13

inputs/lecture_13.tex

@@ -0,0 +1,166 @@
+\lecture{13}{2023-11-30}{}
+
+\begin{remark}[``Constructive'' approach to $\omega_1$ ]
+    There are many well-orders on $\omega$.
+    Let $W$ be the set of all such well orders.
+    For $R, S \in W$,
+    write $R \le S$ if $R$ is isomorphic to
+    an initial segment of $S$.
+    Consider $\faktor{W}{\sim}$,
+    where $R \sim S :\iff R \le S \land S \le R$.
+    Define $\le $ on $\faktor{W}{\sim }$
+    by $[R] \le [S] :\iff R \le S$.
+    Clearly this is well-defined
+    and $<$ is a well-order on $\faktor{W}{\sim}$:
+    Suppose that $\{R_n : n \in \omega\} \subseteq W$
+    is such that $R_{n+1} < R_n$.
+    Then there exist $n_i \in \omega$
+    such that $R_i \cong R_0\defon{x <_{R_0} n_i}$
+    and these form a $<_{R_0}$ strictly decreasing sequence.
+
+    So $(\faktor{W}{\sim})$
+    is a well-ordered set.
+    Every well-order on a countable set is isomorphic
+    to  $(\omega, R)$ for some  $[R] \in \faktor{W}{\sim}$.
+
+    Moreover if $R \in  W$,
+    then
+    \[
+        (\omega; R) \cong (\underbrace{\{[S] \in \faktor{W}{\sim} : [S] < [R]\}}_{I}; <\defon{I}),
+    \] 
+    where the isomorphism is given by
+    \[
+    n \longmapsto [R \defon{\{ m : (m,n) \in R\} }].
+    \] 
+
+    This also shows that every $[R] \in  \faktor{W}{\sim }$
+    has only countably many $<$-predecessors.
+    This then also shows that $(\faktor{W}{\sim}, <)$ itself
+    is not a well-order on a countable set.
+
+    Thus $\otp(\faktor{W}{\sim}, <) = \omega_1$.
+    \todo{move this}
+\end{remark}
+
+\begin{notation}
+    Let $I \neq \emptyset$
+    and let $\{\kappa_i : i \in I\}$
+    be a set of cardinals.
+
+    Then
+    \[
+    \sum_{i \in I} \kappa_i \coloneqq \left|\bigcup_{ i \in I} (\kappa_i \times \{i\})\right|
+    \] 
+    and
+    \[
+    \prod_{i \in I} \coloneqq \left| \bigtimes_{i \in I} \kappa_i\right|,
+    \] 
+    where 
+    \[
+    \bigtimes_{i \in I} A_i \coloneqq \{f : f \text{is a function with domain  $I$ and $f(i) \in A_i$}\}.
+    \] 
+\end{notation}
+\begin{remark}
+    \AxC is equivalent to $\forall i \in I.~A_i \neq \emptyset \implies \bigtimes_{i \in I} A_i \neq \emptyset$.
+\end{remark}
+
+\begin{theorem}[K\H{o}nig]
+    \yalabel{K\H{o}nig's Theorem}{K\H{o}nig}{thm:koenig}
+    Let $I \neq \emptyset$.
+    Let $\{\kappa_i : i \in I\}$,
+    $\{\lambda_i : i \in I\}$
+    be sets of cardinals
+    such that $\kappa_i < \lambda_i$ for all $i \in I$.
+
+    Then
+    \[
+    \sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i.
+    \] 
+\end{theorem}
+\begin{proof}
+    Consider a function $F\colon \bigcup_{i \in I} (\kappa_i \times \{i\}) \to \bigtimes_{i \in I}\lambda_i$.
+    We want to show that $F$ is not surjective.
+
+    For $i \in I$, let $\xi_i$ be the least $\xi < \lambda_i$
+    such that for all $\eta < \kappa_i$
+    \[
+    \underbrace{F((\eta, i))(i)}_{\in \lambda_i} \neq \xi.
+    \] 
+    Such $\xi$ exists, since $\kappa_i < \lambda_i$.
+
+    Let $f \in \bigtimes_{i \in I} \lambda_i$
+    be defined by $i \mapsto \xi_i$.
+
+    Then $f \not\in \ran(F)$.
+\end{proof}
+
+\begin{corollary}
+    For infinite cardinals $\kappa$,
+    it is $\cf(2^{\kappa}) > \kappa$.
+\end{corollary}
+\begin{proof}
+    If $2^{\kappa}$ is a successor cardinal,
+    then $\cf(2^{\kappa}) = 2^{\kappa} > \kappa$,
+    since successor cardinals are regular.
+    
+    Suppose $\cf(2^{k}) \le  \kappa$ is a limit cardinal.
+    Then there is some cofinal $f\colon  \kappa\to 2^{\kappa}$.
+    Write $\kappa_i = f(i)$
+    (replacing $f(i)$ by $|f(i)|^{+}$ we may assume
+    that every $\kappa_i$ is a cardinal).
+
+    For $i \in \kappa$, write $\lambda_i = 2^{k}$.
+    By \yaref{thm:koenig},
+    \[
+    \sup \{\kappa_i : i  < \kappa\} \le \sum_{i \in \kappa} \kappa_i < \prod_{i \in \kappa} \lambda_i = \left( 2^{\kappa} \right)^{\kappa} = 2^{\kappa \cdot \kappa} = 2^{\kappa}
+    \]
+    and $f$ is not cofinal.
+\end{proof}
+
+\begin{fact}
+    Properties of the function $\kappa \mapsto 2^{\kappa}$.
+    \begin{itemize}
+        \item $\mu < \kappa \implies 2^{\mu} \le 2^{\kappa}$
+            (it is independent of $\ZFC$ whether or not this is strictly increasing).
+        \item $\cf(2^{\kappa}) \ge  \kappa^+$.
+    \end{itemize}
+    This is ``all'' you can prove in $\ZFC$.
+\end{fact}
+
+The next goal is to show the following:
+(However the method might be more interesting than the result)
+\begin{theorem}[Silver]
+    If $2^{\aleph_\alpha} = \aleph_{\alpha + 1}$
+    for all $\alpha < \omega_1$,
+    then $2^{\aleph_{\omega_1}} = \aleph_{\omega_1 + 1}$.
+\end{theorem}
+
+
+Relevant concepts to prove this theorem:
+
+\begin{definition}
+    Let $\alpha$ be a limit ordinal.
+    \begin{itemize}
+        \item We say that $A \subseteq \alpha$
+            is \vocab{unbounded} (in $\alpha$),
+            iff for all $\beta < \alpha$,
+            there is some $\gamma \in \alpha$
+            such that $\beta < \gamma$.
+        \item We say that $A \subseteq \alpha$ 
+            is \vocab{closed},
+            iff it is closed with respect to the order topology on $\alpha$,
+            i.e.~for all $\beta < \alpha$,
+            \[
+            \sup(A \cap \beta) \in A.
+            \] 
+        \item $A$ is \vocab{club} (closed unbounded)
+            iff it is closed and unbounded.
+    \end{itemize}
+\end{definition}
+\begin{fact}
+    $A \subseteq \alpha$ being unbounded
+    is equivalent to $f\colon \beta\to \alpha$ being cofinal,
+    where
+    $(\beta, \in ) \overset{f}{\cong} (A, \in )$.
+\end{fact}
+

logic2.tex

@@ -36,6 +36,7 @@
 \input{inputs/lecture_10}
 \input{inputs/lecture_11}
 \input{inputs/lecture_12}
+\input{inputs/lecture_13}
 
 
 \cleardoublepage