$. 255 | Hence $x \in S - A$, which shows the result since $x$ was arbitrary. 256 | } 257 | 258 | \mainprob 259 | \qproof{ 260 | We must show that the above definition of $\fs$ satisfies the three parts of the definition of a $\s$-algebra: 261 | \defsa 262 | 263 | For (a) clearly $\abs{\es} = 0 \leq \al_0$ so that $\es \in \fs$, and $S - S = \es$ so that $\abs{S - S} = \abs{\es} = 0 \leq \al_0$. 264 | Hence $S \in \fs$ as well. 265 | 266 | For (b) suppose that $X \ss S$ and $X \in \fs$. 267 | Then either $\abs{X} \leq \al_0$ or $\abs{S - X} \leq \al_0$. 268 | If $\abs{X} \leq \al_0$ then by Lemma~\ref{lem:aoc:setdm} we have 269 | \gath{ 270 | S - (S - X) = (S - S) \cup (S \cap X) = \es \cup (S \cap X) = S \cap X = X 271 | } 272 | since $X \ss S$. 273 | Therefore $\abs{S - (S - X)} = \abs{X} \leq \al_0$ so that $S - X \in \fs$. 274 | On the other hand, if $\abs{S - X} \leq \al_0$, then obviously $S - X \in \fs$ by definition. 275 | 276 | Lastly, regarding (c), suppose that $\braces{X_n}_{n=0}^\infty$ is a system of sets where each $X_n$ is in $\fs$. 277 | Thus $\abs{X_n} \leq \al_0$ or $\abs{S - X_n} \leq \al_0$ for each natural $n$. 278 | 279 | Now we show that $\bigcup_{n=0}^\infty X_n \in \fs$. 280 | First, if $\abs{X_n} \leq \al_0$ for all natural $n$, then it follows from Theorem~8.1.7 that $\abs{\bigcup_{n=0}^\infty X_n} \leq \al_0$, which of course uses the Axiom of Choice. 281 | If, on the other hand, there is a natural $m$ such that $\abs{X_m} \not\leq \al_0$, then it has to be that $\abs{S - X_m} \leq \al_0$ since $X_m \in \fs$. 282 | Then, since clearly $X_m \ss \bigcup_{n=0}^\infty X_n$, it follows from Lemma~\ref{lem:aoc:smss} that $S - \bigcup_{n=0}^\infty X_n \ss S - X_m$ so that clearly $\abs{S - \bigcup_{n=0}^\infty X_n} \leq \abs{S - X_m} \leq \al_0$. 283 | Thus in all cases we have that either $\abs{\bigcup_{n=0}^\infty X_n} \leq \al_0$ or $\abs{S - \bigcup_{n=0}^\infty X_n} \leq \al_0$ so that $\bigcup_{n=0}^\infty X_n \in \fs$. 284 | 285 | Lastly we show that $\bigcap_{n=0}^\infty X_n \in \fs$ as well. 286 | If it is the case $\abs{S - X_n} \leq \al_0$ for all natural $n$, then clearly we have that $\bigcup_{n=0}^\infty (S - X_n) \leq \al_0$, again by Theorem~8.1.7. 287 | It also follows from Exercise~2.3.11 that $S - \bigcap_{n=0}^\infty X_n = \bigcup_{n=0}^\infty (S - X_n)$ so that we have $\abs{S - \bigcap_{n=0}^\infty X_n} = \abs{\bigcup_{n=0}^\infty (S - X_n)} \leq \al_0$. 288 | Now, on the other hand, if there is a natural $m$ such that $\abs{S - X_m} \not\leq \al_0$, then it has to be that $\abs{X_m} \leq \al_0$ since $X_m \in \fs$. 289 | Since clearly $\bigcap_{n=0}^\infty X_n \ss X_m$, we then have $\abs{\bigcap_{n=0}^\infty X_n} \leq \abs{X_m} \leq \al_0$. 290 | Hence in either case we have that $\abs{\bigcap_{n=0}^\infty X_n} \leq \al_0$ or $\abs{S - \bigcap_{n=0}^\infty X_n} \leq \al_0$ so that $\bigcap_{n=0}^\infty X_n \in \fs$ by definition. 291 | 292 | We have therefore shown parts (a), (b), and (c) so that $\fs$ is a $\s$-algebra as desired. 293 | } 294 | } 295 | 296 | \def\fc{\mathfrak{C}} 297 | \def\ft{\mathfrak{T}} 298 | \exercise{5}{ 299 | Let $\fc$ be any collection of subsets of $S$. 300 | Let $\fs = \bigcap \braces{\ft \where \fc \ss \ft \text{ and $\ft$ is a $\s$-algebra of subsets of $S$}}$. 301 | Prove that $\fs$ is a $\s$-algebra (it is called the \emph{$\s$-algebra generated} by $\fc$). 302 | } 303 | \sol{ 304 | \qproof{ 305 | First, let $T = \braces{\ft \where \fc \ss \ft \text{ and $\ft$ is a $\s$-algebra of subsets of $S$}}$ so that $\fs = \bigcap T$. 