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\subsubsection*{Ordinals}

``$x$ is an ordinal'' means $x$ is transitive and
$(x, \in)$ is a well-order.

We know: being well-founded is not expressible in
first-order logic (see \es{1}).

Because all transitive models satisfy
Foundation, we have that if $M$ is transitive,
then
\[
  M \models \text{$x$ is transitive} \wedge (x,
  \in) \text{ is linearly ordered}
.\]
characterises ordinals. But this is clearly in
$\delz$.

So: being an ordinal is \gls{abs} for transitive
models.

Thus $M \cap \Ord = \{x \in M : M \models
\text{$x$ is an ordinal}\}$. This is transitive,
thus there is $\alpha \in \Ord$ such that $\alpha
= M \cap \Ord$.

Also \gls{abs}:
\begin{itemize}
  \item ``$x$ is a successor ordinal'' ($\exists y
    \in TODO$)
  \item ``$x$ is a limit ordinal''
  \item ``$x$ is a non-zero limit ordinal''
  \item $x = \omega$ TODO
\end{itemize}

\subsubsection*{Cardinals}

``$x$ is a cardinal'' if and only if \[
  \text{$x$ is an
  ordinal} \wedge \forall f, \forall y \in x, f :
  y \to x \implies \text{$f$ is not a surjection}
\]
Note that $\forall f$ is not bounded (while
$\forall y \in x$ is bounded).

Observe: this is $\pio$ and therefore \gls{dabs}.

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item We may not \emph{want} this to be
      \gls{abs}. If it was, we couldn't change
      cardinal behaviour.
    \item We can't obviously bound $\forall f$,
      since the natural bound would be
      \[
        \{h : h \to y \to x\}
      \]
      or
      \[
        \mathcal{P}(y \times x)
      .\]
      These, however, are not (yet??) on our list
      of \gls{abs} concepts.
    \item Not that neither (1) nor (2) is an
      argument, since there could be an equivalent
      formula that is $\delz$.
  \end{enumerate}
\end{remark*}

\subsection{Non-absoluteness}

Assume that $M \models \zfc$ is transitive and
countable. Then
\[
  M \cap \Ord = \alpha < \omega_1
.\]
However, $M \models \zfc$ implies $M \models
\text{there are uncountable cardinals}$.

Let $\beta < \alpha$ be such that $M \models
\text{$\beta$ is the least uncountable cardinal}$.

But $\beta$ is a countable ordinal, so not a
cardinal.

\textbf{Consequence:} All cardinals in $M$ except
$\aleph_0$ are going to be fake.

So ``$x$ is a cardinal'' can't be \gls{abs}.

\begin{note*}
  This also shows that ``$x = \mathcal{P}(y)$''
  cannot be \gls{abs}:

  Take $y$ such that $M \models y =
  \mathcal{P}(\omega)$. Then $y \subseteq
  \mathcal{P}(\omega)$, but is countable since $y
  \subseteq M$.

  Thus $y \neq \mathcal{P}(\omega)$. Therefore
  ``$x = \mathcal{P}(y)$'' is not \gls{abs}.
\end{note*}

Recall the general proof strategy mentioned
before:

\begin{quote}
  If $M$ is a countable transitive set such that $M
  \models \ZFC$, then there is a a countable
  transitive set $N \supseteq M$ such that $N
  \models \ZFC + \neg \CH$.
\end{quote}

\textbf{Question:} Is this really solving the
original problem? i.e. $\Con(\zfc) \implies
\Con(\zfc + \neg \ch)$.

\glsnoundefn{ctm}{countable transitive
model}{countable transitive models}%
It's not obvious that $\Con(\zfc)$ implies that
there is a countable transitive model (ctm) of
\zfc.

\textbf{Answer:} That's not only not obvious, but
fake.

Let's prove that $\Con(\zfc)$ $\not \implies$
there is a \gls{ctm} of \zfc.

