inputs/lecture_17.tex

@@ -1,240 +0,0 @@
-\lecture{17}{2023-12-18}{Large cardinals}
-
-
-\begin{definition}
-    \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 $(V_{\kappa}, \in ) \models\ZFC$.}
-\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 $V_{\alpha}$
-        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 $\cF = \{[f] : f \in {}^{\kappa}V\}$
-    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 $f_n \in {}^\kappa V$
-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 $f_1, \ldots, f_k \in {}^{\kappa} V$,
-    \[
-        (\cF, \tilde{\in}) \models \phi([f_1], \ldots, [f_k])
-        \iff \{\xi < \kappa : (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U.
-    \]
-\end{theorem}
-\begin{proof}
-    Induction on the complexity of $\phi$.
-\end{proof}

logic.sty

@@ -127,7 +127,6 @@
 
 
 \DeclareSimpleMathOperator{CH}
-\DeclareSimpleMathOperator{GCH}
 \DeclareSimpleMathOperator{DC}
 \DeclareSimpleMathOperator{Ord}
 \DeclareSimpleMathOperator{OR} % Ordinals

logic2.tex

@@ -40,7 +40,6 @@
 \input{inputs/lecture_14}
 \input{inputs/lecture_15}
 \input{inputs/lecture_16}
-\input{inputs/lecture_17}
 
 
 \cleardoublepage