306 | Then we must show that $\fs$ meets the definition of a $\s$-algebra: 307 | \defsa 308 | 309 | Regarding (a), consider any $\ft \in T$. 310 | Since $\ft$ is then a $\s$-algebra it follows that both $\es \in \ft$ and $S \in \ft$ by (a). 311 | Then, since $\ft \in T$ was arbitrary, it follows that both $\es$ and $S$ are in $\bigcap T = \fs$. 312 | 313 | For (b) suppose that $X \in \fs = \bigcap T$ so that $X \in \ft$ for all $\ft \in T$. 314 | So consider any such $\ft \in T$ so that clearly $X \in \ft$. 315 | Then, since $\ft$ is then a $\s$-algebra, it follows that $S - X \in \ft$ by (b). 316 | Since $\ft \in T$ was arbitrary, we have that $S - X \in \bigcap T = \fs$. 317 | 318 | Lastly, for part (c) of the definition, suppose that $X_n \in \fs = \bigcap T$ for all $n \in \nats$. 319 | Let $\ft$ be any element of $T$ so that $X_n \in \ft$ for all natural $n$. 320 | Since $\ft$ is a $\s$-algebra, it then follows from (c) that both $\bigcup_{n=0}^\infty X_n$ and $\bigcap_{n=0}^\infty X_n$ are in $\ft$. 321 | Since $\ft \in T$ was arbitrary we have that $\bigcup_{n=0}^\infty X_n$ and $\bigcap_{n=0}^\infty X_n$ are in $\bigcap T = \fs$. 322 | 323 | Hence we have shown all three parts of the definition so that $\fs$ is indeed a $\s$-algebra. 324 | } 325 | } 326 | 327 | \def\defsam{ 328 | \begin{enumerate} 329 | \item[\ilab{i}] $\mu(\es) = 0$, $\mu(S) > 0$. 330 | \item[\ilab{ii}] If $\braces{X_n}_{n=0}^\infty$ is a collection of mutually disjoint sets from $\fs$, then 331 | \gath{ 332 | \mu\parens{\bigcup_{n=0}^\infty X_n} = \sum_{n=0}^\infty \mu(X_n). 333 | } 334 | \end{enumerate} 335 | } 336 | \exercise{6}{ 337 | Fix $a \in S$ and define $\mu$ on $\pset{S}$ by: $\mu(A) = 1$ if $a \in A$, $\mu(A) = 0$ if $a \notin A$. 338 | Show that $\mu$ is a $\s$-additive measure on $S$. 339 | } 340 | \sol{ 341 | \qproof{ 342 | Let $\fs = \pset{S}$, which we know is the largest $\s$-algebra of subsets of $S$. 343 | We must show that $\mu$ as defined above satisfies the properties of $\s$-additive measure on $S$: 344 | \defsam 345 | 346 | To show \ilab{i}, we clearly have that $a \notin \es$ so that by definition $\mu(\es) = 0$. 347 | Also, clearly $a \in S$ so that $\mu(S) = 1 > 0$. 348 | 349 | Regarding \ilab{ii}, suppose that $\braces{X_n}_{n=0}^\infty$ is a collection of mutually disjoint sets in $\fs = \pset{S}$. 350 | 351 | Case: $a \in \bigcup_{n=0}^\infty X_n$. 352 | Then by definition $\mu\parens{\bigcup_{n=0}^\infty X_n} = 1$. 353 | There is also an $n \in \nats$ such that $a \in X_n$, and since the sets $\braces{X_k}_{k=0}^\infty$ are mutually disjoint, it follows that $a \notin X_m$ for any natural $m \neq n$ (since otherwise $X_n$ and $X_m$ would not be disjoint). 354 | Thus we have $\mu(X_n) = 1$ while $\mu(X_m) = 0$ for all natural $m \neq n$. 355 | Hence 356 | \gath{ 357 | \sum_{k=0}^\infty \mu(X_k) = \sum_{k \in \nats} \mu(X_k) = \sum_{k=0}^{n-1} \mu(X_k) + \mu(X_n) + \sum_{k=n+1}^\infty \mu(X_k) 358 | = \sum_{k=0}^{n-1} 0 + 1 + \sum_{k=n+1}^\infty 0 = 1. 359 | } 360 | Thus clearly $\mu\parens{\bigcup_{n=0}^\infty X_n} = \sum_{n=0}^\infty \mu(X_n) = 1$. 361 | 362 | Case: $a \notin \bigcup_{n=0}^\infty X_n$. 363 | Then $\mu\parens{\bigcup_{n=0}^\infty X_n} = 0$ by definition. 364 | It also follows that $a \notin X_n$ for every natural $X_n$ so that $\mu(X_n) = 0$. 365 | Hence clearly 366 | \gath{ 367 | \mu\parens{\bigcup_{n=0}^\infty X_n} = 0 = \sum_{n=0}^\infty 0 = \sum_{n=0}^\infty \mu(X_n). 368 | } 369 | 370 | Thus \ilab{ii} is shown in both cases so that $\mu$ is indeed a $\s$-additive measure on $S$ since we also showed \ilab{i}. 