Merge branch 'main' of https://git.abstractnonsen.se/josia-notes/w23-logic-3

inputs/tutorial_14.tex

@@ -0,0 +1,162 @@
+\tutorial{14}{2024-01-30}{}
+
+\subsection{Sheet 12}
+\nr 1
+% Examinable
+
+% TODO  (there is a more direct way to do it, not using analytic / coanalytic)
+\nr 2
+% Examinable
+
+
+\nr 3
+% somewhat examinable (for 1.0)
+
+\nr 4
+
+% Examinable!
+
+% RECAP
+Let $X$ be a metrizable topological space.
+
+Let $K(X) \coloneqq  \{ K \subseteq  X : \text{ compact}\}$.
+
+The Vietoris topology has a basis given by
+$\{K \subseteq U\}$, $U$ open (type 1)
+and $\{K : K \cap U \neq \emptyset\}$, $U$ open (type 2).
+
+The Hausdorff metric on $K(X)$,
+$d_H(K,L)$ is the smallest  $\epsilon$
+such that $K \subseteq  B_{\epsilon}(L) \land L \subseteq  B_\epsilon(K)$.
+This is equal to the maximal point to set distance,
+$\max_{a \in A} d(a,B)$.
+
+On previous sheets, we checked that $d_H$ is a metric.
+If $X$ is separable, then so is $K(X)$.
+% END RECAP
+
+\begin{fact}
+Let $(X,d)$ be a complete metric space.
+Then so is  $(K(X), d_H)$.
+\end{fact}
+\begin{proof}
+    We need to show that  $(K(X), d_H)$ is complete.
+
+    Let $(K_n)_{ n< \omega}$ be Cauchy in $(K(X), d_H)$.
+    Wlog.~$K_n \neq \emptyset$ for all $n$.
+
+    Let $K = \{ x \in X : \forall  x \in U \overset{\text{open}}{\subseteq} X.~
+    \text{ $X$ intersects $K_n$ for infinitely many $n$}\}$.
+
+    Equivalently,
+    $K = \{x : x \text{ is a cluster point of some subsequence $(x_n)$  with $x_n \in K_n$ for all $K_n$}\}$.
+
+    (A cluster point is a limit of some subsequence).
+
+    \begin{claim}
+        $K_n \to K$.
+    \end{claim}
+    \begin{subproof}
+    Note that $K$ is closed (the complement is open).
+
+
+    \begin{claim}
+        $K \neq \emptyset$.
+    \end{claim}
+    \begin{subproof}
+        As $(K_n)$ is Cauchy,
+        there exists a sequence $(x_n)$ with $x_n \in K_n$
+        such that there exists a subsequence $(x_{n_i})$
+        with $d(x_{n_i}, x_{n_{i+1}}) < \frac{1}{2^{i+1}}$.
+
+        Let $n_0,n_1,\ldots$
+        be such that $d_H(K_a, K_b) < 2^{-i-1}$
+        for $a,b \ge n_i$.
+
+        Pick $x_{n_0} \in  K_{n_0}$.
+        Then let $x_{n_{i+1}} \in  K_{n_{i+1}}$ be such that
+        $d(x_{n_i}, x_{n_{i+1}})$ is minimal.
+
+        Then $x_{n_i} \xrightarrow{i \to \infty} x$
+        and we have $x \in K$.
+    \end{subproof}
+
+    \begin{claim}
+        $K$ is compact.
+    \end{claim}
+    \begin{subproof}
+        We show that $K$ is complete and totally bounded.
+        Since $K$ is a closed subset of a complete
+        space, it is complete.
+
+        So it suffices to show that $K$ is totally bounded.
+        Let $\epsilon > 0$
+        Take $N$ such that $d_H(K_i,K_j) < \epsilon$
+        for all $i,j \ge N$.
+
+        Cover $K_N$ with finitely many  $\epsilon$-balls
+        with centers $z_i$.
+
+        Take $x \in K$.
+        Then the $\epsilon$-ball around $x$ intersects $K_j$
+        for some $j \ge N$, so
+        there exists $z_i$ such that $d(x,z_i) < 3\epsilon$.
+
+        Note that a subset of a bigger space is totally
+        bounded iff it is totally bounded in itself.
+    \end{subproof}
+
+    Now we show that $K_n \to K$
+    in $K(X)$.
+
+    Let  $\epsilon > 0$.
+    Take $N$ such that  for all $m,n \ge N$,
+    $d_H(K_m,K_n) < \frac{\epsilon}{2}$.
+    We'll first show that $\delta(K, K_n) < \epsilon$ for all $n > N$.
+
+    Let $x \in K$.
+    Take $(x_{n_i})$ with $x_{n_i} \in  K_{n_i}, x_{n_i} \to x$.
+    Then for large  $i$,
+    we have $n_i \ge N$ and $d(x_{n_i}, x) < \frac{\epsilon}{2}$.
+    Take $n \ge N$.
+    Then there exists $y_n \in K_n$
+    with $d(y_n, x_{n_i}) < \frac{\epsilon}{2}$.
+    So $d(x,y_n) < \epsilon$.
+
+
+    Now show that $\delta(K_n, K) < \epsilon$ for all $n \ge N$.
+
+    Take $y \in K_n$.
+    Show that $d(y,K) < \epsilon$.
+    To do this, construct a sequence of $y_{n_i} \in K_{n_i}$
+    starting with $y$ such that $d(y_{n_i}, y_{n_{i+1}}) < \frac{\epsilon}{2^{i+2}}$.
+    (same trick as before).
+    \end{subproof}
+
+\end{proof}
+
+\begin{fact}
+    If $X$ is compact metrisable,
+    then so is $K(X)$.
+\end{fact}
+\begin{proof}
+    We have just shown that $X$ is complete.
+    So it suffices to show that it is totally bounded.
+
+    Let $\epsilon > 0$.
+    Cover $X$ with finitely many $\epsilon$-balls.
+    Let $F$ be the set of the centers of these balls.
+
+    Consider $\cP(F) \setminus \{\emptyset\}$.
+    Clearly $\{B_x^{d_H} : x \in \cP(F) \setminus \{\emptyset\} \}$
+    is a finite cover of $K(X)$.
+\end{proof}
+
+
+% TODO complete and totally bounded Sutherland metric and topological spaces
+
+
+
+
+
+

logic3.tex

@@ -73,6 +73,7 @@
 \input{inputs/tutorial_11}
 \input{inputs/tutorial_12b}
 \input{inputs/tutorial_12}
+\input{inputs/tutorial_14}
 
 \section{Facts}
 \input{inputs/facts}