228 lines
7.1 KiB
TeX
228 lines
7.1 KiB
TeX
\tutorial{14}{2024-01-30}{}
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\subsection{Sheet 12}
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\nr 1
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% Examinable
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Let $\LO(\N) \overset{\text{closed}}{\subseteq} 2^{\N\times \N}$ denote the set of linear orders on $\N$.
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Let $S \subseteq \LO(\N)$ be the set of orders having a least
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element and such that every element has an immediate successor.
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\begin{itemize}
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\item $S$ is Borel in $\LO(\N)$:
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Let $M_n \subseteq \LO(\N)$ be the set of orders with minimal element $n$.
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Let $I_{n,m} \subseteq \LO(\N)$ be the set of orders such
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that $m$ is the immediate successor of $n$.
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Clearly $S = \left(\bigcap_n \bigcup_{m\neq n} I_{n,m}\right) \cap \bigcup_n M_n$,
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so it suffices to show that $M_n$ and $I_{n,m}$ are Borel.
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It is $M_n = \bigcap_{m\neq n} \{\prec : m \not\prec n\}$
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and $I_{n,m} = \{\prec: n \prec m\} \cap \bigcap_{i} \{\prec : n \preceq i \preceq m \implies n = i \lor n = m \}$.
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\item Give an example of an element of $S$ which is not well-ordered:
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Consider $\{1 - \frac{1}{n} : n \in \N^+\} \cup \{1 + \frac{1}{n} : n \in \N^{+}\} \subseteq \R$
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with the order $<_\R$.
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This is an element of $S$,
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but $\{x \in S: x \ge 1\}$ has no minimal element,
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hence it is not well-ordered.
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\end{itemize}
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\nr 2
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% Examinable
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Recall the definition of the circle shift flow $(\R / \Z, \Z)$
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with parameter $\alpha \in \R$, $1 \cdot x \coloneqq x + \alpha$.
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\begin{itemize}
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\item If $\alpha \not\in \Q$, then $(\R / \Z, \Z)$ is minimal:
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This is known as \href{https://en.wikipedia.org/wiki/Dirichlet's_approximation_theorem}{Dirichlet's Approximation Theorem}.
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\item Consider $\R/\Z$ as a topological group.
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Any subgroup $H$ of $\R / \Z$ is dense in $\R / \Z$
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or of the form $H = \{ x \in \R / \Z | mx = 0\}$
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for some $m \in \Z$.
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If $H$ contains an irrational element $\alpha$, then
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it is dense by the previous point.
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Suppose that $H \subseteq \Q / \Z$.
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Let $D$ be the set of denominators of elements of $H$
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written as irreducible fractions.
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If $D$ is finite,
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then $H = \{x \in \R / \Z : \mathop{lcm}(D)x = 0\}$.
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Otherwise $H$ is dense, as it contains
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elements of arbitrarily large denominator.
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\end{itemize}
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\nr 3
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% somewhat examinable (for 1.0)
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% TODO
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\begin{enumerate}[(a)]
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\item $(X,T)$ is distal iff it does not have a proximal pair,
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i.e.~$a\neq b$, $c$ such that $t_n \in T$,
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$t_na, t_nb \to c$.
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Equivalently,
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for all $a,b$ there exists an $\epsilon$,
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such that for all $t \in T$, $d(ta,tb) > \epsilon$.
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\item \todo{TODO}% TODO (not too hard)
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% (b)
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% Let $(X,T)$ be distal with a dense orbit,
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% then it is distal minimal.
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% Sheet 8: has dense orbit is Borel
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% Distal flow decomposes into distal minimal flows.
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\end{enumerate}
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\nr 4
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% Examinable!
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% TODO THINK!
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\gist{%
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% RECAP
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Let $X$ be a metrizable topological space
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and let $K(X) \coloneqq \{ K \subseteq X : K \text{ compact}\}$.
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The Vietoris topology has a basis given by
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$\{K \subseteq U\}$, $U$ open (type 1)
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and $\{K : K \cap U \neq \emptyset\}$, $U$ open (type 2).
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The Hausdorff metric on $K(X)$,
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$d_H(K,L)$ is the smallest $\epsilon$
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such that $K \subseteq B_{\epsilon}(L) \land L \subseteq B_\epsilon(K)$.
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This is equal to the maximal point to set distance,
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$\max_{a \in A} d(a,B)$.
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On previous sheets, we checked that $d_H$ is a metric.
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If $X$ is separable, then so is $K(X)$.
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% END RECAP
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}{}
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\begin{fact}
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\label{fact:s12e4}
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Let $(X,d)$ be a complete metric space.
