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\lecture { 18} { 2023-12-18} { Large cardinals}
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\begin { definition}
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\label { def:inaccessible}
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\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,
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then $ V _ \kappa \models \ZFC $ .\footnote { More formally $ ( V _ { \kappa } , \in \defon { V _ \kappa } ) \models \ZFC $ .}
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\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}
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$ \alpha \in \OR $
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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 $ .
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Let $ \leftindex ^ { \kappa } V $ be the class of all function from $ \kappa $ to $ V $ .
For $ f,g \in \leftindex ^ { \kappa } V $ define
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$ f \sim g : \iff \{ \xi < \kappa : f ( \xi ) = g ( \xi ) \} \in U $ .
This is an equivalence relation since $ U $ is a filter.
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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$ } \} $ .%
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\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.
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Now write $ \cF = \{ [ f ] : f \in \leftindex ^ { \kappa } V \} $
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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 $
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such that $ f _ n \in \leftindex ^ \kappa V $
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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}
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For all formulae $ \phi $ and for all $ f _ 1 , \ldots , f _ k \in \leftindex ^ { \kappa } V $ ,
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\[
(\cF , \tilde { \in } ) \models \phi ([f_ 1], \ldots , [f_ k])
\iff \{ \xi < \kappa : (V, \in ) \models \phi (f_ 1(\xi ), \ldots , f_ k(\xi ))\} \in U.
\]
\end { theorem}
\begin { proof}
Induction on the complexity of $ \phi $ .
\end { proof}