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\subsection{Picture of the Universe}

Idea: everything is built up from $\emptyset$
using $\PP$ and $\cup$. Have
\[ V_0 = \emptyset, V_1 = \PP \emptyset =
\{\emptyset\}, V_2 = \PP\PP\emptyset =
\{\emptyset, \{\emptyset\}\}, \ldots \]
and then will have
\[ V_{\oomega} =
\bigcup \{V_0, V_1, V_2, \ldots\}, V_{\oomega + 1}
= \PP V_{\oomega}, \text{etc} \]
It will be \gls{Fnd} that guarantees that every
set appears in a $V_\alpha$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/154befa6c00440a2.png}
\end{center}
\begin{flashcard}[von-neumann-hierarchy]
\glsnoundefn{vnh}{von Neumann hierarchy}{N/A}
\glssymboldefn{vnh}{$V_\alpha$}{$V_\alpha$}
We define sets $V_\alpha$, $\alpha \in \ON$ by
\gls{epsrec}:
\begin{itemize}
  \item \cloze{$\alpha = 0$: $V_0 = \emptyset$}
  \item \cloze{$\alpha = \beta\osucc$: $V_\alpha = \PP
    V_\beta$}
  \item \cloze{$\alpha \neq 0$ \glsref[limord]{limit}:
    $V_\alpha = \bigcup \{V_\gamma \st \gamma \wlt
    \alpha\}$}
\end{itemize}
The sets $V_\alpha$ form \cloze{the \emph{von Neumann
hierarchy}.}
\end{flashcard}

\textbf{Aim:} Every set appears in this hierarchy.

\begin{flashcard}[V-alpha-trans]
\begin{lemma}
  \label{vnh_trans}
  $\vnh_\alpha$ is \gls{transset} for all $\alpha
  \in \ON$.
\end{lemma}

\begin{proof}
  \cloze{By \gls{induction} on $\alpha$.
  \begin{itemize}
    \item $\alpha = 0$: $\vnh_0 = \emptyset$ is
      \gls{transset}.
    \item $\alpha = \beta\osucc$: Let $x \zin \vnh_\alpha
      = \PP \vnh_\beta$. Then $x \subset
      \vnh_\beta$. If $y \zin x$, then $y \zin
      \vnh_\beta$, so by \gls{induction}
      hypothesis, $y \subset \vnh_\beta$
      ($\vnh_\beta$ is \gls{transset}). So every
      $y \zin x$ has $y \zin \PP \vnh_\beta =
      \vnh_\alpha$. Thus $\vnh_\alpha$ is
      \gls{transset}.
    \item $\alpha \neq 0$ \glsref[limord]{limit}:
      If $x \zin \vnh_\alpha$, then $\exists
      \gamma \wlt \alpha$, $x \zin \vnh_\gamma$.
      By induction, $\vnh_\gamma$ is
      \gls{transset}, so $x \subset \vnh_\gamma
      \subset \vnh_\alpha$. So $\vnh_\alpha$ is
      \gls{transset} \qedhere
  \end{itemize}
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[V-alpha-inclusion]
\begin{lemma}
  \label{vnh_chain}
  If $\alpha \wle \beta$, then $\vnh_\alpha
  \subset \vnh_\beta$.
\end{lemma}

\begin{proof}
  \cloze{By \gls{induction} on $\beta$.
  \begin{itemize}
    \item $\beta = 0$: $\alpha \wle \beta$, so
      $\alpha = 0$, so $\vnh_\alpha = \vnh_\beta$.
    \item $\beta = \gamma\osucc$: If $\alpha =
      \beta$ then $\vnh_\alpha = \vnh_\beta$. If
      $\alpha \wlt \beta$, then $\alpha \wle
      \gamma$, so by \gls{induction} hypothesis,
      $\vnh_\alpha \subset \vnh_\gamma$. If $x
      \zin \vnh_\gamma$, then $x \subset
      \vnh_\gamma$ ($\vnh_\gamma$ is
      \gls{transset}), so $x \zin \PP \vnh_\gamma
      \subset \vnh_\beta$. Thus $\vnh_\gamma
      \subset \vnh_{\gamma\osucc} = \vnh_\beta$,
      and hence $\vnh_\alpha \subset \vnh_\beta$.
    \item If $\beta \neq 0$
      \glsref[limord]{limit}: then if $\alpha \wlt
      \beta$ then $\vnh_\alpha \subset \vnh_\beta$
      by definition. \qedhere
  \end{itemize}
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[vnh-exhausts-universe]
\begin{theorem}
  \label{vnh_exhaustion}
  The \gls{vnh} exhausts the set theoretic
  universe $V$, i.e.
  \[ (\pforall x)(\pexists \alpha \in \ON)(x \zin
  \vnh_\alpha) \]
  or
  \[ V = \bigcup_{\alpha \in \ON} \vnh_\alpha .\]
\end{theorem}

\fcscrap{
\begin{note*}
  If $x \zin \vnh_\alpha$ then $x \subset
  \vnh_\alpha$ (by \cref{vnh_trans}). If $x
  \subset \vnh_\alpha$ then $x \zin \PP
  \vnh_\alpha = \vnh_{\alpha \opl 1}$.

