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193 lines
6.4 KiB
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193 lines
6.4 KiB
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\lecture{13}{2023-11-30}{}
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\gist{%
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\begin{remark}[``Constructive'' approach to $\omega_1$ ]
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There are many well-orders on $\omega$.
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Let $W$ be the set of all such well-orders.
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For $R, S \in W$,
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write $R \le S$ iff $R$ is isomorphic to
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an initial segment of $S$.
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Consider $\faktor{W}{\sim}$,
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where $R \sim S :\iff R \le S \land S \le R$.
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Define $\le $ on $\faktor{W}{\sim }$
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by $[R] \le [S] :\iff R \le S$.
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Clearly this is well-defined
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and $<$ is a well-order on $\faktor{W}{\sim}$:
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Suppose that $\{R_n : n \in \omega\} \subseteq W$
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is such that $R_{n+1} < R_n$.
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Then there exist $n_i \in \omega$
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such that $R_i \cong R_0\defon{\{x : x <_{R_0} n_i\}}$
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and these form a $<_{R_0}$ strictly decreasing sequence.
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So $(\faktor{W}{\sim})$
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is a well-ordered set.
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Every well-order on a countable set is isomorphic
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to $(\omega, R)$ for some $[R] \in \faktor{W}{\sim}$.
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Moreover if $R \in W$,
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then
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\[
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(\omega; R) \cong (\underbrace{\{[S] \in \faktor{W}{\sim} : [S] < [R]\}}_{I}; <\defon{I}),
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\]
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where the isomorphism is given by
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\[
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n \longmapsto [R \defon{\{ m : (m,n) \in R\} }].
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\]
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This also shows that every $[R] \in \faktor{W}{\sim }$
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has only countably many $<$-predecessors.
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This then also shows that $(\faktor{W}{\sim}, <)$ itself
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is not a well-order on a countable set.
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Thus $\otp(\faktor{W}{\sim}, <) = \omega_1$.
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\end{remark}
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}{}
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\gist{%
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\begin{notation}
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Let $I \neq \emptyset$
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and let $\{\kappa_i : i \in I\}$
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be a set of cardinals.
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Then
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\[
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\sum_{i \in I} \kappa_i \coloneqq \left|\bigcup_{ i \in I} (\kappa_i \times \{i\})\right|
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\]
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and
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\[
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\prod_{i \in I} \coloneqq \left| \bigtimes_{i \in I} \kappa_i\right|,
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\]
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where
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\[
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\bigtimes_{i \in I} A_i \coloneqq \{f : f \text{ is a function}, \dom(f) = I, \forall i.~f(i) \in A_i\}.
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\]
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\end{notation}
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\begin{remark}
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\AxC is equivalent to $\forall i \in I.~A_i \neq \emptyset \implies \bigtimes_{i \in I} A_i \neq \emptyset$.
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\end{remark}
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}{}
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\begin{theorem}[K\H{o}nig]
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\yalabel{K\H{o}nig's Theorem}{K\H{o}nig}{thm:koenig}
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Let $I \neq \emptyset$.
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Let $\{\kappa_i : i \in I\}$,
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$\{\lambda_i : i \in I\}$
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be sets of cardinals
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such that $\kappa_i < \lambda_i$ for all $i \in I$.
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Then
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\[
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\sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i.
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\]
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\end{theorem}
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\begin{proof}
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\gist{%
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Consider a function $F\colon \bigcup_{i \in I} (\kappa_i \times \{i\}) \to \bigtimes_{i \in I}\lambda_i$.
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We want to show that $F$ is not surjective.
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For $i \in I$, let $\xi_i$ be the least $\xi < \lambda_i$
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such that for all $\eta < \kappa_i$
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\[
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\underbrace{F((\eta, i))(i)}_{\in \lambda_i} \neq \xi.
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\]
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Such $\xi$ exists, since $\kappa_i < \lambda_i$.
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Let $f \in \bigtimes_{i \in I} \lambda_i$
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be defined by $i \mapsto \xi_i$.
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Then $f \not\in \ran(F)$.
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}{%
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\begin{itemize}
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\item Consider $F\colon \bigcup_{i \in I}(\kappa_i \times \{i\}) \to \bigtimes_{i \in I} \lambda_i$. Want: $F$ is not surjective.