371 | } 372 | } 373 | 374 | \exercise{7}{ 375 | For $A \ss S$ let $\mu(A) = 0$ if $A = \es$, $\mu(A) = \infty$ otherwise. 376 | Show that $\mu$ is a $\s$-additive measure on $S$. 377 | } 378 | \sol{ 379 | \qproof{ 380 | Let $\fs = \pset{S}$, which we know is the largest $\s$-algebra of subsets of $S$. 381 | We must show that $\mu$ as defined above satisfies the properties of $\s$-additive measure on $S$: 382 | \defsam 383 | 384 | For \ilab{i} we clearly have $\mu(\es) = 0$ by definition and $\mu(S) = \infty >0$ since $S$ is nonempty, which follows from the fact that $\pset{S}$ is a $\s$-algebra of subsets of $S$. 385 | 386 | Regarding \ilab{ii}, suppose that $\braces{X_n}_{n=0}^\infty$ is a collection of disjoint sets in $\fs = \pset{S}$. 387 | 388 | Case: $\bigcup_{n=0}^\infty X_n = \es$. 389 | Then by definition $\mu\parens{\bigcup_{n=0}^\infty X_n} = 0$, and it also has to be that $X_n = \es$ for all $n \in \nats$ so that $\mu(X_n) = 0$. 390 | Therefore 391 | \gath{ 392 | \sum_{n=0}^\infty \mu(X_n) = \sum_{n=0}^\infty 0 = 0 = \mu\parens{\bigcup_{n=0}^\infty X_n}. 393 | } 394 | 395 | Case: $\bigcup_{n=0}^\infty X_n \neq \es$. 396 | Then by definition $\mu\parens{\bigcup_{n=0}^\infty X_n} = \infty$. 397 | It also follows that $X_n \neq \es$ for at least one $n \in \nats$ so that $\mu(X_n) = \infty$. 398 | We then have 399 | \ali{ 400 | \sum_{k=0}^\infty \mu(X_k) &= \sum_{k=0}^{n-1} \mu(X_k) + \mu(X_n) + \sum_{k=n+1}^\infty \mu(X_k) 401 | = \sum_{k=0}^{n-1} \mu(X_k) + \infty + \sum_{k=n+1}^\infty \mu(X_k) \\ 402 | &= \infty = \mu\parens{\bigcup_{n=0}^\infty X_n} 403 | } 404 | since each $\mu(X_m) \in \braces{0, \infty}$ for all natural $m \neq n$. 405 | 406 | Thus \ilab{ii} is shown in both cases so that $\mu$ is indeed a $\s$-additive measure on $S$ as desired. 407 | } 408 | } 409 | 410 | \def\unxn{\bigcup_{n=0}^\infty X_n} 411 | \exercise{8}{ 412 | For $A \ss S$ let $\mu(A) = \abs{A}$ if $A$ is finite, $\mu(A) = \infty$ if $A$ is infinite. 413 | $\mu$ is a $\s$-additive measure on $S$; it is called the \emph{counting measure} on $S$. 414 | } 415 | \sol{ 416 | \qproof{ 417 | Let $\fs = \pset{S}$, which we know is the largest $\s$-algebra of subsets of $S$. 418 | We must show that $\mu$ as defined above satisfies the properties of $\s$-additive measure on $S$: 419 | \defsam 420 | 421 | For \ilab{i} we have that $\mu(\es) = \abs{\es} = 0$ since $\es$ is finite. 422 | If $S$ is finite then $\mu(S) = \abs{S} > 0$ since $S$ is nonempty. 423 | If $S$ is infinite then $\mu(S) = \infty > 0$ as well so that in either case $\mu(S) > 0$ as desired. 424 | 425 | To show \ilab{ii} suppose that $\braces{X_n}_{n=0}^\infty$ is a collection of disjoint sets in $\fs = \pset{S}$. 426 | 427 | Case: There is an $m \in \nats$ where $X_m$ is infinite. 428 | Then obviously $\mu(X_m) = \infty$ by definition. 429 | We also clearly have that $\unxn$ is infinite since $X_m \ss \unxn$, and hence $\mu(\unxn) = \infty$. 430 | Then 431 | \gath{ 432 | \sum_{n=0}^\infty \mu(X_n) = \sum_{n=0}^{m-1} \mu(X_n) + \mu(X_m) + \sum_{n=m+1}^\infty \mu(X_n) 433 | = \sum_{n=0}^{m-1} \mu(X_n) + \infty + \sum_{n=m+1}^\infty \mu(X_n) = \infty 434 | } 435 | since $\mu(X_n) \in \nats \cup \braces{\infty}$ for every $n \neq m$. 436 | Therefore $\mu(\unxn) = \sum_{n=0}^\infty \mu(X_n) = \infty$ as desired. 437 | 438 | Case: $X_n$ is finite for every $n \in \nats$. 439 | Clearly then $\mu(X_n) = \abs{X_n}$ for every natural $n$. 440 | First, if there is a natural $N$ such that $X_n = \es$ for all $n > N$, then clearly $\unxn = \bigcup_{n=0}^N X_n$, i.e. the union is finite. 