\item A cardinal $\kappa$ is called \vocab{weakly inaccessible}
iff $\kappa$ is uncountable,\footnote{dropping this we would get that $\aleph_0$ is inaccessible}
regular and $\forall\lambda < \kappa.~\lambda^+ < \kappa$.
\item A cardinal $\kappa$ is \vocab[inaccessible!strongly]{(strongly) inaccessible}
iff $\kappa$ is uncountable, regular
and $\forall\lambda < \kappa.~2^{\lambda} < \kappa$.
\end{itemize}
\end{definition}
\begin{remark}
Since $2^{\lambda}\ge\lambda^+$,
strongly inaccessible cardinals are weakly inaccessible.
If $\GCH$ holds, the notions coincide.
\end{remark}
\begin{theorem}
If $\kappa$ is inaccessible,
then $V_\kappa\models\ZFC$.\footnote{More formally $(V_{\kappa}, \in)\models\ZFC$.}
\end{theorem}
\begin{proof}
Since $\kappa$ is regular, \AxRep works.
Since $2^{\lambda} < \kappa$,
\AxPow works.
The other axioms are trivial.
\todo{Exercise}
\end{proof}
\begin{corollary}
$\ZFC$ does not prove the existence of inaccessible
cardinals, unless $\ZFC$ is inconsistent.
\end{corollary}
\begin{proof}
If $\ZFC$ is consistent,
it can not prove that it is consistent.
In particular, it can not prove the existence of a model of $\ZFC$.
\end{proof}
\begin{definition}[Ulam]
A cardinal $\kappa > \aleph_0$ is \vocab{measurable}
iff there is an ultrafilter $U$ on $\kappa$,
such that $U$ is not principal\footnote{%
i.e.~$\{\xi\}\not\in U$ for all $\xi < \kappa$%
}
and
if $\theta < \kappa$
and $\{X_i : i < \theta\}\subseteq U$,
then $\bigcap_{i < \theta} X_i \in U$
\end{definition}
\begin{goal}
We want to prove
that if $\kappa$ is measurable,
then $\kappa$ is inaccessible
and there are $\kappa$ many
inaccessible cardinals below $\kappa$
(i.e.~$\kappa$ is the $\kappa$\textsuperscript{th} inaccessible).
\end{goal}
\begin{theorem}
The following are equivalent:
\begin{enumerate}
\item$\kappa$ is a measurable cardinal.
\item There is an elementary embedding%
\footnote{Recall: $j\colon V \to M$
is an \vocab{elementary embedding} iff $j''V =\{j(x) : x \in V\}\prec M$,
i.e.~for all formulae $\phi$ and $x_1,\ldots,x_k \in V$,
$V \models\phi(x_1,\ldots,x_u)\iff M \models\phi(j(x_1),\ldots,j(x_u))$.%
}
$j\colon V \to M$ with $M$
transitive
such that $j\defon{\kappa}=\id$,
$j(\kappa)\neq\kappa$.
\end{enumerate}
\end{theorem}
\begin{proof}
2. $\implies$ 1.:
Fox $j\colon V \to M$.
Let $U =\{X \subseteq\kappa : \kappa\in j(X)\}$.
We need to show that $U$ is an ultrafilter:
\begin{itemize}
\item Let $X,Y \in U$.
Then $\kappa\in j(X)\cap j(Y)$.
We have $M \models j(X \cap Y)= j(X)\cap j(Y)$,
and thus $j(X \cap Y)= j(X)\cap j(Y)$.
It follows that $X \cap Y \in U$.
\item Let $X\in U$
and $X \subseteq Y \subseteq\kappa$.
Then $\kappa\in j(X)\subseteq j(Y)$
by the same argument,
so $Y \in U$.
\item We have $j(\emptyset)=\emptyset$
(again $M \models j(\emptyset)$ is empty),
hence $\emptyset\not\in U$.
\item$\kappa\in U$ follows from $\kappa\in j(\kappa)$.
This is shown as follows:
\begin{claim}
For every ordinal $\alpha$,
$j(\alpha)$ is an ordinal such that $j(\alpha)\ge\alpha$.
\end{claim}
\begin{subproof}
$\alpha\in\Ord$
can be written as
\[\forall x \in\alpha .~\forall y \in x.~y \in\alpha
\land\forall x \in\alpha .~ \forall y \in\alpha.~(x \in y \lor x = y \lor y \in x).
\]
So if $\alpha$ is an ordinal,
then $M \models\text{``$j(\alpha)$ is an ordinal''}$
in the sense above.
Therefore $j(\alpha)$ really is an ordinal.
If the claim fails,
we can pick the least $\alpha$ such that $j(\alpha) < \alpha$.
Then $M \models j(j(\alpha)) < j(\alpha)$,
i.e. $j(j(\alpha)) < j(\alpha)$
contradicting the minimality of $\alpha$.
\end{subproof}
Therefore as $j(\kappa)\neq\kappa$,
we have $j(\kappa) > \kappa$,
i.e.~$\kappa\in j(\kappa)$.
\item$U$ is an ultrafilter:
Let $X \subseteq\kappa$.
Then $\kappa\in j(\kappa)= j(X \cup(\kappa\setminus X))= j(X)\cup j(\kappa\setminus X)$.
So $X \in U$ or $\kappa\setminus X \in U$.
Let $\theta < \kappa$
and $\{X_i : i < \theta\}\subseteq U$.
Then $\kappa\in j(X_i)$ for all $i < \theta$,
hence
\[
\kappa\in\bigcap_{i < \theta} j(X_i)
= j\left( \bigcap_{i < \theta} X_i \right) \in U.
\]
This holds since $j(\theta)=\theta$ (as $\theta < \kappa$),
so $j(\langle X_i : i < \theta\rangle)=\langle j(X_i) : i < \theta\rangle$.
Also if $\xi < \kappa$,
then $j(\{\xi\})=\{\xi\}$
so $\kappa\not\in j(\{\xi\})$
and $\{\xi\}\not\in U$.
\end{itemize}
1. $\implies$ 2.
Fix $U$.
Let ${}^{\kappa}V$ be the class of all function from $\kappa$ to $V$.
For $f,g \in{}^{\kappa}V$ define
$f \sim g :\iff\{\xi < \kappa : f(\xi)= g(\xi)\}\in U$.
This is an equivalence relation since $U$ is a filter.
Write $[f]=\{g \in{}^\kappa V : g \sim ~ f \land g \in V_\alpha\text{ for the least $\alpha$ such that there is some $h \in V_\alpha$ with $h \sim f$}\}$.%
\footnote{This is know as \vocab{Scott's Trick}.
Note that by defining equivalence classes
in the usual way (i.e.~without this trick),
one ends up with proper classes:
For $f\colon\kappa\to V$,
we can for example change $f(0)$ to be an arbitrary $V_{\alpha}$