241 lines
9.1 KiB
TeX
241 lines
9.1 KiB
TeX
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\lecture{17}{2023-12-18}{Large cardinals}
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\begin{definition}
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\begin{itemize}
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\item A cardinal $\kappa$ is called \vocab{weakly inaccessible}
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iff $\kappa$ is uncountable,\footnote{dropping this we would get that $\aleph_0$ is inaccessible}
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regular and $\forall \lambda < \kappa.~\lambda^+ < \kappa$.
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\item A cardinal $\kappa$ is \vocab[inaccessible!strongly]{(strongly) inaccessible}
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iff $\kappa$ is uncountable, regular
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and $\forall \lambda < \kappa.~2^{\lambda} < \kappa$.
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\end{itemize}
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\end{definition}
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\begin{remark}
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Since $2^{\lambda} \ge \lambda^+$,
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strongly inaccessible cardinals are weakly inaccessible.
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If $\GCH$ holds, the notions coincide.
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\end{remark}
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\begin{theorem}
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If $\kappa$ is inaccessible,
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then $V_\kappa \models \ZFC$.\footnote{More formally $(V_{\kappa}, \in ) \models\ZFC$.}
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\end{theorem}
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\begin{proof}
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Since $\kappa$ is regular, \AxRep works.
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Since $2^{\lambda} < \kappa$,
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\AxPow works.
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The other axioms are trivial.
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\todo{Exercise}
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\end{proof}
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\begin{corollary}
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$\ZFC$ does not prove the existence of inaccessible
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cardinals, unless $\ZFC$ is inconsistent.
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\end{corollary}
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\begin{proof}
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If $\ZFC$ is consistent,
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it can not prove that it is consistent.
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In particular, it can not prove the existence of a model of $\ZFC$.
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\end{proof}
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\begin{definition}[Ulam]
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A cardinal $\kappa > \aleph_0$ is \vocab{measurable}
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iff there is an ultrafilter $U$ on $\kappa$,
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such that $U$ is not principal\footnote{%
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i.e.~$\{\xi\} \not\in U$ for all $\xi < \kappa$%
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}
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and
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if $\theta < \kappa$
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and $\{X_i : i < \theta\} \subseteq U$,
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then $\bigcap_{i < \theta} X_i \in U$
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\end{definition}
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\begin{goal}
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We want to prove
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that if $\kappa$ is measurable,
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then $\kappa$ is inaccessible
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and there are $\kappa$ many
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inaccessible cardinals below $\kappa$
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(i.e.~$\kappa$ is the $\kappa$\textsuperscript{th} inaccessible).
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\end{goal}
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\begin{theorem}
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The following are equivalent:
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\begin{enumerate}
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\item $\kappa$ is a measurable cardinal.
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\item There is an elementary embedding%
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\footnote{Recall: $j\colon V \to M$
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is an \vocab{elementary embedding} iff $j''V = \{j(x) : x \in V\} \prec M$,
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i.e.~for all formulae $\phi$ and $x_1,\ldots,x_k \in V$,
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$V \models\phi(x_1,\ldots,x_u) \iff M \models\phi(j(x_1),\ldots,j(x_u))$.%
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}
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$j\colon V \to M$ with $M$
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transitive
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such that $j\defon{\kappa} = \id$,
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$j(\kappa) \neq \kappa$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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2. $\implies$ 1.:
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Fox $j\colon V \to M$.
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Let $U = \{X \subseteq \kappa : \kappa \in j(X)\}$.
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We need to show that $U$ is an ultrafilter:
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\begin{itemize}
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\item Let $X,Y \in U$.
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Then $\kappa \in j(X) \cap j(Y)$.
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We have $M \models j(X \cap Y) = j(X) \cap j(Y)$,
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and thus $j(X \cap Y) = j(X) \cap j(Y)$.
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It follows that $X \cap Y \in U$.
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\item Let $X\in U$
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and $X \subseteq Y \subseteq \kappa$.
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Then $ \kappa \in j(X) \subseteq j(Y)$
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by the same argument,
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so $Y \in U$.
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\item We have $j(\emptyset) = \emptyset$
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(again $M \models j(\emptyset)$ is empty),
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hence $\emptyset\not\in U$.
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\item $\kappa \in U$ follows from $\kappa \in j(\kappa)$.
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This is shown as follows:
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\begin{claim}
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For every ordinal $\alpha$,
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$j(\alpha)$ is an ordinal such that $j(\alpha) \ge \alpha$.
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\end{claim}
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\begin{subproof}
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$\alpha \in \Ord$
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can be written as
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\[\forall x \in \alpha .~\forall y \in x.~y \in \alpha
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\land \forall x \in \alpha .~ \forall y \in \alpha.~(x \in y \lor x = y \lor y \in x).
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\]
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So if $\alpha$ is an ordinal,
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then $M \models \text{``$j(\alpha)$ is an ordinal''}$
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in the sense above.
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Therefore $j(\alpha)$ really is an ordinal.
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If the claim fails,
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we can pick the least $\alpha$ such that $j(\alpha) < \alpha$.
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Then $M \models j(j(\alpha)) < j(\alpha)$,
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i.e. $j(j(\alpha)) < j(\alpha)$
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contradicting the minimality of $\alpha$.
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\end{subproof}
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Therefore as $j(\kappa) \neq \kappa$,
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we have $j(\kappa) > \kappa$,
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i.e.~$\kappa \in j(\kappa)$.
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\item $U$ is an ultrafilter:
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Let $X \subseteq \kappa$.
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Then $\kappa \in j(\kappa) = j(X \cup (\kappa \setminus X)) = j(X) \cup j(\kappa \setminus X)$.