441 | It then follows that $\bigcup_{n=0}^N X_n$ is finite by Theorem~4.2.7 so that $\mu\parens{\bigcup_{n=0}^N X_n} = \abs{\bigcup_{n=0}^N X_n}$. 442 | Then, since the sets $\braces{X_n}_{n=0}^N$ are mutually disjoint, we have 443 | \gath{ 444 | \mu\parens{\unxn} = \mu\parens{\bigcup_{n=0}^N X_n} = \abs{\bigcup_{n=0}^N X_n} 445 | = \sum_{n=0}^N \abs{X_n} = \sum_{n=0}^N \mu(X_n) = \sum_{n=0}^\infty \mu(X_n) 446 | } 447 | by the definition of cardinal addition, noting that clearly the last step follows from the fact that $\mu(X_n) = \abs{\es} = 0$ for all $n > N$. 448 | 449 | On the other hand, if there is no such $N$, then it follows that, for every $N \in \nats$, there is a natural $n > N$ such that $X_n \neq \es$. 450 | At this point we need two facts from real analysis, supposing that $\angles{a_n}_{n=0}^\infty$ is a real sequence: 451 | \begin{enumerate} 452 | \item By definition, the sequence converges to a real $a$ if, for every real $\e > 0$, there is an $N \in \nats$ such that $\abs{a_n - a} < \e$ for every natural $n \geq N$. 453 | \item If the infinite series $\sum_{n=0}^\infty a_n$ converges (to a finite value) then the sequence itself must converge to zero. 454 | \end{enumerate} 455 | We shall show that $\sum_{n=0}^\infty \mu(X_n)$ diverges by showing that the sequence $\angles{\mu(X_n)}_{n=0}^\infty$ does \emph{not} converge to zero (i.e. the contrapositive of 2). 456 | So let $\e = 1/2$, noting that clearly $\e = 1/2 > 0$. 457 | Then consider any natural $N$ so that there is a natural $n > N$ such that $X_n \neq \es$. 458 | It then follows that, since $X_n$ is finite but nonempty, $\abs{\mu(X_n) - 0} = \abs{\mu(X_n)} = \abs{\abs{X_n}} = \abs{X_n} \geq 1 \geq 1/2 = \e$. 459 | Therefore we have shown 460 | \gath{ 461 | \exists \e > 0 \forall N \in \nats \exists n \geq N \parens{\abs{\mu(X_n) - 0} \geq \e} \\ 462 | \lnot \forall \e > 0 \exists N \in \nats \forall n \geq N \parens{\abs{\mu(X_n) - 0} < \e}, 463 | } 464 | which shows by definition that $\angles{\mu(X_n)}_{n=0}^\infty$ does not converge to zero. 465 | Hence $\sum_{n=0}^\infty \mu(X_n)$ diverges so that by convention $\sum_{n=0}^\infty \mu(X_n) = \infty$. 466 | 467 | Lastly, we show that $\unxn$ must be infinite. 468 | Suppose to the contrary that $\unxn$ is finite. 469 | We then construct a function $f : \unxn \to \nats$ as follows: for each $x \in \unxn$ there is a unique $m \in \nats$ where $x \in X_m$. 470 | Clearly such an $m$ exists since $x \in \unxn$, and it is unique because the sets $\braces{X_n}_{n=0}^\infty$ are mutually disjoint (if it was not unique then there would be distinct $n$ and $m$ where $x \in X_n$ and $x \in X_m$ so that $X_n \cap X_m \neq \es$). 471 | We then simply set $f(x) = m$. 472 | 473 | It then follows from Theorem~4.2.5 that $\ran(f)$ is finite since $\dom(f) = \unxn$ is. 474 | Since $\ran(f)$ is then a finite set of natural numbers, it has greatest natural number $N$. 475 | But we know that there is an $m > N$ such that $X_m \neq \es$ so that there is an $x \in X_m$. 476 | It then follows that clearly $x \in \unxn$ and that $f(x) = m$. 477 | However, then $m$ would be in $\ran(f)$ so that $m \leq N$ since $N$ is the greatest element of $\ran(f)$. 478 | But we already know that $m > N$, which is a contradiction. 479 | So it has to be that $\unxn$ is in fact infinite as desired. 480 | 481 | We therefore have $\mu\parens{\unxn} = \infty = \sum_{n=0}^\infty \mu(X_n)$ and hence ii) has been shown in every case and sub-case so that $\mu$ is indeed a $\s$-additive measure on $S$ by definition. 