Exercise 8.2.1

Question

% Custom definitions (IntroToSetTheory)
% Source section: sec_8_2.tex

% Common sets
\def
ats{{\boldsymbol{N}}}
\def\ints{{\boldsymbol{Z}}}
\defats{{\boldsymbol{Q}}}
\def\prats{ats^+}
\defeals{{\boldsymbol{R}}}
\def\es{\varnothing}

% Common variables/symbols
\def\vphi{\varphi}
\def\a{\alpha}
\def\b{\beta}
\def\g{\gamma}
\def\d{\delta}
\def\e{\varepsilon}
\def\z{\zeta}
\def\k{\kappa}
\def\l{\lambda}
\def
{
u}
\def{ho}
\def\s{\sigma}
\def	{	au}
\def\x{\xi}
\def\w{\omega}
\def\W{\Omega}
\def\al{\aleph}

% Logic
\def\bic{\leftrightarrow}

ewcommand{\propP}[1]{\prop{P}(#1)}
\def\lxor{\veebar}

% For set-buider notation
\def\where{\mid}

% Other stuff
\def\dom{\mathrm{dom}\,}
\defan{\mathrm{ran}\,}

% Cardinality shortcuts
\def\cnats{{\al_0}}
\def\ccont{{2^\cnats}}

% Equivalence classes

ewcommand\eclass[2]{\squares{#1}_{#2}}

% Shortcuts that make writing easier

ewcommand\parens[1]{\left( #1 ight)}

ewcommand\squares[1]{\left[ #1 ight]}

ewcommand\braces[1]{\left\{ #1 ight\}}

ewcommand\angles[1]{\left\langle #1 ightangle}

ewcommand\ceil[1]{\left\lceil #1 ightceil}

ewcommand\floor[1]{\left\lfloor #1 ightfloor}

ewcommand\abs[1]{\left| #1 ight|}

ewcommand\dabs[1]{\left\| #1 ight\|}

ewcommand\vect[1]{\mathrm{\mathbf{#1}}}

ewcommand\conj[1]{\overline{#1}}

ewcommand\pset[1]{\mathcal{P}\left(#1ight)}

ewcommand\inv[1]{#1^{-1}}

ewcommand\prop[1]{\mathbf{#1}}
\defest{estriction}

ewcommand	et[2]{{^{#1}#2}}

% Families of set
\def\famF{\mathcal{F}}

% These are needed so that half-open intervals do not cause auto-indentation issues due to unmatched brackets

ewcommand\clop[1]{[#1)}

ewcommand\ilab[1]{#1)}

% Other miscellaneous stuff
\def\ss{\subseteq}
\def\pss{\subset}
\def\Seq{\mathrm{Seq}}
\def\prece{\preccurlyeq}
\def\sd{\bigtriangleup}

% Environment shortcuts

ewcommand\gath[1]{\begin{gather*} #1 \end{gather*}}

ewcommand\ali[1]{\begin{align*} #1 \end{align*}}

ewcommand\qproof[1]{\begin{proof} #1 \end{proof}}

ewcommand\bicproof[2]{
  $(	o)$ #1

$(\leftarrow)$ #2
}

ewcommand\seteqproof[2]{
  $(\ss)$ #1

$(\supseteq)$ #2
}

% Environment for indenting nested paragraphs (useful in case trees)

ewenvironment{indpar}
{
  \begin{adjustwidth}{1cm}{}
    }{
  \end{adjustwidth}
}

% Exercise, Theorem, and solution shortcuts

ewcommand\exercise[2]{{
      enewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections
      \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering
      \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset
      \setcounter{question}{#1-1} % Manually set the question number since we always know the exercise number
      \question{#2} % The actual exam class question
    }}

ewcommand\exerciseapp[3]{
  \setqf{#2}
  \exercise{#1}{#3}
  \setqf{}
}

ewcommand	heorem[2]{{
      enewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections
      \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering
      \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset
      \setcounter{question}{#1-1} % Manually set the question number since we always know the theorem number
      \question{#2} % The actual exam class question
    }}

ewcommand	heoremapp[3]{
  \setqf{#2}
  	heorem{#1}{#3}
  \setqf{}
}

ewcommand\sol[1]{
  \begin{solution}

#1
  \end{solution}
}

% Main problem and theorem labels
\def\mainprob{	extbf{Main Problem.}}
\def\mainthrm{	extbf{Main Theorem.}}

% Saves some typing
\def\cbthrm{Cantor-Bernstein Theorem}

% Book title
\def\booktitle{
  Introduction to Set Theory \
  Third Edition, Revised and Expanded \
  by Karel Hrbacek and Thomas Jech
}

