diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index 7a44b32..71ff02f 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -51,7 +51,7 @@ We will very rarely use ordinal arithmetic. \begin{definition} Let $\alpha$, $\beta$ be ordinals. - We say that $f\colon \alpha \to \beta$ is \vocab{cofinal} + We say that $f\colon \alpha \to \beta$ is \vocab{cofinal} iff for all $\xi < \beta$, there is some $\eta < \alpha$ such that $f(\eta) \ge \xi$. \end{definition} diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex new file mode 100644 index 0000000..7bd0c83 --- /dev/null +++ b/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} + diff --git a/logic2.tex b/logic2.tex index a951d71..70d8614 100644 --- a/logic2.tex +++ b/logic2.tex @@ -36,6 +36,7 @@ \input{inputs/lecture_10} \input{inputs/lecture_11} \input{inputs/lecture_12} +\input{inputs/lecture_13} \cleardoublepage