  \glssymboldefn{vnhrank}{rank}{rank}
  If $\exists
  \alpha \in \ON$, $x \subset \vnh_\alpha$ then
  define the \emph{rank of $x$} to be $\vnhrank(x)$,
  the least $\alpha \in \ON$ such that $x \subset
  \vnh_\alpha$.
\end{note*}
}

\begin{proof}
  \cloze{We will show $(\pforall x)(\pexists \alpha \in
  \ON)(x \subset \vnh_\alpha)$ by \gls{epsind}.
  Fix $x$ and assume for each $y \zin x$, $y
  \subset \vnh_\alpha$ for some $\alpha \in \ON$,
  so for all $y \zin x$, $y \subset
  \vnh_{\vnhrank(y)}$. Let
  \[ \alpha = \sup\{\vnhrank(y)\osucc \st y \zin
  x\} ,\]
  which is a set by \gls{Rep}. We'll show $x
  \subset \vnh_\alpha$. If $y \zin x$, then $y
  \subset \vnh_{\vnhrank(y)}$, so $y \zin \PP
  \vnh_{\vnhrank(y)} = \vnh_{\vnhrank(y)\osucc}
  \subset \vnh_\alpha$ (where the final $\subset$
  is using \cref{vnh_chain}). This shows $x \subset
  \vnh_\alpha$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[rank-computation-coro-vnh]
\begin{corollary}
  For every set $x$,
  \[ \vnhrank(x) = \cloze{\sup\{\vnhrank(y)\osucc \st y
  \zin x\}} \]
\end{corollary}

\begin{proof}
  \phantom{}
  \cloze{
  \begin{enumerate}[$\le$:]
    \item[$\le$:] Follows from proof of
      \cref{vnh_exhaustion}.
    \item[$\ge$:] We first show that $x \zin
      \vnh_\alpha \implies \vnhrank(x) \wlt
      \alpha$.
      \begin{itemize}
        \item $\alpha = 0$ is true.
        \item $\alpha = \beta\osucc$: $x \zin \PP
          \vnh_\beta$, so $x \subset \vnh_\beta$,
          so $\vnhrank(x) \wle \beta \wlt \alpha$
        \item $\alpha \neq 0$
          \glsref[limord]{limit}: $x \zin
          \vnh_\alpha \implies \exists \gamma \wlt
          \alpha$ with $x \zin \vnh_\gamma$, so
          $\vnhrank(x) \wlt \gamma \wlt \alpha$.
      \end{itemize}
      Now let $\alpha = \vnhrank(x)$. Then $x
      \subset \vnh_\alpha$, so for $y \zin x$, $y
      \zin \vnh_\alpha$ and so $\vnhrank(y) \wlt
      \alpha$. Hence
      \[ \sup\{\vnhrank(y)\osucc \st y \zin x\}
      \wle \alpha . \qedhere \]
  \end{enumerate}
  }
\end{proof}
\end{flashcard}

\begin{example*}
  $\vnhrank(\alpha) = \alpha$ for all $\alpha \in
  \ON$. By \gls{induction}:
  \begin{align*}
    \vnhrank(\alpha)
    &= \sup\{\vnhrank(\beta)\osucc \st \beta \wlt
    \alpha\} \\
    &= \sup\{\beta\osucc \st \beta \wlt \alpha\}
    &&\text{(induction hypothesis)}
    \\
    &= \alpha
  \end{align*}
\end{example*}

\newpage
\section{Cardinal Arithmetic}

\glssymboldefn{ssize}{cong}{cong}
Look at the size of sets. We write $x \ssize y$ to
mean
\[ (\pexists f)(f : x \to y \pand \text{``$f$ is a
bijection''}) .\]
This is an equivalence relation \gls{cls}. The
equivalence classes are proper \glspl{cls} (except
$\{\emptyset\}$).

\glssymboldefn{card}{card $x$}{card $x$}
How do we pick a representative from each
equivalence class? We seek for each set $x$, a set
$\card x$ such that
\[ (\pforall x)(\pforall y)(\card x = \card y \iff
x \ssize y) \]

In \gls{ZFC} this is easy: given a set $x$, $x$
can be \glsref[wellord]{well-ordered}, so $x
\ssize \OT(x)$, i.e. $x \ssize \alpha$ for some
$\alpha \in \ON$. Can define $\card x$ to be the
least $\alpha \in \ON$ such that $x \ssize
\alpha$.

In \gls{ZF} (due to D. S. Scott): define the
\emph{essential rank} as follows:
\[ \essrank(x) = \text{least
$\alpha$ such that $\exists y \subset \vnh_\alpha$
with $y \ssize x$} .\]
Note $\essrank(x) \wle
\vnhrank(x)$. Define
\[ \card x = \{y \subset \vnh_{\essrank(x)} \st y
\ssize x\} .\]
