lecture 11

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}
+

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 $2^{\aleph_0} = \aleph_1$.
+