From aa7f3de504fd8213d137225766822d635619247f Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 19 Dec 2023 13:37:02 +0100 Subject: [PATCH] lecture 10 --- inputs/tutorial_10.tex | 201 +++++++++++++++++++++++++++++++++++++++++ logic3.tex | 1 + 2 files changed, 202 insertions(+) create mode 100644 inputs/tutorial_10.tex diff --git a/inputs/tutorial_10.tex b/inputs/tutorial_10.tex new file mode 100644 index 0000000..92158d7 --- /dev/null +++ b/inputs/tutorial_10.tex @@ -0,0 +1,201 @@ +\tutorial{10}{2023-12-19}{Sheet 9} +\subsection{Sheet 9} + +\nr 1 +$(X, \tau') \xrightarrow{x \mapsto x} (X, \tau)$ is Borel +(by one of the equivalent definitions of being Borel). +Thus $\cB(X, \tau) \subseteq \cB(X, \tau')$ +(by the other equivalent definition of being Borel). +Let $U \subseteq (X, \tau')$ be Borel. +$\id\defon{U}$ is injective, +hence $U$ is Borel in $(X, \tau)$ by Lusin-Suslin. + + + +\paragraph{Related stuff} + +\begin{fact} + Let $X,Y$ be Polish. + $f\colon X \to Y$ + is Borel iff its graph $\Gamma_f$ is Borel. +\end{fact} +\begin{proof} + Take a countable open base + $V_0, V_1, \ldots$ of $Y$. + Then $\Gamma_f = \{(x,y) : \forall n < \omega.~f(x) \in V_n \implies y \in V_n\}$ + (because the space is Hausdorff). + If $f$ is Borel, + then clearly the RHS is Borel + since + \begin{IEEEeqnarray*}{rCl} + &&\{(x,y) : \forall n < \omega.~f(x) \in V_n \implies y \in V_n\}\\ + &=& \bigcap_{n < \omega} (f^{-1}(V_n)^{c}Y \cup f^{-1}(V_n) \times V_n\}\\ + \end{IEEEeqnarray*} + + On the other hand suppose that $\Gamma_f$ is Borel. + Then + \[ + f^{-1}(B) = \pi_X(X \times B \cap \Gamma_f) + \] + is analytic.\footnote{Note that the projection of a Borel set is not necessarily Borel. + Moreover note that we only used that $\Gamma_f$ is analytic.} + On the other hand + \[ + f^{-1}(B)^c = f^{-1}(B^c) + \] + is analytic + and we know that $\Sigma_1^1 \cap \Pi_1^1 = \cB$ + by the \yaref{cor:lusinseparation}. +\end{proof} +In fact we have shown +\begin{fact} + The following are equivalent + \begin{itemize} + \item $f$ is Borel, + \item $\Gamma_f$ is Borel, + \item $\Gamma_f$ is analytic. + \end{itemize} +\end{fact} + + + +\nr 2 + +\begin{definition} + Let $X$ be a topological space. + Let $K(X)$ be the set of all compact subspaces of $X$. + The \vocab{Vietoris Topology}, $\tau_V$, on $K(X)$ + is the topology with basic open sets + \[ + [U_0; U_1, \ldots, U_n] = \{K \in K(X) : K \subseteq U_0 \land \forall 1 \le i \le n .~K \cap U_i \neq \emptyset\} + \] + for $U_i \overset{\text{open}}{\subseteq} X$. +\end{definition} +\begin{definition} + Let $(X,d)$ be a matric space with $d \le 1$. + We define a metric $d_H$ on $K(X)$ as follows: + $d_H(\emptyset, \emptyset) \coloneqq 0$, $d_H(K, \emptyset) \coloneqq 1$ for $K \neq \emptyset$ + and + \[ + d_H(K_0, K_1) \coloneqq \max \{\max_{x \in K_0} d(x,K_1), \max_{x \in K_1} d(x,K_0)\} + \] + for $K_0, K_1 \neq \emptyset$. +\end{definition} +\begin{fact} + $d_H$ is indeed a metric. +\end{fact} +\begin{proof} + Let $\delta(K, L) \coloneqq \max_{x \in K} d(x,L)$. + It suffices to show + $\delta(X,Z) \le \delta(X,Y) + \delta(Y,Z)$, + since then + \begin{IEEEeqnarray*}{rCl} + d_H(X,Z) &\le & \max \{\delta(X,Y) + \delta(Y,Z), \delta(Z,Y) + \delta(Y,X)\}\\ + &\le & d_H(X,Y) + d_H(Y,Z). + \end{IEEEeqnarray*} + + Using the fact that $d(\cdot , Z)$ is uniformly continuous, + specifically + \[ + |d(x,Z) - d(y,Z)| \le d(x,y), % TODO REF SHEET 1 + \] + we get + \begin{IEEEeqnarray*}{lrCl} + &d(x,Z) &\le & d(x,y) + d(y, Z)\\ + & &\le & d(x,y) + \delta(Y,Z)\\ + \implies& d(x,Z) - \delta(Y,Z) &\le & d(x,Y)\\ + \implies& d(x,Z) &\le & \delta(X,Y) + \delta(Y,Z)\\ + \implies & \delta(X,Z) &\le & \delta(X,Y) + \delta(Y,Z). + \end{IEEEeqnarray*} +\end{proof} + +\begin{itemize} + \item We have + \begin{IEEEeqnarray*}{rCl} + d_H(K_0, K_1) < \epsilon & \iff & \max \{\max_{x \in K_0}d(x, K_1), \max_{x \in K_1} d(x,K_0)\} < \epsilon\\ + &\iff& \max_{x \in K_0} d(x, K_1) < \epsilon \land \max_{x \in K_1} d(x, K_0) < \epsilon\\ + &\iff& K_0 \subseteq B_{\epsilon}(K_1) \land K_1 \subseteq B_\epsilon(K_0). + \end{IEEEeqnarray*} + \item Note that a subbase of $\tau_V$ + is given by $[U]$ and $\langle U \rangle \coloneqq [X;U]$ for $U \overset{\text{open}}{\subseteq} X$. + + Let $K \in [U]$. + Then $d(\cdot , U^c)\colon U \to \R_{\ge 0}$ + is always non-zero and continuous. + So $d(K,U^c)$ attains a minimum $\epsilon > 0$. + Then $B_{\epsilon}^H_\epsilon(K) \subseteq U$, + so $[U]$ is open in $\tau_V$. + + Let $K \in \langle U \rangle$. + Take some $k \in K \cap U$. + Then there is some $\epsilon > 0$ + such that $B_\epsilon(k) \subseteq U$. + Then $K \in B_{\epsilon}^H(K) \subseteq \langle U \rangle$. + + \todo{Other direction} + % $\tau_H \subseteq \tau_V$ + + \item Consider a countable dense subset of $X$. + Let $\cK$ be the set of finite subsets of that countable dense subset. + Then $\cK \subseteq K(X)$ is dense: + Take $K \in K(X)$ an let $\epsilon > 0$. + $K$ can be covered with finitely many $\epsilon$-balls + with centers from the countable dense subsets. + Let $K' \in \cK$ be the set of the centers. + Then $d_H(K, K') \le \epsilon$. +\end{itemize} + +\nr 3 +\begin{itemize} + \item By transfinite induction we get that $\alpha$ is an ordinal, + since $\prec$ is well-founded and the supremum of a sets + of ordinals is an ordinal. + Since $\rho_{\prec}\colon X \to \alpha$ + is a surjection, it follows that $\alpha \le |X|$, + i.e.~$\alpha < |X|^+$. + \item By induction on $\rho_{\prec_X}(x)$ we show that + $\rho_{\prec_{X}}(x) \le \rho_{\prec_Y}(f(x))$. + For $0$ this is trivial. + Suppose that $\rho_{\prec_{X}}(x) = \alpha$ + and the statement was shown for all $\beta < \alpha$. + Then + \begin{IEEEeqnarray*}{rCl} + \rho_{\prec_Y}(f(x)) &=& \sup \{\rho_{\prec_Y}(y') + 1 | y' \prec f(x)\}\\ + &\ge& \sup \{\rho_{\prec_Y}(f(x')) + 1 | f(x') \prec f(x)\}\\ + &\ge & \sup \{\rho_{\prec_Y}(f(x')) + 1 | x' \prec x\}\\ + &\ge & \sup \{\rho_{\prec_X}(x') + 1 | x' \prec x\}\\ + &=& \rho_{\prec_X}(x). + \end{IEEEeqnarray*} + \item Infinite branches of $T_\prec$ correspond + to infinite descending chain of $\prec$, + hence $T_{\prec}$ is well-founded iff $\prec$ is well-founded. + + % Unwarp$^{\text{tm}}$ definitions. + Suppose that $\prec$ is well-founded. + Note that $\rho_T(s)$ depends only on the last element of $s$, + as for $s, s' \in T$ with the same last element, + we have $s \concat x \in T \iff s' \concat x \in T$. + + Let $s = (s_0, \ldots, s_n)$. + Let us show that $\rho_T(s) = \rho_{\prec}(s_n)$. + We use induction on $\rho_T(s)$. + For leaves this is immediate. + From the last exercise sheet we know that + \[ + \rho_T(s) = \sup \{\rho_T(s \concat a) + 1 | s \concat a \in T\}. + \] + Hence + \begin{IEEEeqnarray*}{rCl} + \rho_T(s) &=& \sup \{\rho_T(s \concat a) + 1 | s \concat a \in T\}\\ + &=& \sup \{\rho_{\prec}(a) + 1 | s \concat a \in T\}\\ + &=& \sup \{\rho_{\prec}(a) + 1 | a \prec s_n\}\\ + &=& \rho_\prec(s_n). + \end{IEEEeqnarray*} +\end{itemize} + + + +\nr 4 + +\todo{Next time} + diff --git a/logic3.tex b/logic3.tex index f0d961a..52f3cab 100644 --- a/logic3.tex +++ b/logic3.tex @@ -61,6 +61,7 @@ \input{inputs/tutorial_07} \input{inputs/tutorial_08} \input{inputs/tutorial_09} +\input{inputs/tutorial_10} \section{Facts} \input{inputs/facts}