Why? Note that $\Con(\zfc)$, or $\Con(T)$ for any
$T$ is $\delz$. So, it's absolute for transitive
models.

So if $M$ is a \gls{ctm} of \zfc, then
$\Con(\zfc)$ is true, so by \gls{abs}ness, $M
\models \Con(\zfc)$. So $M \models \zfc +
\Con(\zfc)$. This contradicts G\"odel's
Incompleteness Theorem.

We can get a proof of
\[
  \Con(\zfc) \implies \Con(\zfc + \neg \ch)
\]
via a trick (\es{1}).

\begin{fclemmastar}[Cohen Lemma]
  Assuming:
  - $T \subseteq \zfc$
  Then:
  there is finite $T^* \subseteq \zfc$ such that
  if $M$ is a \gls{ctm} of $T^*$, then there is $N
  \supseteq M$ such that $N$ is a \gls{ctm} of $T
  + \neg \ch$.
\end{fclemmastar}

This reduces the problem to:

Find \glspl{ctm} of $T^*$ for sufficiently large
finite $T^* \subseteq \zfc$.

\begin{fcdefnstar}[Hierarchy]
  \glsnoundefn{hier}{hierarchy}{hierarchies}%
  We call an assignment $\alpha \mapsto Z_\alpha$
  a \emph{hierarchy} if
  \begin{enumerate}[(i)]
    \item $Z_\alpha$ is a transitive set
    \item $\Ord \cap Z_\alpha = \alpha$
    \item $\alpha < \beta \implies Z_\alpha
      \subseteq Z_\beta$
    \item $\lambda \text{ limit} \implies
      Z_\lambda = \bigcup_{\alpha < \lambda}
      Z_\alpha$
  \end{enumerate}

  If $\{Z_\alpha : \alpha \in \Ord\}$ is a
  hierarchy, we can define $Z \defeq
  \bigcup_{\alpha \in \Ord} Z_\alpha$.
  This is a proper class as $\Ord \subseteq Z$. We
  also define $\rho_Z(x) \defeq \min \{\alpha : x
  \in Z_\alpha\}$, a notion of \emph{$Z$-rank}.
\end{fcdefnstar}

Paradigmatic example: von Neumann hierarchy
$V_\alpha$, and $V$ is the entire universe.

\begin{fcthm}[Levy Reflection Theorem]
  \label{lrt}
  % [L\'evy Reflection Theorem]
  Assuming:
  - $Z$ is a hierarchy
  - $\varphi$ is a formula
  Then:
  there are unboundedly many $\theta$ such that
  $\varphi$ is absolute between $Z_\theta$ and
  $Z$.
\end{fcthm}

\begin{fcprop}[Tarski-Vaught Test]
  \label{tvt}
  Assuming:
  - $\mathcal{M}$ is a substructure of $\mathcal{N}$
  Then:
  \begin{iffc}
    \lhs
    $\mathcal{M}$ is an elementary substructure
    \rhs
    for any formula $\phi(v, \ol{w})$ and $\ol{a}
    \in M$, if there is $b \in N$ such that
    $\mathcal{N} \models \phi(b, \ol{a})$, then
    there is $c \in M$ such that $\mathcal{N}
    \models \phi(c, \ol{a})$.
  \end{iffc}
\end{fcprop}

\noproof

TVT${}_\Phi$:

Let $M \subseteq N$ and $\Phi$ be a collection of
formulas closed under subformulas. Then the
following are equivalent:
\begin{enumerate}[(i)]
  \item All formulas in $\Phi$ are absolute
    between $M$ and $N$.
  \item For all $\varphi \in \Phi$, the
    \emph{TV-condition} holds: if $\varphi =
    \exists x \psi$, then for all $\ol{y} \in M$
    if there is $a \in N$ sucht hat $N \models
    \psi(a, \ol{y})$, then there is $b \in M$ such
    that $N \models \psi(b, \ol{y})$.
\end{enumerate}