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Then so is $(K(X), d_H)$.
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\end{fact}
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\begin{refproof}{fact:s12e4}
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We need to show that $(K(X), d_H)$ is complete.
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Let $(K_n)_{ n< \omega}$ be Cauchy in $(K(X), d_H)$.
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Wlog.~$K_n \neq \emptyset$ for all $n$.
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Let $K = \{ x \in X : \forall x \in U \overset{\text{open}}{\subseteq} X.~
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\text{ $U \cap K_n \neq \emptyset$ for infinitely many $n$}\}$.
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Equivalently,
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$K = \{x : x \text{ is a cluster point of some subsequence $(x_n)$ with $x_n \in K_n$ for all $K_n$}\}$.
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(A cluster point is a limit of some subsequence).
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\begin{claim}
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\label{fact:s12e4:c1}
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$K_n \to K$.
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\end{claim}
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\begin{refproof}{fact:s12e4:c1}
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Note that $K$ is closed (the complement is open).
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\begin{claim}
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$K \neq \emptyset$.
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\end{claim}
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\begin{subproof}
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As $(K_n)$ is Cauchy,
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there exists a sequence $(x_n)$ with $x_n \in K_n$
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such that there exists a subsequence $(x_{n_i})$
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with $d(x_{n_i}, x_{n_{i+1}}) < \frac{1}{2^{i+1}}$.
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Let $n_0,n_1,\ldots$
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be such that $d_H(K_a, K_b) < 2^{-i-1}$
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for $a,b \ge n_i$.
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Pick $x_{n_0} \in K_{n_0}$.
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Then let $x_{n_{i+1}} \in K_{n_{i+1}}$ be such that
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$d(x_{n_i}, x_{n_{i+1}})$ is minimal.
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Then $x_{n_i} \xrightarrow{i \to \infty} x$
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and we have $x \in K$.
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\end{subproof}
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\begin{claim}
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$K$ is compact.
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\end{claim}
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\begin{subproof}
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We show that $K$ is complete and totally bounded.
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Since $K$ is a closed subset of a complete
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space, it is complete.
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So it suffices to show that $K$ is totally bounded.
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Let $\epsilon > 0$.
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Take $N$ such that $d_H(K_i,K_j) < \epsilon$
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for all $i,j \ge N$.
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Cover $K_N$ with finitely many $\epsilon$-balls
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with centers $z_i$.
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Take $x \in K$.
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Then the $\epsilon$-ball around $x$ intersects $K_j$
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for some $j \ge N$, so
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there exists $z_i$ such that $d(x,z_i) < 3\epsilon$.
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Note that a subset of a bigger space is totally
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bounded iff it is totally bounded in itself.
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\end{subproof}
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Now we show that $K_n \to K$
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in $K(X)$.
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Let $\epsilon > 0$.
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Take $N$ such that for all $m,n \ge N$,
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$d_H(K_m,K_n) < \frac{\epsilon}{2}$.
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We'll first show that $\delta(K, K_n) < \epsilon$ for all $n > N$.
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Let $x \in K$.
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Take $(x_{n_i})$ with $x_{n_i} \in K_{n_i}, x_{n_i} \to x$.
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Then for large $i$,
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we have $n_i \ge N$ and $d(x_{n_i}, x) < \frac{\epsilon}{2}$.
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Take $n \ge N$.
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Then there exists $y_n \in K_n$
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with $d(y_n, x_{n_i}) < \frac{\epsilon}{2}$.
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So $d(x,y_n) < \epsilon$.
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Now show that $\delta(K_n, K) < \epsilon$ for all $n \ge N$.
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Take $y \in K_n$.
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Show that $d(y,K) < \epsilon$.
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To do this, construct a sequence of $y_{n_i} \in K_{n_i}$
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starting with $y$ such that $d(y_{n_i}, y_{n_{i+1}}) < \frac{\epsilon}{2^{i+2}}$.
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(same trick as before).
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\end{refproof}
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\end{refproof}
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\begin{fact}
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If $X$ is compact metrisable,
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then so is $K(X)$.
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\end{fact}
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\begin{proof}
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We have just shown that $X$ is complete.
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So it suffices to show that it is totally bounded.
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Let $\epsilon > 0$.
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Cover $X$ with finitely many $\epsilon$-balls.
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Let $F$ be the set of the centers of these balls.
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Consider $\cP(F) \setminus \{\emptyset\}$.
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Clearly $\{B_x^{d_H} : x \in \cP(F) \setminus \{\emptyset\} \}$
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is a finite cover of $K(X)$.
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\end{proof}
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% TODO complete and totally bounded Sutherland metric and topological spaces
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