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\begin{flashcard}[motowski-thm]
\begin{theorem}[Mostowki's Collapsing Theorem]
  \label{motowski}
  \cloze{Let $r$ be a \gls{wfdd}, \gls{extensional}
  relation on a set $a$. Then there is a
  \gls{transset} $b$ and a bijection $f : a \to b$
  such that
  \[ (\pforall x, y \zin a)(x r y \iff f(x) \zin
  f(y)) .\]
  Moreover, $(b, f)$ is unique.}
\end{theorem}
  
\begin{proof}
  \cloze{By \gls{rrec} on $a$, there's a \gls{funccls}
  such that
  \[ \pforall x \in a \quad f(x) = \{f(y) \st y
  \zin a \pand y r x\} .\]
  Note that $f$ is a function, not just a
  \gls{funccls}, since $\{(x, f(x)) \st x \zin
  a\}$ is a set by \gls{Rep}. Then
  \[ b = \{f(x) \st x \in a\} \]
  is a set by \gls{Rep}. Now we check:
  \begin{itemize}
    \item $b$ is \gls{transset}:
      let $z \zin b$ and $w \zin z$. There's a $x \zin
      a$ such that $z = f(x)$, and so there's $y \zin
      a$ such that $y r x$ and $w = f(y) \zin b$. 
    \item $f$ is surjective (true by definition of
      $b$).
    \item $\pforall x, y \zin a$, $x r y \pimpl
      f(x) \zin f(y)$ is true by definition of
      $f$.
    \item It remains to show that $f$ is
      injective. It will then follow that
      $\pforall x, y \zin a$, $f(x) \zin f(y)
      \pimpl x r y$. Indeed, if $f(x) \zin f(y)$,
      then $f(x) = f(z)$ for some $z \zin a$ with
      $z r y$. Since $f$ is injective, $x = z$, so
      $x r y$.
      We will show
      \[ (\pforall x \zin a)\ub{(\pforall y \zin a)(f(x) =
      f(y) \pimpl x = y)}_{\text{``$f$ is injective at
      $x$''}} \]
      by \gls{rind}. Fix $x \zin a$ and assume that
      $f$ is injective at $s$ whenever $s \zin a$ and
      $s r x$. Assume $f(x) = f(y)$ for some $y \zin
      a$, i.e.
      \[ \{f(s) \st s \zin \pand s r x\} = \{f(t) \st
      t \zin a \pand t r y\} .\]
      Since $f$ is injective at every $s \zin a$ with
      $s r x$, it follows that
      \[ \{s \zin a \st s r x\} = \{t \zin a \st t r
      y\} .\]
      By \glsref[extensional]{extensionality} for $r$,
      it follows that $x =
      y$.
  \end{itemize}
  Now we check that $(b, f)$ is unique. Assume
  that $(b, f)$ and $(b', f')$ both satisfy the
  theorem. We prove
  \[ (\pforall x \zin a)(f(x) = f'(x)) \]
  by \gls{rind}. Fix $x \zin a$ and assume $f(y) =
  f'(y)$ whenever $y \zin a$ and $y r x$. If $z
  \zin f(x)$, then $z \zin b$ ($b$
  \gls{transset}), so $z = f(y)$ for some $y \zin
  a$ with $y r x$. Then $z = f(y) = f'(y)$
  (induction hypothesis). Then $z = f'(y) \zin
  f'(x)$. Similarly, if $z \zin f'(x)$ thne $z
  \zin f(x)$. By \gls{Ext}, $f(x) = f'(x)$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[ordinal-defn-for-real-this-time]
\begin{definition*}[Ordinal (set theoretic)]
  \glsnoundefn{setordinal}{ordinal}{ordinals}
  \cloze{An \emph{ordinal} is a \gls{transset} which is
  \glsref[wellord]{well-ordered} by $\zin$
  (equivalently, \glsref[linord]{linearly ordered}
  since $\zin$ is \gls{wfdd} by \gls{Fnd}).}
\end{definition*}
\end{flashcard}

\begin{note*}
  Let $a$ be a set and $r$ be a \gls{wellord} on
  $a$. Then $r$ is \gls{wfdd} and
  \gls{extensional} (if $x, y \zin a$ and $\pnot x
  = y$ then $x r y$ or $r y x$, but not both).

  By \nameref{motowski}, there exists a
  \gls{transset} $b$ and a bijection $f : a \to b$
  such that $x r y \iff f(x) \zin f(y)$, i.e.
  $f(a, r) \to (b, \zin)$ is an \gls{ordisism}. So
  $b$ is an \gls{setordinal}.

  So by \nameref{motowski}, every
  \glsref[wellord]{well-ordered} set is
  \gls{ordisic} to a unique \gls{setordinal},
  called the order-type of $x$.

