Exercise 7.1.3

Question

% Custom definitions (IntroToSetTheory)
% Source section: sec_7_1.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}}
For any set $A$, there is a mapping of $\pset{A 	imes A}$ onto $h(A)$.
    [Hint: Define $f(R) =$ the ordinal isomorphic to $R$, if $R \ss A 	imes A$ is a well-ordering of its field; $f(R) = 0$ otherwise.]

Dan Whitman · Accepted Answer

% Custom definitions (IntroToSetTheory)
% Source section: sec_7_1.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:aleph:pset:sslt}
    For a set $A$ and ordinal $\a$, if $\a$ is equipotent to a subset of $A$ then $\a < h(A)$.
  \end{lem}
  \qproof{
    Suppose that $\a$ is equipotent to $X \ss A$ but that $\a \geq h(A)$.
    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.
    On the other hand if $\a > h(A)$ then let $f$ be a bijection from $\a$ to $X$.
    Then, since $h(A) < \a$ we have that $h(A) \in \a$ and $h(A) \ss \a$ since ordinals are transitive.
    It then follows that $f est h(A)$ is a bijection from $h(A)$ to $f[h(A)] \ss X \ss A$.
    Hence again $h(A)$ is equipotent to a subset of $A$, contradicting the definition of the Hartogs number.
    So it has to be that $\a < h(A)$ as desired.
  }

\mainprob
  \qproof{
    We define a function $f : \pset{A 	imes A} 	o h(A)$.
    So for any $R \in \pset{A 	imes A}$ clearly $R \ss A 	imes A$ so that $R$ is a relation on $A$.
    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 $\a$.
    We then set
    $$
      f(R) = \begin{cases}
        \a & 	ext{$R$ is a well-ordering of some $X \ss A$}    \
        0  & 	ext{$R$ is not a well-ordering of any $X \ss A$} \
      \end{cases}.
    $$

First we show that $f(R)$ really is in $h(A)$ for any $R \in \pset{A 	imes A}$.
    So for any such $R$, if $R$ is not a well-ordering of some $X \ss A$ then clearly $f(R) = 0$.
    Then, since clearly $\es \ss A$ and $\es$ is equipotent to 0, it follows that $0 < h(A)$ by Lemma~ef{lem:aleph:pset:sslt}.
    Hence $f(R) = 0 \in h(A)$.\
    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 $f(R) = \a$.
    Since this means that $\a$ is equipotent to $X \ss A$, it again follows from Lemma~ef{lem:aleph:pset:sslt} that $\a < h(A)$ so that $f(R) = \a \in h(A)$.

To show that $f$ is surjective consider any $\a \in h(A)$ so that $\a < h(A)$.
    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 $X \ss A$.
    Then let $R$ be the well-ordering of $X$ such that $(X,R)$ is isomorphic to $\a$.
    Clearly $R$ is a relation on $X$ and therefore also a relation on $A$ since $X \ss A$.
    Thus $R \ss A 	imes A$ so that $R \in \pset{A 	imes A}$.
    Clearly also $f(R) = \a$ and since $\a$ was arbitrary this shows that $f$ is surjective.
  }

Note that this does not mean that $h(A) \leq \abs{\pset{A 	imes A}}$ unless the Axiom of Choice is employed.

Exercise 7.1.3

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Exercise 7.1.3

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