From 4fc85f84732c65b548a2e7ce69bcdee799f3412a Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Thu, 23 Nov 2023 15:38:28 +0100 Subject: [PATCH] lecture 11 --- inputs/lecture_10.tex | 40 +++++++++ inputs/lecture_11.tex | 195 ++++++++++++++++++++++++++++++++++++++++++ 2 files changed, 235 insertions(+) create mode 100644 inputs/lecture_10.tex create mode 100644 inputs/lecture_11.tex diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex new file mode 100644 index 0000000..542af7a --- /dev/null +++ b/inputs/lecture_10.tex @@ -0,0 +1,40 @@ +\lecture{10}{}{} % Mirko + +Applications of induction and recursion: +\begin{fact} + For every set $x$ there is a transitive set $t$ + such that $x \in t$. +\end{fact} +\begin{proof} + Take $R = \in $. + We want a function $F$ with domain $\omega$ + such that $F(0) = \{x\}$ + and $F(n+1) = \bigcup F(n)$. + Once we have such a function, + $\{x\} \cup \bigcup \ran(F)$ + is a set as desired. + + \todo{insert formal application of recursion theorem} +\end{proof} + +\begin{notation} + Let $\OR$ denote the class of all ordinals + and $V$ the class of all sets. +\end{notation} +\begin{lemma} + There is a function $F\colon \OR \to V$ + such that $F(\alpha) = \bigcup \{\cP(F(\beta)): \beta < \alpha\}$. +\end{lemma} +\begin{proof} + \todo{TODO} +\end{proof} +\begin{notation} + Usually, one write $V_\alpha$ for $F(\alpha)$. + They are called the \vocab{rank initial segments} of $V$. +\end{notation} +\begin{lemma} + If $x$ is any set, then there is some $\alpha \in \OR$ + such that $x \in V_\alpha$, + i.e.~$V = \bigcup \{V_{\alpha} : \alpha \in \OR\}$. +\end{lemma} + diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex new file mode 100644 index 0000000..d8b597b --- /dev/null +++ b/inputs/lecture_11.tex @@ -0,0 +1,195 @@ +\lecture{11}{2023-11-23}{Cardinals} + +\subsection{Cardinals} + +Consequence of the Mostowski collapse: + +If $<$ is a well-order on a set $a$ +then there is some transitive $b$ +with $(b, \in\defon{b}) \cong (a, <)$. + +\begin{definition} + Let $a$ be any set. + The \vocab{cardinality} of $a$ + denoted by $\overline{\overline{a}}$, $|a|$ or $\card(a)$, + is the smallest ordinal $\alpha$ + such that there is some bijection $f\colon \alpha \to a$. + + An ordinal $\alpha$ is called a \vocab{cardinal}, + iff there is some set $a$ with $|a| = \alpha$ + (equivalently, $|\alpha| = \alpha$). +\end{definition} + +We often write $\kappa, \lambda, \ldots$ for cardinals. + +\begin{lemma} + For every cardinal $\kappa$, + there is come cardinal $\lambda > \kappa$. +\end{lemma} +\begin{proof} + Consider the powerset of $\kappa$. + We know that there is no surjection $\kappa \twoheadrightarrow \cP(\kappa)$. + Hence $\kappa < |2^{\kappa}|$. +\end{proof} + +\begin{definition} + For each cardinal $\kappa$, + $\kappa^+$ denotes the least cardinal $\lambda > \kappa$. +\end{definition} +\begin{warning} + This has nothing to do with the ordinal successor of $\kappa$. +\end{warning} +\begin{lemma} + Let $X$ be any set of cardinals. + Then $\sup X$ is a cardinal. +\end{lemma} +\begin{proof} + If there is some $\kappa \in X$ with $\lambda \le \kappa$ + for all $\lambda \in X$, + then $\kappa = \sup(X)$ is a cardinal. + + Let us now assume that for all $\kappa \in X$ + there is some $\lambda \in X$ + with $\lambda > \kappa$. + Suppose that $\sup(X)$ is no a cardinal + and write $\mu = |\sup(X)|$. + Then $\mu \in \sup(X)$, + since $\sup(X)$ is an ordinal. + However $\sup(X)$ is the least ordinal + larger than all $\alpha \in X$, + so there is $\lambda \in X$ + with $\lambda > \mu$. + However, there exists $\mu \twoheadrightarrow \sup(X)$, + hence also $\mu \twoheadrightarrow \lambda$ + (which is in contradiction to $\lambda$ being a cardinal). +\end{proof} +We may now use the recursion theorem +to define a sequence $\langle \aleph_\alpha : \alpha \in \OR \rangle$ +with the following properties: +\begin{IEEEeqnarray*}{rCl} + \aleph_0 &=& \omega,\\ + \aleph_{\alpha+1} &=& (\aleph_\alpha)^+,\\ + \aleph_{\lambda} &=& \sup \{\aleph_\alpha : \alpha < \lambda\}. +\end{IEEEeqnarray*} +Each $\aleph_\alpha$ is a cardinal. +Also, a trivial induction show that $\alpha \le \aleph_\alpha$. +In particular $|\alpha| \le \aleph_{\alpha}$. +Therefore the $\aleph_\alpha$ are all the infinite cardinals: +If $a$ is any infinite set, then $|a| \le \aleph_{|a|}$, +so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$. + +\begin{notation} + Sometimes we write $\omega_\alpha$ for $\aleph_\alpha$ + (when viewing it as an ordinal). +\end{notation} + +\begin{notation} + Let ${}^a b \coloneqq \{f : f \text{ is a function}, \dom(f) = \alpha, \ran(f) \subseteq b\}$. +\end{notation} + +\begin{definition}[Cardinal arithmetic] + Let $\kappa$, $\lambda$ be cardinals. + Define + \begin{IEEEeqnarray*}{rCl} + \kappa + \lambda &\coloneqq & |\{0\} \times \kappa \cup \{1\} \times \lambda|,\\ + \kappa \cdot \lambda &\coloneqq & |\kappa \times \lambda|,\\ + \kappa^{\lambda} &\coloneqq & |{}^{\lambda}\kappa|. + \end{IEEEeqnarray*} +\end{definition} +\begin{warning} + This is very different from ordinal arithmetic! +\end{warning} + +\begin{theorem}[Hessenberg] + \label{thm:hessenberg} + For all $\alpha$ we have + \[ + \aleph_\alpha \cdot \aleph_\alpha = \aleph_\alpha. + \] + +\end{theorem} +\begin{corollary} + For all $\alpha, \beta$ it is + \[ + \aleph_\alpha + \aleph_\beta = \aleph_\alpha \cdot \aleph_\beta = \max \{\aleph_\alpha, \aleph_\beta\}. + \] +\end{corollary} +\begin{proof} + Wlog.~$\alpha \le \beta$. + Trivially $\aleph_\alpha \le \aleph_\beta$. + It is also clear that + \[ + \aleph_{\beta} \le \aleph_\alpha + \aleph_{\beta} \le \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta. + \] +\end{proof} +\begin{refproof}{thm:hessenberg} + Define a well-order $<^\ast$ on $\OR \times \OR$ by setting + \[ + (\alpha,\beta) <^\ast (\gamma,\delta) + \] + iff + \begin{itemize} + \item $\max(\alpha,\beta) < \max(\gamma,\delta)$ or + \item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha < \gamma$ or + \item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha = \gamma$ and $\beta < \delta$. + \end{itemize} + + It is clear that this is a well-order. + + There is an isomorphism + \[ + (\OR, <) \cong^{\Gamma^{-1}} (\OR \times \OR, <^\ast). + \] + $\Gamma$ is called the \vocab{Gödel pairing function}. + \begin{claim} + For all $\alpha$ it is + $\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha})) = \aleph_\alpha$, + i.e. + \[ + \aleph_\alpha = \{\xi: \exists \eta, \eta' < \aleph_\alpha.~\xi = \Gamma((\eta,\eta'))\}. + \] + \end{claim} + \begin{subproof} + We use induction of $\alpha$. + The claim is trivial for $\alpha = 0$. + Now let $\alpha > 0$ and suppose the claim + to be true for all $\beta < \alpha$. + It is easy to see that + \[ + \ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha}) \supseteq \aleph_\alpha, + \] + as otherwise $\Gamma\defon{\aleph_\alpha \times \alepha_\alpha}: \alepha_{\alpha} \times \alepha_\alpha \to \eta$ + would be a bijection for some $\eta < \alepha_\alpha$, + but $\alepha_\alpha$ is a cardinal. + + Suppose that $\ran(\Gamma\defon{\aleph_\alpha \times \alepha_\alpha}) \supsetneq \aleph_\alpha$. + Then there exist $\eta, \eta' < \aleph_\alpha$ with + \[ + \Gamma((\eta, \eta')) = \alepha_\alpha. + \] + + So $\Gamma\defon{\{(\gamma,\delta) : (\gamma,\delta) <^\ast (\eta, \eta'\}}$ + is bijective onto $\alepha_\alpha$. + If $(\gamma,\delta) <^\ast (\eta, \eta')$, + then $\max \{\gamma,\delta\} \le \max \{\eta, \eta'\}$. + Say $\eta \le \eta' < \alepha_\alpha$ + and let $\aleph_\alpha = |\eta'|$. + There is a surjection + \[f\colon \underbrace{(\eta +1)}_{ \le \aleph_\beta} \times \underbrace{(\eta' + 1)}_{\sim \aleph_\beta} \twoheadrightarrow \aleph_\alpha.\] + + This gives rise to a surjection $f^\ast \colon \aleph_\beta \times \aleph_\beta \to \aleph_\alpha$. + The inductive hypothesis then produces a surjection + $f^\ast\colon \aleph_\beta \to \aleph_\alpha \lightning$. + \end{subproof} +\end{refproof} +However, exponentiation of cardinals is far from trivial: +\begin{observe} + $2^{\kappa} = |\cP(\kappa)|$, + since ${}^{\kappa} \{0,1\} \leftrightarrow\cP(\kappa)$. + + Hence by Cantor $2^{\kappa}\ge \kappa^+$. +\end{observe} +This is basically all we can say. + +The \vocab{continuum hypothesis} states that $2^{\aleph_0} = \aleph_1$. +