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\begin{enumerate}[(1)]
  \setcounter{enumi}{7}
  \item \begin{flashcard}[ax-of-replacement]
      \glsnoundefn{Rep}{(Rep)}{N/A}
      \prompt{\cloze{(8) }}\textbf{\cloze{Axiom of
      Replacement} (Rep):} \cloze{\gls{Inf}
      says that there exist sets containing $0,
      1, 2, 3, \ldots$. Are there sets containing
      $\emptyset, \PP \emptyset, \PP \PP
      \emptyset, \ldots$? There's a function-like
      object that sends $0 \mapsto \emptyset$, $1
      \mapsto \PP\emptyset$, $2 \mapsto
      \PP\PP\emptyset, \ldots$. Need an axiom that
      says that the image of a set under a
      function-like object is a set.
      The axiom is:
      \[ (\pforall t_1) \cdots (\pforall t_n)
      \bigg[(\pforall x)(\pforall y)(\pforall z)((p
      \pand \subst p[z / y]) \pimpl y = z)\qquad \]
      \[ \qquad\pimpl
      (\pforall x)(\pexists y)(\pforall z)(z \zin
      y \iff (\pexists u)(u \zin x \pand \subst
      p[u / x, z / y]))\bigg] \]
      for any \gls{form} $p$ with $\FV(p) = \{x,
      y, t_1, \ldots, t_n\}$.

      We will explain the reasoning below, by
      discussing function-classes.
      }
    \end{flashcard}
\end{enumerate}

\subsubsection*{Digression on classes}

\begin{flashcard}[class-defn]
\begin{definition*}[Class]
  \glsnoundefn{cls}{class}{classes}
  \cloze{A \emph{class} is a subset $C$ of a \gls{struc}
  $V$ of the language of \gls{ZF} such that there is
  a \gls{form} $p$ with $\FV(p) = \{x\}$ such that
  $p_V = C$, i.e. $x \in C$ if and only if $p(x)$
  holds in $V$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  $V$ is a \gls{cls}: for example take $p$ to be
  $x = x$. The set of sets of size $1$ is a
  \gls{cls}: for example, take $p$ to be
  $(\pexists y)(x = \{y\})$.
\end{example*}

\begin{flashcard}[proper-class-defn]
\begin{definition*}[Proper class]
  \glsadjdefn{propc}{proper}{\gls{cls}}
  \cloze{Say the \gls{cls} is a set if $(\pexists
  y)(\pforall x)(x \zin y \iff p)$ holds in $V$.
  If $C$ is not a set, we say $C$ is a
  \emph{proper class}.}
\end{definition*}
\end{flashcard}

\begin{example*}
  $V$ is a \gls{propc} \gls{cls} (Russell's
  paradox).
\end{example*}

\begin{flashcard}[function-class-defn]
\begin{definition*}[Function class]
  \glsnoundefn{funccls}{function-class}{function-classes}
  \cloze{A \emph{function-class} is a subset $G$ of $V
  \times V$ such that there's a \gls{form} $p$
  with \gls{freev} \glspl{var} $\FV(p) = \{x, y\}$
  such that
  \[ (\pforall x)(\pforall y)(\pforall z)((p \pand
  \subst p[z / y]) \pimpl y = z) \]
  holds in $V$ and $G = p_V$, i.e. $(x, y) \in G$
  if and only if $p(x, y)$ holds in $V$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  $G = \{(x, \{x\}) \st x \in V\}$ is the
  \gls{funccls} mapping $x \mapsto \{x\}$, and is
  given by $p = (y = \{x\})$.
\end{example*}

\begin{enumerate}[(1)]
  \setcounter{enumi}{8}
  \item \begin{flashcard}[ax-of-foundation]
      \glsnoundefn{Fnd}{(Fnd)}{N/A}
      \prompt{\cloze{(9) }}\textbf{\cloze{Axiom of
      Foundation} (Fnd):} \cloze{We want to avoid
      pathological behaviour like $x \zin x$, i.e.
      $\{x\}$ has no $\zin$-minimal member, or $x
      \zin y \pand y \zin x$ (in which case $\{x,
      y\}$ has no $\zin$-minimal member).
      \gls{Fnd} says that every non-empty set has
      an $\zin$-minimal member:
      \[ (\pforall x)(\pnot x = \emptyset \pimpl
      (\pexists y)(y \zin x \pand (\pforall z)(z
      \zin x \pimpl \pnot z \zin y))) \]
      }
    \end{flashcard}
\end{enumerate}

The above axioms and axiom-schemes (1)-(9) form
\gls{ZF}. The \gls{aoc} (AC) is not
included:
\[ (\pforall x)((\pforall y)(y \zin x \pimpl \pnot
y = \emptyset) \pimpl (\pexists f)((f : X \to
\bigcup x) \pand (\pforall y)(y \zin x \pimpl f(y)
\zin y))) .\]
\glsnoundefn{ZFC}{ZFC}{N/A}
We write ZFC for \gls{ZF} + \glsref[aoc]{AC}. For
the rest of this chapter we work within \gls{ZF}.