482 | } 483 | } 484 | -------------------------------------------------------------------------------- /shared/cor_aleph_wonlegt.tex: -------------------------------------------------------------------------------- 1 | \def\aA{\abs{A}} 2 | \def\aB{\abs{B}} 3 | \begin{cor}\label{cor:aleph:wonlegt} 4 | If $A$ and $B$ are well ordered sets then $\aA \nleq \aB$ if and only if $\aB < \aA$. 5 | \end{cor} 6 | \qproof{ 7 | ($\to$) Suppose that $\aA \nleq \aB$. 8 | Then it follows from Lemma~\ref{lem:aleph:wolege} above that $\aB \leq \aA$. 9 | Suppose that $\aB = \aA$. 10 | Then there is a bijection $f$ from $B$ to $A$. 11 | But then clearly $\inv{f}$ is also a bijection and therefore injective. 12 | Hence by definition $\aA \leq \aB$, a contradiction. 13 | So it cannot be that $\aB = \aA$. 14 | Hence $\aB < \aA$ by definition as desired. 15 | 16 | ($\leftarrow$) We show this by proving the contrapositive. 17 | So suppose that $\aA \leq \aB$. 18 | Also suppose that $\aB \leq \aA$ so that by Lemma~\ref{lem:aleph:wolege} above $\aA = \aB$. 19 | Thus we have shown that 20 | \gath{ 21 | \aB \leq \aA \to \aA = \aB \\ 22 | \aB \nleq \aA \lor \aA = \aB \\ 23 | \lnot \parens{\aB \leq \aA \land \aA \neq \aB} \\ 24 | \lnot (\aB < \aA), 25 | } 26 | thereby showing the contrapositive. 27 | } 28 | -------------------------------------------------------------------------------- /shared/lem_aleph_infwo_succ.tex: -------------------------------------------------------------------------------- 1 | \begin{lem}\label{lem:aleph:infwo:succ} 2 | If $\a$ is an infinite ordinal then $\abs{\a} = \abs{\a+1}$, i.e. $\a$ and $\a+1$ are equipotent. 3 | \end{lem} 4 | \qproof{ 5 | First we note that since $\a$ is infinite we have $\a + 1 > \a \geq \w$. 6 | We then construct a bijection from $\a+1$ to $\a$. 7 | So define $f: \a+1 \to \a$ by 8 | $$ 9 | f(\b) = 10 | \begin{cases} 11 | \b+1 & \b < \w \\ 12 | \b & \b \geq \w \text{ and } \b \neq \a \\ 13 | 0 & \b = \a 14 | \end{cases} 15 | $$ 16 | for $\b \in \a+1$. 17 | 18 | First we show that $f$ is injective. 19 | So consider any $\b$ and $\g$ in $\a+1$ where $\b \neq \g$. 20 | Without loss of generality we can assume that $\b < \g$. 21 | We then have the following: 22 | 23 | Case: $\b < \w$. 24 | Then clearly $f(\b) = \b + 1 < \w$ since $\b < \w$ and $\w$ is a limit ordinal, but we also clearly have that $0 < \b + 1 = f(\b)$. 25 | Now, if also $\g < \w$ then clearly $f(\b) = \b + 1 < \g + 1 = f(\g)$ since $\b < \g$. 26 | If $\g \geq \w$ and $\g \neq \a$ then we have $f(\b) < \w \leq \g = f(\g)$. 27 | Lastly if $\g = \a$ then we have $f(\g) = 0 < f(\b)$. 28 | 29 | Case: $\b \geq \w$ and $\b \neq \a$. 30 | Here since $\b < \g$ we have $\w \leq \b < \g$. 31 | Thus if also $\g \neq \a$ then clearly we have $f(\b) = \b < \g = f(\g)$. 32 | On the other hand if $\g = \a$ then $f(\g) = 0 < \w \leq \b = f(\b)$. 33 | 34 | Thus in every case we have $f(\b) \neq f(\g)$, thereby showing that $f$ is injective. 35 | We note that the case in which $\b = \a$ is impossible since $\a$ is the greatest element of $\a+1$ but $\g > \b$ and $\g \in \a+1$. 36 | 37 | Next we show that $f$ is surjective. 38 | So consider any $\b \in \a$. 39 | 40 | Case: $\b < \w$. 41 | If $\b = 0$ then clearly $f(\a) = 0 = \b$. 42 | On the other hand if $0 < \b < \w$ then $\b$ is a successor ordinal, say $\b = \g+1$, so that $\g < \b < \w$ hence clearly $\g \in \a+1$ and $f(\g) = \g+1 = \b$. 43 | 44 | Case: $\b \geq \w$. 45 | Then since $\b \in \a$ we have $\b < \a < \a+1$ so that $\b \neq \a$ but $\b \in \a+1$. 46 | Then clearly $f(\b) = \b$. 