228 lines
7.1 KiB
TeX
228 lines
7.1 KiB
TeX
\tutorial{14}{20240130}{}




\subsection{Sheet 12}


\nr 1


% Examinable




Let $\LO(\N) \overset{\text{closed}}{\subseteq} 2^{\N\times \N}$ denote the set of linear orders on $\N$.




Let $S \subseteq \LO(\N)$ be the set of orders having a least


element and such that every element has an immediate successor.


\begin{itemize}


\item $S$ is Borel in $\LO(\N)$:




Let $M_n \subseteq \LO(\N)$ be the set of orders with minimal element $n$.


Let $I_{n,m} \subseteq \LO(\N)$ be the set of orders such


that $m$ is the immediate successor of $n$.




Clearly $S = \left(\bigcap_n \bigcup_{m\neq n} I_{n,m}\right) \cap \bigcup_n M_n$,


so it suffices to show that $M_n$ and $I_{n,m}$ are Borel.


It is $M_n = \bigcap_{m\neq n} \{\prec : m \not\prec n\}$


and $I_{n,m} = \{\prec: n \prec m\} \cap \bigcap_{i} \{\prec : n \preceq i \preceq m \implies n = i \lor n = m \}$.


\item Give an example of an element of $S$ which is not wellordered:




Consider $\{1  \frac{1}{n} : n \in \N^+\} \cup \{1 + \frac{1}{n} : n \in \N^{+}\} \subseteq \R$


with the order $<_\R$.


This is an element of $S$,


but $\{x \in S: x \ge 1\}$ has no minimal element,


hence it is not wellordered.




\end{itemize}




\nr 2


% Examinable


Recall the definition of the circle shift flow $(\R / \Z, \Z)$


with parameter $\alpha \in \R$, $1 \cdot x \coloneqq x + \alpha$.




\begin{itemize}


\item If $\alpha \not\in \Q$, then $(\R / \Z, \Z)$ is minimal:




This is known as \href{https://en.wikipedia.org/wiki/Dirichlet's_approximation_theorem}{Dirichlet's Approximation Theorem}.




\item Consider $\R/\Z$ as a topological group.


Any subgroup $H$ of $\R / \Z$ is dense in $\R / \Z$


or of the form $H = \{ x \in \R / \Z  mx = 0\}$


for some $m \in \Z$.






If $H$ contains an irrational element $\alpha$, then


it is dense by the previous point.




Suppose that $H \subseteq \Q / \Z$.


Let $D$ be the set of denominators of elements of $H$


written as irreducible fractions.


If $D$ is finite,


then $H = \{x \in \R / \Z : \mathop{lcm}(D)x = 0\}$.


Otherwise $H$ is dense, as it contains


elements of arbitrarily large denominator.


\end{itemize}






\nr 3


% somewhat examinable (for 1.0)


% TODO




\begin{enumerate}[(a)]


\item $(X,T)$ is distal iff it does not have a proximal pair,


i.e.~$a\neq b$, $c$ such that $t_n \in T$,


$t_na, t_nb \to c$.




Equivalently,


for all $a,b$ there exists an $\epsilon$,


such that for all $t \in T$, $d(ta,tb) > \epsilon$.






\item \todo{TODO}% TODO (not too hard)


% (b)


% Let $(X,T)$ be distal with a dense orbit,


% then it is distal minimal.


% Sheet 8: has dense orbit is Borel


% Distal flow decomposes into distal minimal flows.


\end{enumerate}






\nr 4




% Examinable!


% TODO THINK!




\gist{%


% RECAP


Let $X$ be a metrizable topological space


and let $K(X) \coloneqq \{ K \subseteq X : K \text{ compact}\}$.




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}


\label{fact:s12e4}


Let $(X,d)$ be a complete metric space.


Then so is $(K(X), d_H)$.


\end{fact}


\begin{refproof}{fact:s12e4}


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{ $U \cap K_n \neq \emptyset$ 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}


\label{fact:s12e4:c1}


$K_n \to K$.


\end{claim}


\begin{refproof}{fact:s12e4:c1}


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 $d_H(K_a, K_b) < 2^{i1}$


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{refproof}


\end{refproof}




\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
