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Warm-up: let $(M, \in) \models \zfc$. Find
countable $N \subseteq M$ such that $(N, \in)
\prec (M, \in)$.

Suppose $\ol{p} = (p_0, \ldots, p_n) \in M$ and $M
\models \exists y, \psi(y, \ol{p})$. Let $w(\psi,
\ol{p})$ be a witness for this:
\[
  M \models \psi(w(\psi, \ol{p}), \ol{p})
\]
(if necessary, use Axiom of Choice).

(if $M \models \neg \exists y, \psi(y, \ol{p})$,
then let $w(\psi, \ol{p}) = \emptyset$).

Set:
\begin{align*}
  N_0
  &\defeq \emptyset \\
  N_{i + 1}
  &\defeq \{w(\psi, \ol{p}) : \text{$\psi$ formula and
  $\ol{p} \in N_1^{<\omega}$}\} \\
  N
  &\defeq \bigcup_{i \in \omega} N_i
\end{align*}
Note:
\begin{enumerate}[(1)]
  \item $N$ is countable.
  \item $M \prec N$ by \nameref{tvt}.
\end{enumerate}

\begin{remark*}
  In general, even if $M$ is transitive, $N$ is
  not.

  For example, if $\omega_1 \in M$, then
  \[
    \exists x, (\text{$x$ is the least countable
    ordinal})
  \]
  is true in $M$.
  \[
    w(\psi, \emptyset) = \omega_1
  .\]
  So $\omega_1 \in N$. But $\omega_1 \subseteq N$,
  since $N$ is countable.
\end{remark*}

Relevant later!

Also see \es{1}.

\begin{proof}[Proof of \nameref{lrt}]
  Fix $\varphi$ and let $\Phi$ be its collection
  of subformulas. This is a finite set!

  Need to show: $\forall \alpha, \exists \theta >
  \alpha$ such that $Z_\theta \models \varphi \iff
  Z \models \varphi$.

  For each $\psi \in \Phi$ and $\ol{p} = (p_0,
  \ldots, p_n)$, write
  \begin{align*}
    o(\psi, \ol{p})
    &\defeq \begin{cases}
      \text{least $\alpha$ such that $\exists y \in
      Z_\alpha$ with $Z \models \psi(y, \ol{p})$}
      & \text{if it exists} \\
      0 & \text{otherwise}
    \end{cases} \\
    o(\ol{p})
    &\defeq \max_{\psi \in \Phi} o(\psi, \ol{p})
    \\
    \theta_0
    &\defeq \alpha + 1 \\
    \theta_{i + 1}
    &\defeq \sup \{o(\ol{p}) : \ol{p} \in
    Z_{\theta_i}^{< \omega}\} \\
    \theta
    &\defeq \sup_{i \in \omega} \theta_i
  \end{align*}
  Then \nameref{tvt} implies that $Z_\theta$ and
  $Z$ agree on $\varphi$.
\end{proof}

\begin{fccorostar}[]
  If $T \subseteq \zfc$ is finite, then there is
  \cloze{$M$ transitive such that $M \models T$.}
\end{fccorostar}

\begin{proof}
  Let $\varphi \defeq \bigwedge_{\psi \in T}
  \psi$. Since $\zfc \vdash \varphi$, we have that
  $\varphi$ is true. By \nameref{lrt}, we can find
  $\theta$ such that $V_{\theta} \models \varphi$.
  Note $V_\theta$ is transitive.
\end{proof}

Remark about the proof of \nameref{lrt}:

Can you do the same if $\Phi$ is infinite?

Of course not: otherwise we colud prove that there
exists $\theta$ such that $V_\theta \models \zfc$,
and hence get $\Con(\zfc)$.

The problem is the case distinction in the
definition of $o(\psi, \ol{p})$: it requires to
check whether $\exists y, \psi$ is true.

\textbf{Next goal:} Obtain some $M \subseteq
V_\theta$ countable such that $M \models \varphi$
and $M$ is transitive.

TODO

\begin{fcthmstar}[Mostowski's Collapsing Theorem]
  Let $r$ be a relation on a set $a$ that is
  well-founded and extensional. Then there exists
  a transitive set $b$, adn a bijection $f : a \to
  b$ such that $(\forall x, y \in a)(x~r~y \iff
  f(x) \in f(y))$. Moreover, $b$ and $f$ are
  unique.
\end{fcthmstar}

\noproof

\begin{proof}
  See \courserefcustom{LST}{Logic and Set Theory}.
\end{proof}

\begin{fccorostar}[]
  For every $T \subseteq \zfc$ finite, there is a
  \gls{ctm} of $T$.
\end{fccorostar}

\begin{proof}
  Without loss of generality that $T$ contains
  the axiom of extensionality.
  Form $M \models T$ transitive by \nameref{lrt}.

  Use warm-up to obtain $N \prec M$ countable.
  This is extensional and well-founded, so by
  Mostowski find $W$ transitive such that
  \[
    (W, \in) \cong (N, \in)
  .\]
  Then $W \models T$ adn $|W| = ||$, so $W$ is
  countable.
\end{proof}

The next few lectures will be spent proving
$\Con(\zfc + \ch)$ using G\"odel's constructible
universe.

Absoluteness is preserved under transfinite
recursion.

Let $F, G, H$ be three operations.
\begin{align*}
  R(0, \ol{x})
  &\defeq F(\ol{x}) \\
  R(\alpha + 1, \ol{x})
  &\defeq G(\alpha, R(\alpha, \ol{x}), \ol{x}) \\
  \tag{$*$}
  \label{lec4eq}
  R(\lambda, \ol{x})
  &\defeq H(\lambda, \{R(\alpha, \ol{x}) : \alpha
  < \lambda\}, \ol{x})
\end{align*}

\begin{proof}
  Attempts: set functions satisfying the
  \eqref{lec4eq}.
  \begin{enumerate}[L1]
    \item All attempts agree on their common
      domain.
    \item $\forall \alpha, \exists r$ attempt such
      that $(\alpha, \ol{x}) \in \dom(r)$.
  \end{enumerate}
  $R(\alpha, \ol{x}) \defeq y$ if and only if
  there exists attempt $r$ with $(\alpha, \ol{x})
  \in \dom(r)$ and $r(\alpha, \ol{x}) = y$.
\end{proof}

Note that for $F, G, H$ fixed, there is a finite
fragment $T_{F, G, H} \subseteq \zfc$ that proves
the recursion theorem instance for $F, G, H$.

\begin{fcthmstar}[]
  If $T \supseteq T_{F, G, H}$ and $F, G, H$ are
  absolute for transitive models of $T$, then so
  is $R$ defined by \eqref{lec4eq}.
\end{fcthmstar}

\noproof