w23-logic-2/inputs/lecture_18.tex

\lecture{18}{2023-12-18}{Large cardinals}

\begin{definition}
    \label{def:inaccessible}
    \begin{itemize}
        \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 $(V_{\kappa}, \in\defon{V_\kappa}) \models\ZFC$.}
\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}
                $\alpha \in \OR$
                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 $\leftindex^{\kappa}V$ be the class of all function from $\kappa$ to $V$.
    For $f,g \in \leftindex^{\kappa}V$ 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 \leftindex^\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 $V_{\alpha}$
        and get another element of $[f]$.
    }
    For any two such equivalence classes $[f], [g]$
    define
    \[[f] \tilde{\in} [g] :\iff \{\xi < \kappa : f(\xi) \in  g(\xi)\} \in U.\]

    This is independent of the choice of the representatives,
    so it is well-defined.
    Now write $\cF = \{[f] : f \in \leftindex^{\kappa}V\}$
    and look at $(\cF, \tilde{\in})$.

    The key to the construction is \yaref{thm:los} (see below).
    Given \yaref{thm:los},
    we may define an elementary embedding
    $\overline{j}\colon (V, \in ) \to (\cF, \tilde{\in })$
    as follows:

Let $\overline{j}(x) = [c_x]$,
where $c_x \colon \kappa \to  \{x\}$ is the constant function
with value $x$.

Then
\begin{IEEEeqnarray*}{rCl}
    (V, \in ) \models \phi(x_1,\ldots,x_k)&\iff&
    \{\xi < \kappa: (V, \in ) \models \phi(c_{x_1}(\alpha), \ldots, c_{x_k}(\alpha))\} \in U\\
        &\overset{\yaref{thm:los}}{\iff}&
            (\cF, \tilde{\in }) \models \phi(\overline{j}(x_1), \ldots, \overline{j}(x_k)).
\end{IEEEeqnarray*}

Let us show that $(\cF, \tilde{\in })$
is well-founded.
Otherwise there is $\langle f_n : n < \omega \rangle$
such that $f_n \in \leftindex^\kappa V$
and $[f_{n+1}] \tilde{\in } [f_n]$ for all $n < \omega$.

Then $X_n \coloneqq \{\xi < \kappa : f_{n+1}(\xi) \in  f_n(\xi)\} \in U$,
so $\bigcap X_n \in U$.
Let $\xi_0 \in \bigcap X_n$.
Then $f_0(\xi_0) \ni f_1(\xi_0) \ni f_2(\xi_0) \ni \ldots$ $\lightning$.

Note that $\tilde{\in }$ is set-like,
therefore by the \yaref{lem:mostowski}
there is some transitive $M$ with $(\cF, \tilde{\in }) \overset{\sigma}{\cong} (M, \in )$.

We can now define an elementary embedding $j\colon  V \to  M$
by $j \coloneqq  \sigma \circ \overline{j}$.

It remains to show that $\alpha < \kappa \implies j(\alpha) = \alpha$.
This can be done by induction:
Fix $\alpha$. We already know $j(\alpha) \ge \alpha$.
Suppose $\beta \in j(\alpha)$.
Then $\beta = \sigma([f])$ for some $f$
and $\sigma([f]) \in \sigma([c_{\alpha}])$,
i.e.~$[f] \tilde{\in} [c_\alpha]$.
Thus $\{\xi < \kappa : f(\xi) \in \underbrace{c_{\alpha}(\xi)}_{\alpha}\} \in U$.
Hence there is some $\delta < \alpha$
such that
\[
    X_\delta \coloneqq  \{\xi < \kappa : f(\xi) = \delta\} \in U,
\]
as otherwise $\forall \delta < \alpha. ~  \kappa \setminus X_\delta \in U$,
i.e.~$\emptyset = (\bigcap_{\delta < \alpha} \kappa \setminus X_{\delta}) \cap X \in U \lightning$.
We get $[f] = [c_{\delta}]$,
so $\beta = \sigma([f]) = \delta([c_{\delta}]) = j(\delta) = \delta$,
where for the last equality we have applied the induction hypothesis.
So $j(\alpha) \le \alpha$.

% It is also easy to show $j(\kappa) > \kappa$.
\end{proof}

\begin{theorem}[\L o\'s]
    \yalabel{\L o\'s's Theorem}{\L o\'s}{thm:los}
    For all formulae $\phi$ and for all $f_1, \ldots, f_k \in \leftindex^{\kappa} V$,
    \[
        (\cF, \tilde{\in}) \models \phi([f_1], \ldots, [f_k])
        \iff \{\xi < \kappa : (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U.
    \]
\end{theorem}
\begin{proof}
    Induction on the complexity of $\phi$.
\end{proof}