47 | 48 | Hence in all cases there is a $\g \in \a+1$ such that $f(\g) = \b$ so that $f$ is injective. 49 | Therefore we have shown that $f$ is a bijection so that by definition $\a+1$ and $\a$ are equipotent. 50 | } 51 | -------------------------------------------------------------------------------- /shared/lem_aleph_initle.tex: -------------------------------------------------------------------------------- 1 | \begin{lem}\label{lem:aleph:initle} 2 | For initial ordinals $\a$ and $\b$, if $\abs{\a} \leq \abs{\b}$, then $\a \leq \b$. 3 | \end{lem} 4 | \qproof{ 5 | Suppose that $\abs{\a} \leq \abs{\b}$ but that $\a > \b$. 6 | Then clearly $\b$ is isomorphic (and therefore equipotent) to an initial segment of $\a$ so that $\abs{\b} \leq \abs{\a}$. 7 | Then by the \cbthrm{} we have $\abs{\a} = \abs{\b}$. 8 | However since $\a$ is an initial ordinal and $\b < \a$ it cannot be that $\abs{\a} = \abs{\b}$. 9 | Thus we have a contradiction so that it must be that $\a \leq \b$ as desired. 10 | } 11 | -------------------------------------------------------------------------------- /shared/lem_aleph_initlimit.tex: -------------------------------------------------------------------------------- 1 | \begin{lem}\label{lem:aleph:initlimit} 2 | Every infinite initial ordinal is a limit ordinal. 3 | \end{lem} 4 | \qproof{ 5 | Suppose that $\a$ is an infinite initial ordinal and that it a successor so that $\a = \b+1$. 6 | It was shown in Lemma~\ref{lem:aleph:infwo:succ} that $\abs{\b} = \abs{\b+1} = \abs{\a}$, but since clearly $\b < \a$ this contradicts the fact that $\a$ is an initial ordinal. 7 | Hence $\a$ must be a limit ordinal. 8 | } 9 | -------------------------------------------------------------------------------- /shared/lem_aleph_initltwa.tex: -------------------------------------------------------------------------------- 1 | \begin{lem}\label{lem:aleph:initltwa} 2 | For any ordinal $\a$ and any infinite initial ordinal $\W$ where $\W < \w_\a$, there is a $\g < \a$ such that $\W = \w_\g$. 3 | \end{lem} 4 | \qproof{ 5 | We show this by induction on $\a$. 6 | For $\a = 0$ we have $\w_\a = \w_0 = \w$ so that there is no infinite initial ordinal $\W$ such that $\W < \w_\a = \w$. 7 | Hence the hypothesis is vacuously true. 8 | Now suppose that, for every infinite initial ordinal $\W < \w_\a$, there is a $\g < \a$ such that $\W = \w_\g$. 9 | Consider any infinite initial ordinal $\W < \w_{\a+1}$. 10 | Then $\W < \w_{\a+1} = h(\w_\a)$ so that $\W$ is equipotent to some subset of $\w_\a$ by the definition of the Hartogs number. 11 | From this it clearly follows that $\abs{\W} \leq \abs{\w_\a}$ and hence $\W \leq \w_\a$ by Lemma~\ref{lem:aleph:initle} since both $\W$ and $\w_\a$ are initial ordinals. 12 | If $\W = \w_\a$ then we are finished but if $\W < \w_\a$ then by the induction hypothesis there is a $\g < \a$ such that $\W = \w_\g$ so that we are also finished. 13 | 14 | Now suppose that $\a$ is a nonzero limit ordinal and that for every $\b < \a$ and infinite initial ordinal $\W < \w_\b$ there is a $\g < \b$ such that $\W = \w_\g$. 15 | Consider then any infinite initial ordinal $\W < \w_\a$. 16 | Then since $\w_\a = \sup\braces{\w_\b \where \b < \a}$ it follows that $\W$ is not an upper bound of $\braces{\w_\b \where \b < \a}$ so that there is a $\b < \a$ such that $\W < \w_\b$. 17 | But then by the induction hypothesis there is a $\g < \b$ such that $\W = \w_\g$. 18 | This completes the transfinite induction. 