Exercise 5.2.6

Question

% Custom definitions (IntroToSetTheory)
% Source section: sec_5_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}}
The cardinality of the set of all discontinuous functions is $2^\ccont$.
    [Hint: Using Exercise~2.5, show that $\abs{eals^eals - C} = 2^\ccont$ whenever $\abs{C} \leq \ccont$.]

Dan Whitman · Accepted Answer

% Custom definitions (IntroToSetTheory)
% Source section: sec_5_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:card:contminus}
    If $B$ is a set with $|B| = 2^\ccont$ and $A$ is a subset of $B$ with $|A| \leq \ccont$ then $|B - A| = 2^\ccont$.
  \end{lem}
  \qproof{
    The proof is analogous to that of Theorem~5.2.4.
    So suppose that $C$ is a set with $|C| = 2^\ccont$.
    Let $B = C 	imes C$ so that by Exercise~5.2.5 we have
    $$
      |B| = |C 	imes C| = 2^\ccont \cdot 2^\ccont = 2^\ccont.
    $$
    Also suppose that $A \subseteq B$ where $|A| = \ccont$.
    Now define a set
    $$
      P = \braces{x \in C \where \exists y \in C ((x,y) \in A)}.
    $$
    Clearly then $|P| \leq |A| = \ccont$.
    Since also $|C| = 2^\ccont$ but $P \subseteq C$ it follows that there is an $x_0 \in C$ where $x_0 
otin P$.
    If we let $X = \braces{x_0} 	imes C$ then any $(x,y) \in X$ is not in $A$ so that $(x,y) \in C	imes C - A = B - A$.
    Hence $X \subseteq B - A$ but also since there is an obvious bijection between $X$ and $C$ we have
    $$
      2^\ccont = |C| = |X| \leq |B-A|.
    $$
    Since also clearly $B - A \subseteq B$ we also have that
    $$
      |B-A| \leq |B| = 2^\ccont.
    $$
    Hence by the \cbthrm{} $|B-A| = 2^\ccont$ as desired.
  }

\mainprob
  \qproof{
    By Lemma~5.2.7 $|eals^eals| = 2^\ccont$.
    Also by Theorem~5.2.6a the set $C$ of all continuous $f: eals 	o eals$ has cardinality of $\ccont$.
    Thus clearly the set of all discontinuous functions from $eals 	o eals$ is simply
    $$
      D = eals^eals - C.
    $$
    But then by Lemma~ef{lem:card:contminus} above we have that
    $$
      |D| = |eals^eals| = 2^\ccont
    $$
    as desired.
  }

Exercise 5.2.6

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Exercise 5.2.6

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