\textbf{Aim:} to describe the set-theoretic
universe, i.e. any \gls{model} $V$ of \gls{ZF}.

\begin{flashcard}[transitive-set-defn]
\begin{definition*}[Transitive set]
  \glsadjdefn{transset}{transitive}{set}
  \cloze{We say a set $x$ is \emph{transitive} if every
  membet of $x$ is a member of $x$. So ``$x$ is
  transitive'' is shorthand for
  \[ (\pforall y)((\pexists z)(z \zin x \pand y
  \zin z) \pimpl y \zin x) .\]
  Equivalently, $\bigcup x \subset x$.}
\end{definition*}
\end{flashcard}

\begin{note*}
  This is \emph{not} the same as saying that
  $\zin$ is a \gls{ltrans} relation on $x$.
\end{note*}

\begin{example*}
  $\omega$ is \gls{transset}. We need to show that
  $x \subset \omega$ for all $x \zin \omega$. Form
  the set $z = \{y \zin \omega \st y \subset
  \omega\}$. Check $z$ is a successor set, so $z =
  \omega$. Similarly,
  \[ \{x \in \omega \st \text{``$x$ is
  \gls{transset}''}\} \]
  is a successor set ($\bigcup x^+ = x$) so it is
  $\omega$. So every element of $\omega$ is a
  \gls{transset} set.
\end{example*}

\begin{flashcard}[transitive-closure-lemma]
\begin{lemma}
  Every set $x$ is contained in a \gls{transset}
  set, i.e.
  \[ (\pforall x)(\pexists y)(\text{``$y$ is
  transitive''} \pand x \subset y) .\]
\end{lemma}

\fcscrap{
\begin{remark*}
  \glssymboldefn{TC}{TC$(x)$}{TC$(x)$}
  The intersection of \gls{transset} sets if
  \gls{transset}, so $x$ is contained in a
  smallest \gls{transset} set, called the
  \emph{transitive closure} of $x$, denoted by
  $\TC(x)$.
\end{remark*}
}

\cloze{\textbf{Idea:} If $x \subset y$, $y$
\gls{transset}, then $\bigcup x \subset y$ and so
$\bigcup \bigcup x \subset y$, $\bigcup \bigcup
\bigcup x \subset y$, \ldots. Want to form
\[ \bigcup\left\{x, \bigcup x, \bigcup \bigcup x,
\ldots\right\} \]
Is this a set? Yes, by \gls{Rep}. We need a
\gls{funccls} $0 \mapsto x$, $1 \mapsto \bigcup
x$, $2 \mapsto \bigcup \bigcup x$, \ldots.}

\begin{proof}
  \cloze{Say ``$f$ is an attempt'' to mean:
  \begin{align*}
    \text{``$f$ is a function''} \pand \text{``$\dom f
    \in \omega$''} \pand \text{``$f(0) = x$''} \\
    \pand (\pforall m)(\pforall n)[((m \in \dom f)
    \pand (n \in \dom f) \pand (n = m^+)) \pimpl
    (f(n) = \bigcup f(m))]
  \end{align*}
  We prove by $\omega$-induction that:
  \[ (\pforall f)(\pforall g)(\pforall
  n)((\text{``$f$ is an attempt''} \pand
  \text{``$g$ is an attempt''} \pand (n \zin \dom f
  \cap \dom g)) \pimpl (f(n) = g(n))) \tag{$*$}
  \label{lec18_l200} \]
  and
  \[ (\pforall n)(n \zin \omega \pimpl (\pexists
  f)(\text{``$f$ is an attempt''} \pand n \in \dom f))
  \tag{$**$} \label{lec18_l204} \]
  Define a \gls{funccls} via the formula $p(y,
  z)$:
  \[ (\pexists f)(\text{``$f$ is an attempt''}
  \pand f(y) = z) .\]
  By \eqref{lec18_l200} we do have
  \[ (\pforall y)(\pforall z)(\pforall w)((p \pand
  \subset p[w / z]) \pimpl w = z) .\]
  By \gls{Rep} can form $w = \{z \st (\pexists
  y)(y \zin \omega \pand p(y, z))\}$ ($w = \{x,
  \bigcup x, \bigcup \bigcup x, \ldots\}$) and by
  \gls{Un} can form $t = \bigcup w$. Then $x
  \subset t$, since $x \zin w$ ($\{(0, x)\}$ is an
  attempt). Given $a \zin t$, we have $z \zin w$,
  $a \zin z$. There's an attempt $f$ and $n \zin w$
  such that $z = f(n)$. By \eqref{lec18_l204},
  there's an attempt $g$ with $n^+ \in \dom g$.
  Then $n \in \dom g$, so
  \[ \bigcup z = \bigcup f(n)
  \stackrel{\eqref{lec18_l200}}{=} \bigcup g(n) =
  g(n^+) \zin w \]
  hence $a \subset t$.}
\end{proof}
\end{flashcard}