% Environment aliases used in the source sections

ewenvironment{lem}{\begin{lemma}}{\end{lemma}}

ewenvironment{defin}{\begin{definition}}{\end{definition}}

ewenvironment{cor}{\begin{corollary}}{\end{corollary}}

ewenvironment{thrm}{\begin{theorem}}{\end{theorem}}

% Override indpar to avoid changepage/adjustwidth dependency
enewenvironment{indpar}{\begin{quote}}{\end{quote}}
Without using the Axiom of Choice, prove that the two definitions of closure points are equivalent if $A$ is an open set.
    [Hint: $X_n$ is open, so $X_n \cap ats 
eq \es$, and $ats$ can be well-ordered.]

Dan Whitman · Accepted Answer

% Custom definitions (IntroToSetTheory) % Source section: sec_8_2.tex % Common sets \def ats{{\boldsymbol{N}}} \def\ints{{\boldsymbol{Z}}} \def ats{{\boldsymbol{Q}}} \def\prats{ ats^+} \def eals{{\boldsymbol{R}}} \def\es{\varnothing} % Common variables/symbols \def\vphi{\varphi} \def\a{\alpha} \def\b{\beta} \def\g{\gamma} \def\d{\delta} \def\e{\varepsilon} \def\z{\zeta} \def\k{\kappa} \def\l{\lambda} \def { u} \def { ho} \def\s{\sigma} \def { au} \def\x{\xi} \def\w{\omega} \def\W{\Omega} \def\al{\aleph} % Logic \def\bic{\leftrightarrow} ewcommand{\propP}[1]{\prop{P}(#1)} \def\lxor{\veebar} % For set-buider notation \def\where{\mid} % Other stuff \def\dom{\mathrm{dom}\,} \def an{\mathrm{ran}\,} % Cardinality shortcuts \def\cnats{{\al_0}} \def\ccont{{2^\cnats}} % Equivalence classes ewcommand\eclass[2]{\squares{#1}_{#2}} % Shortcuts that make writing easier ewcommand\parens[1]{\left( #1 ight)} ewcommand\squares[1]{\left[ #1 ight]} ewcommand\braces[1]{\left\{ #1 ight\}} ewcommand\angles[1]{\left\langle #1 ight angle} ewcommand\ceil[1]{\left\lceil #1 ight ceil} ewcommand\floor[1]{\left\lfloor #1 ight floor} ewcommand\abs[1]{\left| #1 ight|} ewcommand\dabs[1]{\left\| #1 ight\|} ewcommand\vect[1]{\mathrm{\mathbf{#1}}} ewcommand\conj[1]{\overline{#1}} ewcommand\pset[1]{\mathcal{P}\left(#1 ight)} ewcommand\inv[1]{#1^{-1}} ewcommand\prop[1]{\mathbf{#1}} \def est{ estriction} ewcommand et[2]{{^{#1}#2}} % Families of set \def\famF{\mathcal{F}} % These are needed so that half-open intervals do not cause auto-indentation issues due to unmatched brackets ewcommand\clop[1]{[#1)} ewcommand\ilab[1]{#1)} % Other miscellaneous stuff \def\ss{\subseteq} \def\pss{\subset} \def\Seq{\mathrm{Seq}} \def\prece{\preccurlyeq} \def\sd{\bigtriangleup} % Environment shortcuts ewcommand\gath[1]{\begin{gather*} #1 \end{gather*}} ewcommand\ali[1]{\begin{align*} #1 \end{align*}} ewcommand\qproof[1]{\begin{proof} #1 \end{proof}} ewcommand\bicproof[2]{ $( o)$ #1 $(\leftarrow)$ #2 } ewcommand\seteqproof[2]{ $(\ss)$ #1 $(\supseteq)$ #2 } % Environment for indenting nested paragraphs (useful in case trees) ewenvironment{indpar} { \begin{adjustwidth}{1cm}{} }{ \end{adjustwidth} } % Exercise, Theorem, and solution shortcuts ewcommand\exercise[2]{{ enewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset \setcounter{question}{#1-1} % Manually set the question number since we always know the exercise number \question{#2} % The actual exam class question }} ewcommand\exerciseapp[3]{ \setqf{#2} \exercise{#1}{#3} \setqf{} } ewcommand heorem[2]{{ enewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset \setcounter{question}{#1-1} % Manually set the question number since we always know the theorem number \question{#2} % The actual exam class question }} ewcommand heoremapp[3]{ \setqf{#2} heorem{#1}{#3} \setqf{} } ewcommand\sol[1]{ \begin{solution} #1 \end{solution} } % Main problem and theorem labels \def\mainprob{ extbf{Main Problem.}} \def\mainthrm{ extbf{Main Theorem.