19 | } 20 | -------------------------------------------------------------------------------- /shared/lem_aleph_wolege.tex: -------------------------------------------------------------------------------- 1 | \def\aA{\abs{A}} 2 | \def\aB{\abs{B}} 3 | \begin{lem}\label{lem:aleph:wolege} 4 | For well ordered sets $A$ and $B$ either $\aA \leq \aB$ or $\aB \leq \aA$ (or both in which case $\aA = \aB$). 5 | \end{lem} 6 | \qproof{ 7 | By Theorem~6.1.3 we have: 8 | 9 | Case: $A$ and $B$ are isomorphic. 10 | Let $f$ be the isomorphism from $A$ to $B$. 11 | Then clearly $f$ is a bijection so that $\aA = \aB$. 12 | Also since $f$ is injective $\aA \leq \aB$. 13 | Clearly also $\inv{f}$ is bijective from $B$ to $A$ so that $\aB \leq \aA$ as well. 14 | 15 | Case: $A$ is isomorphic to an initial segment of $B$. 16 | Then if $f$ is the isomorphism clearly $f$ is an injective function from $A$ to $B$ so that $\aA \leq \aB$. 17 | 18 | Case: $B$ is isomorphic to an initial segment of $A$. 19 | Then if $f$ is the isomorphism clearly $f$ is an injective function from $B$ to $A$ so that $\aB \leq \aA$. 20 | 21 | Since these cases are exhaustive by Theorem~6.1.3 clearly the result has been shown. 22 | 23 | Note that this did not require the Axiom of Choice. 24 | } 25 | -------------------------------------------------------------------------------- /shared/lem_aoc_notin.tex: -------------------------------------------------------------------------------- 1 | \begin{lem}\label{lem:aoc:notin} 2 | For any set $A$, there is a $b \notin A$. 3 | \end{lem} 4 | \qproof{ 5 | Let $X = \braces{\a \in A \where \a \text{ is an ordinal number}}$. 6 | Then by Theorem~6.2.6e there is an ordinal $\a$ such that $\a \notin X$. 7 | It also has to be that $\a \notin A$ since, if it were, then $\a$ would be in $X$ since it is an ordinal number, which would be a contradiction. 8 | } 9 | -------------------------------------------------------------------------------- /shared/lem_ord_expin_sups.tex: -------------------------------------------------------------------------------- 1 | \begin{lem}\label{lem:ord:expin:sups} 2 | Suppose that $\g$ is a nonzero limit ordinal and $\braces{\a_\n}$ and $\braces{\b_\n}$ for $\n < \g$ are two transfinite sequences. 3 | Also suppose that $\a_\n \leq \b_\n$ for every $\n < \g$. 4 | Then 5 | $$ 6 | \sup_{\n < \g} \a_\n \leq \sup_{\n < \g} \b_\n. 7 | $$ 8 | \end{lem} 9 | \qproof{ 10 | First let $A = \braces{\a_\n \where \n < \g}$ and $B = \braces{\b_\n \where \n < \g}$ be the ranges of the sequences so that we must show that $\sup{A} \leq \sup{B}$. 11 | Now consider any $\a \in A$ so that $\a = \a_\n$ for some $\n < \g$. 12 | We then have that 13 | $$ 14 | \a = \a_\n \leq \b_\n \leq \sup{B} 15 | $$ 16 | so that $\sup{B}$ is an upper bound of $A$ since $\a$ was arbitrary. 17 | It then follows from the least upper bound property that $\sup{A} \leq \sup{B}$ as desired. 18 | } 19 | -------------------------------------------------------------------------------- /shared/lem_ord_lub.tex: -------------------------------------------------------------------------------- 1 | \begin{lem}\label{lem:ord:lub} 2 | If $A$ is a set of ordinals then ordinal $\a = \sup{A}$ if and only if $\a$ is the least upper bound of $A$, i.e. $\a$ is an upper bound of $A$ and $\b$ is not an upper bound of $A$ for any $\b < \a$. 3 | \end{lem} 4 | \qproof{ 5 | ($\to$) First suppose that $\a = \sup{A}$. 6 | Then by the remarks following the proof of Theorem~6.2.6 in the text $\a$ is an upper bound of $A$ and if $\b$ is an upper bound of $A$ then $\a \leq \b$. 7 | This last statement is simply the contrapositive of the statement that $\b < \a$ implies that $\b$ is \emph{not} an upper bound of $A$ and hence is logically equivalent. 