228 lines
7.1 KiB

\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.
\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 well-ordered:
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 well-ordered.
\nr 2
% Examinable
Recall the definition of the circle shift flow $(\R / \Z, \Z)$
with parameter $\alpha \in \R$, $1 \cdot x \coloneqq x + \alpha$.
\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.
\nr 3
% somewhat examinable (for 1.0)
\item $(X,T)$ is distal iff it does not have a proximal pair,
i.e.~$a\neq b$, $c$ such that $t_n \in T$,
$t_na, t_nb \to c$.
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.
\nr 4
% Examinable!
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 $K(X)$.
Let $(X,d)$ be a complete metric space.
Then so is $(K(X), d_H)$.
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 $n$}\}$.
$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).
$K_n \to K$.
Note that $K$ is closed (the complement is open).
$K \neq \emptyset$.
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 $x \in K$.
$K$ is compact.
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.
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).
If $X$ is compact metrisable,
then so is $K(X)$.
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 $K(X)$.
% TODO complete and totally bounded Sutherland metric and topological spaces