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So $X \in U$ or $\kappa \setminus X \in U$.
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Let $\theta < \kappa$
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and $\{X_i : i < \theta\} \subseteq U$.
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Then $\kappa \in j(X_i)$ for all $i < \theta$,
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hence
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\[
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\kappa \in \bigcap_{i < \theta} j(X_i)
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= j\left( \bigcap_{i < \theta} X_i \right) \in U.
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\]
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This holds since $j(\theta) = \theta$ (as $\theta < \kappa$),
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so $j(\langle X_i : i < \theta \rangle) = \langle j(X_i) : i < \theta \rangle$.
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Also if $\xi < \kappa$,
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then $j(\{\xi\}) = \{\xi\}$
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so $\kappa \not\in j(\{\xi\})$
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and $\{\xi\} \not\in U$.
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\end{itemize}
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1. $\implies$ 2.
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Fix $U$.
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Let ${}^{\kappa}V$ be the class of all function from $\kappa$ to $V$.
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For $f,g \in {}^{\kappa}V$ define
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$f \sim g :\iff \{\xi < \kappa : f(\xi) = g(\xi)\} \in U$.
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This is an equivalence relation since $U$ is a filter.
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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 $h \sim f$}\}$.%
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\footnote{This is know as \vocab{Scott's Trick}.
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Note that by defining equivalence classes
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in the usual way (i.e.~without this trick),
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one ends up with proper classes:
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For $f\colon \kappa \to V$,
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we can for example change $f(0)$ to be an arbitrary $V_{\alpha}$
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and get another element of $[f]$.
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}
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For any two such equivalence classes $[f], [g]$
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define
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\[[f] \tilde{\in} [g] :\iff \{\xi < \kappa : f(\xi) \in g(\xi)\} \in U.\]
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This is independent of the choice of the representatives,
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so it is well-defined.
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Now write $\cF = \{[f] : f \in {}^{\kappa}V\}$
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and look at $(\cF, \tilde{\in})$.
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The key to the construction is \yaref{thm:los} (see below).
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Given \yaref{thm:los},
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we may define an elementary embedding
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$\overline{j}\colon (V, \in ) \to (\cF, \tilde{\in })$
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as follows:
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Let $\overline{j}(x) = [c_x]$,
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where $c_x \colon \kappa \to \{x\}$ is the constant function
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with value $x$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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(V, \in ) \models \phi(x_1,\ldots,x_k)&\iff&
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\{\xi < \kappa: (V, \in ) \models \phi(c_{x_1}(\alpha), \ldots, c_{x_k}(\alpha))\} \in U\\
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&\overset{\yaref{thm:los}}{\iff}&
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(\cF, \tilde{\in }) \models \phi(\overline{j}(x_1), \ldots, \overline{j}(x_k)).
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\end{IEEEeqnarray*}
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Let us show that $(\cF, \tilde{\in })$
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is well-founded.
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Otherwise there is $\langle f_n : n < \omega \rangle$
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such that $f_n \in {}^\kappa V$
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and $[f_{n+1}] \tilde{\in } [f_n]$ for all $n < \omega$.
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Then $X_n \coloneqq \{\xi < \kappa : f_{n+1}(\xi) \in f_n(\xi)\} \in U$,
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so $\bigcap X_n \in U$.
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Let $\xi_0 \in \bigcap X_n$.
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Then $f_0(\xi_0) \ni f_1(\xi_0) \ni f_2(\xi_0) \ni \ldots$ $\lightning$.
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Note that $\tilde{\in }$ is set-like,
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therefore by the \yaref{lem:mostowski}
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there is some transitive $M$ with $(\cF, \tilde{\in }) \overset{\sigma}{\cong} (M, \in )$.
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We can now define an elementary embedding $j\colon V \to M$
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by $j \coloneqq \sigma \circ \overline{j}$.
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It remains to show that $\alpha < \kappa \implies j(\alpha) = \alpha$.
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This can be done by induction:
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Fix $\alpha$. We already know $j(\alpha) \ge \alpha$.
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Suppose $\beta \in j(\alpha)$.
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Then $\beta = \sigma([f])$ for some $f$
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and $\sigma([f]) \in \sigma([c_{\alpha}])$,
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i.e.~$[f] \tilde{\in} [c_\alpha]$.
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Thus $\{\xi < \kappa : f(\xi) \in \underbrace{c_{\alpha}(\xi)}_{\alpha}\} \in U$.
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Hence there is some $\delta < \alpha$
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such that
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\[
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X_\delta \coloneqq \{\xi < \kappa : f(\xi) = \delta\} \in U,
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\]
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as otherwise $\forall \delta < \alpha. ~ \kappa \setminus X_\delta \in U$,
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i.e.~$\emptyset = (\bigcap_{\delta < \alpha} \kappa \setminus X_{\delta}) \cap X \in U \lightning$.
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We get $[f] = [c_{\delta}]$,
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so $\beta = \sigma([f]) = \delta([c_{\delta}]) = j(\delta) = \delta$,
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where for the last equality we have applied the induction hypothesis.
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So $j(\alpha) \le \alpha$.
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% It is also easy to show $j(\kappa) > \kappa$.
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\end{proof}
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\begin{theorem}[\L o\'s]
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\yalabel{\L o\'s's Theorem}{\L o\'s}{thm:los}
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For all formulae $\phi$ and for all $f_1, \ldots, f_k \in {}^{\kappa} V$,
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\[
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(\cF, \tilde{\in}) \models \phi([f_1], \ldots, [f_k])
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\iff \{\xi < \kappa : (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U.
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\]
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\end{theorem}
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\begin{proof}
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Induction on the complexity of $\phi$.
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\end{proof}
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