8 | 9 | ($\leftarrow$) We show that an ordinal $\a$ with the least upper bound property for $A$ is unique, which suffices to show the result since if $\b$ has this property then $\b = \sup{A}$ since $\sup{A}$ does as well (by what was just shown above) and the ordinal having this property is unique. 10 | 11 | So suppose that ordinals $\a$ and $\b$ both have the least upper bound property for $A$ but that $\a \neq \b$. 12 | Without loss of generality we can assume then that $\a < \b$. 13 | But then, since $\b$ has the least upper bound property, $\a$ cannot be an upper bound of $A$, which contradicts the fact that $\a$ also has the least upper bound property! 14 | Hence it has to be that $\a = \b$, which shows the uniquness. 15 | } 16 | -------------------------------------------------------------------------------- /solutions.tex: -------------------------------------------------------------------------------- 1 | \input{prelude} 2 | 3 | % Import all our custom macros 4 | \input{macros} 5 | 6 | \title{ 7 | \booktitle \\ 8 | \ \\ 9 | Solutions Manual \\ 10 | \ \\ 11 | \ \\ 12 | by Dan Whitman 13 | } 14 | 15 | % So pages will break inside long equation environments 16 | \allowdisplaybreaks 17 | 18 | % Lemmas/Corollaries environments 19 | \newtheorem{lem}{Lemma}[subsubsection] 20 | \newtheorem{defin}[lem]{Definition} 21 | \newtheorem{cor}[lem]{Corollary} 22 | \newtheorem{thrm}[lem]{Theorem} 23 | 24 | \begin{document} 25 | 26 | % Setup section (chapter) format 27 | \titleformat{\section} % command 28 | [hang] % shape 29 | {\bfseries \Large} % format 30 | {Chapter \thesection} % label 31 | {12pt} % sep 32 | {} % before-code 33 | {} % after-code 34 | 35 | % Setup subsection format 36 | \titleformat{\subsection} % command 37 | [hang] % shape 38 | {\bfseries \large} % format 39 | {\S \thesubsection} % label 40 | {12pt} % sep 41 | {} % before-code 42 | {} % after-code 43 | 44 | % Question label format 45 | \newcommand\setqf[1]{\qformat{\hspace{0.5cm} \textbf{Exercise \thesubsection.\thequestion #1} \hfill \vrule depth 1em width 0pt}} 46 | \setqf{} 47 | 48 | % Title page 49 | \begin{titlingpage} 50 | \maketitle 51 | \end{titlingpage} 52 | 53 | \newpage 54 | \input{license.tex} 55 | 56 | \newpage 57 | 58 | \tableofcontents 59 | \newpage 60 | 61 | % Apparently need this to get page number on the first real page 62 | \cfoot{Page \thepage} 63 | \setcounter{page}{1} 64 | 65 | % Start content 66 | \begin{questions} 67 | 68 | % Chapters (sections are used since the exam class has no chapters) 69 | \setcounter{section}{1-1} 70 | 71 | \section{Sets} 72 | \setcounter{subsection}{3-1} 73 | \input{sections/sec_1_3.tex} 74 | \input{sections/sec_1_4.tex} 75 | 76 | \setcounter{section}{5-1} 77 | 78 | \section{Cardinal Numbers} 79 | \input{sections/sec_5_1} 80 | \input{sections/sec_5_2} 81 | 82 | \section{Ordinal Numbers} 83 | \input{sections/sec_6_1} 84 | \input{sections/sec_6_2} 85 | \input{sections/sec_6_3} 86 | \input{sections/sec_6_4} 87 | \input{sections/sec_6_5} 88 | \input{sections/sec_6_6} 89 | 90 | \section{Alephs} 91 | \input{sections/sec_7_1} 92 | \input{sections/sec_7_2} 93 | 94 | \section{The Axiom of Choice} 95 | As in the text, asterisks indicate exercises for which the Axiom of Choice is needed. 96 | \input{sections/sec_8_1} 97 | \input{sections/sec_8_2} 98 | 99 | \section{Arithmetic of Cardinal Numbers} 100 | \input{sections/sec_9_1} 101 | 102 | \end{questions} 103 | \end{document} 104 | --------------------------------------------------------------------------------