  \glssymboldefn{ON}{ON}{ON}
  We let $\ON$ denote the \gls{cls} of
  \glspl{setordinal} (given by the \gls{form}
  ``$x$ is an \gls{setordinal}''). It is a proper
  \gls{cls} by \nameref{burati_forti}.
\end{note*}

\begin{flashcard}[set-theory-ordinal-props]
\begin{proposition}
  \prompt{(Some set theoretic facts for ordinals): }
  Let $\alpha, \beta \in \ON$, and let $a$ be a
  set of \glspl{setordinal}. Then:
  \begin{enumerate}[(i)]
    \item Every member of $\alpha$ is an
      \gls{setordinal}.
    \item $\beta \zin \alpha \iff \beta < \alpha$
      ($\beta$ is \gls{ordisic} to a proper
      \gls{is} of $\alpha$)
    \item $\alpha \zin \beta$ or $\alpha = \beta$
      or $\beta \zin \alpha$
    \item $\alpha\osucc = \alpha \cup \{\alpha\}$
      (i.e. the set theoretic meaning and ordinal
      meanings for ${}^+$ agree).
    \item $\bigcup a$ is an \gls{setordinal} and
      $\bigcup a = \sup a$.
  \end{enumerate}
\end{proposition}

\fcscrap{
\begin{remark*}
  (ii) says that $\alpha$ really \emph{is} the set
  of ordinals $\wlt \alpha$.
  (iii) says that $\zin$ \glsref[linord]{linearly
  orders} the class $\ON$. (iv) resolves the clash
  of notation $x^+$ in \cref{sec2} and
  \cref{sec5}. (v) now shows that any set of
  \glsref[wellord]{well-ordered} sets has an upper
  bound.
\end{remark*}
}

\begin{proof}
  \phantom{}
  \cloze{
  \begin{enumerate}[(i)]
    \item Let $\gamma \zin \alpha$. Then $\gamma
      \subset \alpha$ (since $\alpha$ is
      \gls{transset}) and hence $\zin$
      \glsref[linord]{linearly orders} $\gamma$.
      Given $\eta \zin \delta, \delta \zin \gamma$,
      then $\delta \zin \alpha$ and so $\eta \zin
      \alpha$ ($\alpha$ \gls{transset}). Since
      $\zin$ is \gls{transset} on $\alpha$, we
      have $\eta \zin \gamma$. So $\gamma$ is a
      \gls{transset}, so $\gamma$ is an
      \gls{setordinal}.
    \item If $\beta \zin \alpha$, then $I_\beta =
      \{\gamma \zin \alpha \st \gamma \zin \beta\}
      = \beta$, so $\beta \wlt \alpha$. Any proper
      \gls{is} of $\alpha$ is of the form
      $I_\gamma$ for some $\gamma \zin \alpha$. So
      $\beta \wlt \alpha \implies \beta \zin
      \alpha$.
    \item We know $\beta \wlt \alpha$ or $\beta =
      \alpha$ or $\alpha \wlt \beta$ is true. Then
      done by (ii).
    \item Let $\beta = \alpha \cup \{\alpha\}$
      (successor of $\alpha$). If $\gamma \zin
      \beta$ then either $\gamma = \alpha \subset
      \beta$ or $\gamma \zin \alpha$, so $\gamma
      \subset \alpha \subset \beta$. Thus $\beta$
      is \gls{transset}, \glsref[linord]{linearly
      ordered} by $\zin$ (by (iii)) and $\alpha$
      is the greatest element. So $\beta =
      \alpha\osucc$ in the sense of \cref{sec2}.
    \item $\bigcup a$ is a union of \gls{transset}
      sets, hence \gls{transset}. Every member of
      $\bigcup a$ is an \gls{setordinal}, so
      $\bigcup a$ is
      \glsref[linord]{linearly ordered} by $\zin$
      by (iii). If $\gamma \zin a$, then $\gamma
      \subset \bigcup a$, so either $\gamma =
      \bigcup a$, or $\gamma \zin \bigcup a$
      (by (ii)), i.e. $\gamma \wle \bigcup
      \alpha$. If $\gamma \wle \delta$ for all
      $\gamma \zin \alpha$, then $\gamma = \delta$
      or $\gamma \zin \delta$ for $\gamma \zin
      a$, i.e. $\gamma \subset \delta$ (using
      (ii)). So $\bigcup a \subset \delta$, i.e.
      $\bigcup a \wle \delta$. \qedhere
  \end{enumerate}}
\end{proof}
\end{flashcard}

\begin{example*}
  $0 = \emptyset \in \ON$, hence $n \zin \ON$ for
  all $n \zin \omega$ (by $\omega$-induction).
  $\omega$ is \gls{transset}, so $\bigcup \omega
  \subset \omega$. If $n \zin \omega$, then $n \in
  n\osucc \zin \omega$, so $n \zin \bigcup
  \omega$. So $\omega = \bigcup \omega$ is an
  \gls{setordinal} and $\omega = \sup \omega$.
\end{example*}