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\item Let $\xi_i$ minimal such that $\underbrace{F((\eta,i))(i)}_{\in \lambda_i} \neq \xi$
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for all $\eta < \kappa_i$.
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\item $(i \mapsto \xi_i) \not\in \ran(F)$.
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\end{itemize}
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}
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\end{proof}
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\begin{corollary}
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\label{cor:cfpow}
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For infinite cardinals $\kappa$,
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it is $\cf(2^{\kappa}) > \kappa$.
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\end{corollary}
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\begin{proof}
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\gist{%
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If $2^{\kappa}$ is a successor cardinal,
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then $\cf(2^{\kappa}) = 2^{\kappa} > \kappa$,
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since successor cardinals are regular.
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Suppose $\cf(2^{\kappa}) \le \kappa$ is a limit cardinal.
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Then there is some cofinal $f\colon \kappa\to 2^{\kappa}$.
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Write $\kappa_i = f(i)$
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(replacing $f(i)$ by $|f(i)|^{+}$ we may assume
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that every $\kappa_i$ is a cardinal).
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For $i \in \kappa$, write $\lambda_i = 2^{\kappa}$.
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By \yaref{thm:koenig},
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\[
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\sup \{\kappa_i : i < \kappa\} \le \sum_{i \in \kappa} \kappa_i < \prod_{i \in \kappa} \lambda_i = \left( 2^{\kappa} \right)^{\kappa} = 2^{\kappa \cdot \kappa} = 2^{\kappa}
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\]
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and $f$ is not cofinal.
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}{%
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\begin{itemize}
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\item Trivial for successors ($\cf(2^{\kappa}) \overset{\text{regular}}{=} 2^{\kappa} > \kappa$).
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\item Suppose $2^\kappa$ is a limit, $\exists f\colon \kappa \to 2^{\kappa}$ cofinal.
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Wlog.~$f(i) \in \Card$.
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But
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\[
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\sup \{f(i)) : i < \kappa\} \le \sum_{i \in \kappa} f(i)
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\overset{\yaref{thm:koenig}}{<} \prod_{i \in \kappa} 2^{\kappa}
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= (2^{\kappa})^{\kappa} = 2^{\kappa}. ~ ~\qedhere
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\]
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\end{itemize}
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}
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\end{proof}
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\begin{fact}
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Properties of the function $\kappa \mapsto 2^{\kappa}$.
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\begin{itemize}
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\item $\mu < \kappa \implies 2^{\mu} \le 2^{\kappa}$
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(it is independent of $\ZFC$ whether or not this is strictly increasing).
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\item $\cf(2^{\kappa}) \ge \kappa^+$.
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\end{itemize}
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This is ``all'' you can prove in $\ZFC$.
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\end{fact}
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The next goal is to show the following:
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(However the method might be more interesting than the result)
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\begin{theorem}[Silver]
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\yalabel{Silver's Theorem (case of $\aleph_{\omega_1}$)}{Silver ($\aleph_{\omega_{1}}$)}{thm:silver1}
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If $2^{\aleph_\alpha} = \aleph_{\alpha + 1}$
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for all $\alpha < \omega_1$,
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then $2^{\aleph_{\omega_1}} = \aleph_{\omega_1 + 1}$.
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\end{theorem}
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Relevant concepts to prove this theorem:
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\begin{definition}
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Let $\alpha$ be a limit ordinal.
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\begin{itemize}
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\item We say that $A \subseteq \alpha$
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is \vocab{unbounded} (in $\alpha$),
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iff for all $\beta < \alpha$,
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there is some $\gamma \in A$
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such that $\beta < \gamma$.
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\item We say that $A \subseteq \alpha$
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is \vocab{closed},
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iff it is closed with respect to the order topology on $\alpha$,
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i.e.~for all $\beta < \alpha$,
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\[
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\sup(A \cap \beta) \in A \cup \{0\} .
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\]
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\item $A$ is \vocab{club} (\emph{cl}osed \emph{un}bounded)
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iff it is closed and unbounded.
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\end{itemize}
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\end{definition}
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The interesting case is that $\alpha$ is a regular uncountable cardinal.
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\begin{fact}
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$A \subseteq \alpha$ being unbounded
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is equivalent to $f\colon \beta\to \alpha$ being cofinal,
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where
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$(\beta, \in ) \overset{f}{\cong} (A, \in )$.
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\end{fact}
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