}} % Saves some typing \def\cbthrm{Cantor-Bernstein Theorem} % Book title \def\booktitle{ Introduction to Set Theory \ Third Edition, Revised and Expanded \ by Karel Hrbacek and Thomas Jech } % Environment aliases used in the source sections ewenvironment{lem}{\begin{lemma}}{\end{lemma}} ewenvironment{defin}{\begin{definition}}{\end{definition}} ewenvironment{cor}{\begin{corollary}}{\end{corollary}} ewenvironment{thrm}{\begin{theorem}}{\end{theorem}} % Override indpar to avoid changepage/adjustwidth dependency enewenvironment{indpar}{\begin{quote}}{\end{quote}} \begin{lem}\label{lem:aoc:intopen} If $A$ and $B$ are open subsets of $ eals$ then $A \cap B$ is open. \end{lem} \qproof{ First note that, if $A \cap B = \es$, then this is vacuously open. Otherwise, consider any $x \in A \cap B$. 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 $y \in eals$. 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 $y \in eals$ Let $\d = \min\braces{\d_1, \d_2}$ so that $\d \leq \d_1$ and $\d \leq \d_2$. Then consider any $y \in eals$ where $\abs{y - x} < \d$. Then we have $\abs{y - x} < \d \leq \d_1$ so that $y \in A$. Similarly, $\abs{y - x} < \d \leq \d_2$ so that $y \in B$ as well. Hence $y \in A \cap B$. Since $y$ was arbitrary and $\d > 0$, this shows that $A \cap B$ is open. } \begin{lem}\label{lem:aoc:openrat} If $A$ is a nonempty open subset of $ eals$ then $A \cap ats eq \es$. \end{lem} \qproof{ Suppose that $A \ss eals$ is nonempty and open. Then there is an $x \in A$ and, since $A$ is open, there is a real $\d > 0$ such that $(x-\d, x+\d) \ss A$. Now, since $\d > 0$, we clearly have \gath{ -\d < 0 < \d \ x-\d < x+\d. } Since $ ats$ is dense in $ eals$ with respect to order, there is a $q \in ats$ such that $x-\d < q < x+\d$. Then we have $q \in (x-\d, x+\d)$ so that also $q \in A$ since $(x-\d, x+\d) \ss A$. Thus $q \in ats$ and $q \in A$ so that $q \in A \cap ats$ as desired. } \mainprob \qproof{ A proof of this equivalence is presented in the text as Theorem~8.2.1. Recall that, in the proof that (b) implies (a), $a \in eals$ is the closure point of $A \ss eals$ and $X_n = \braces{x \in A \where \abs{x-a} < 1/n}$, and we know from (b) that each $X_n$ is nonempty. Note that we assume that $X_0 = A \cap (-\infty, \infty) = A \cap eals = A$. Per the remarks after this proof, it suffices to show that the system of nonempty sets $\braces{X_n}_{n \in ats}$ has a choice function when $A$ is open. First we show that each $X_n$ is an open set. Clearly $X_0 = A$ is open, so consider any natural $n > 0$ and let $I_n = (a-1/n, a+1/n)$. We claim that $X_n = A \cap I_n$. To this end we have \ali{ x \in X_n &\bic x \in A \land \abs{x-a} < 1/n \ &\bic x \in A \land -1/n < x - a < 1/n \ &\bic x \in A \land a-1/n < x < a+1/n \ &\bic x \in A \land x \in I_n \ &\bic x \in A \cap I_n. } for any real $x$, which of course shows that $X_n = A \cap I_n$. Then, since $A$ is open and $I_n$ is clearly an open interval, it follows from Lemma~ ef{lem:aoc:intopen} that $X_n$ is open as well. Now, since each $X_n$ is open and nonempty, it follows that $X_n \cap ats eq \es$ from Lemma~ ef{lem:aoc:openrat}. Then, since $ ats$ is countable, it can clearly be well-ordered. So choose a well-ordering of $ ats$. Since $X_n \cap ats$ is clearly a nonempty subset of $ ats$, it then has a least element $x_n$ according to our well-ordering. We then define a function $g$ on $\braces{X_n}_{n \in ats}$ by $g(X_n) = x_n$, which is clearly a choice function. }

Exercise 8.2.1

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Exercise 8.2.1

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