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\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}

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\input{inputs/lecture_10} \input{inputs/lecture_10}
\input{inputs/lecture_11} \input{inputs/lecture_11}
\input{inputs/lecture_12} \input{inputs/lecture_12}
\input{inputs/lecture_13}
\cleardoublepage \cleardoublepage