inputs/lecture_10.tex

@@ -1,40 +0,0 @@
-\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

@@ -1,195 +0,0 @@
-\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 $2^{\aleph_0} = \aleph_1$.
-

logic.sty

@@ -129,7 +129,6 @@
 \DeclareSimpleMathOperator{CH}
 \DeclareSimpleMathOperator{DC}
 \DeclareSimpleMathOperator{Ord}
-\DeclareSimpleMathOperator{OR} % Ordinals
 \DeclareSimpleMathOperator{trcl}
 \DeclareSimpleMathOperator{tcl}
 \newcommand{\concat}{{}^\frown}

logic2.tex

@@ -33,8 +33,6 @@
 \input{inputs/lecture_07}
 \input{inputs/lecture_08}
 \input{inputs/lecture_09}
-% TODO \input{inputs/lecture_10}
-\input{inputs/lecture_11}
 
 
 \cleardoublepage