w23-logic-2/inputs/lecture_13.tex

\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} (\emph{cl}osed \emph{un}bounded)
            iff it is closed and unbounded.
    \end{itemize}
\end{definition}
The interesting case is that $\alpha$ is a regular uncountable cardinal.
\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}