Josia Pietsch 0a14244eb3
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2024-01-23 20:56:14 +01:00

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Continuation of sheet 8, exercise 4.
% Whiteboard https://wbo.ophir.dev/boards/Dphc7eylJJcIA0WbQsn7jzec1domqyx51gXb5qe6rzw-#263,0,0.5
Let $X$ be a compact metric space.
For $K \subseteq X$ compact and $U \overset{\text{open}}{\subseteq} X$
S_{K,U} \coloneqq \{f \in \cC(X,X): f(K) \subseteq U\}.
The \vocab{compact open topology} on $\cC(X,X)$
is the topology that has $S_{K,U}$ as a subbase.
If $X$ is compact,
then the compact open topology
is the topology induced by the uniform metric $d_\infty$.
Take some $S_{K,U}$. We need to show that this can be written
as a union of open $d_{\infty}$-balls.
Let $f_0 \in S_{K,U}$.
Consider the continuous function $d(-, U^c)$.
Since $f_0(K)$ is compact,
there exists $\epsilon \coloneqq \min d(f_0(K), U^c)$
and $B_{\epsilon}(f_0) \subseteq S_{K,U}$.
On the other hand, consider $B_{\epsilon}(f_0)$ for some $\epsilon > 0$
and $f_0 \in \cC(X,X)$.
As $f_0$ is uniformly continuous,
there exists $\delta > 0$ such that $d(x,x') < \delta \implies d(f_0(x), f_0(x')) < \frac{\epsilon}{3}$.
Cover $X$ with finitely many $\delta$-balls $B_\delta(a_1), \ldots, B_{\delta}(a_k)$.
\[f_0(\overline{B_{\delta}(a_i)}) \subseteq \overline{f_0(B_{\delta}(a_i)} \subseteq \overline{B_{\frac{\epsilon}{3}}(f_0(a_i))} \subseteq B_{\frac{\epsilon}{2}}(f_0(a_i)).\]
For $i \le k$, let $S_i \coloneqq S_{\overline{B_{\delta}(a_i)}, B_{\frac{\epsilon}{2}}(f_0(a_i))}$.
Take $\bigcap_{i \le k} S_i$. This is open
in the compact open topology and
$B_{\epsilon}(f_0) \subseteq \bigcap_{i \le k} S_i$.
$f \in \cC(X,X)$ is surjective
iff for all basic open $\emptyset\neq U \subseteq X$
there exists a basic open $\emptyset \neq V \subseteq X$
with $f(\overline{V}) \subseteq U$.
Note that we can write this as a $G_\delta$-condition.
Take $B_\epsilon(f(x_0))\subseteq U$.
Then there exists $\delta > 0$
such that $f(B_{\delta}(x_0)) \subseteq B_{\frac{\epsilon}{2}}(f(x_0))$
hence $f(\overline{B_{\delta}(x_0)}) \subseteq B_\epsilon(f(x_0))$.
For the other direction take $y \in X$.
We want to find a preimage.
For every $B_{\frac{1}{n}}(y)$,
there exists a basic open set $V_n$ with $f(\overline{V}) \subseteq B_{\frac{1}{n}}(y)$.
Take $x_n \in V_n$.
Since $X$ is compact, it is sequentially compact,
so there exists a converging subsequence.
Wlog.~$x_n \to x$,
so $f(x_n) \to f(x) = y$.
$f \in \cC(X,X)$ is injective iff
for all basic open $U$,$V$
with $\overline{U} \cap \overline{V} = \emptyset$
we have $f(\overline{U}) \cap f(\overline{V}) = \emptyset$.
This is a $G_\delta$-condition,
since we can write it as
\bigcap_{U,V} S_{\overline{U}, f(\overline{V})^c}.
$\implies$ is trivial.
$\impliedby$ follows since for all pairs $x,y \in X$,
we can find $x \in U$, $y \in V$ such that $\overline{U} \cap \overline{V} = \emptyset$.
Hence $\Homeo(X,X)$ is $G_\delta$.
In particular it is a Polish space.
Let $D$ be the set of $\Z$-flows with dense orbit.
$f \in D$ $\iff$
for all basic open $U,V \subseteq X$,
there exists $n \in \Z$
such that $f^n(U) \cap V \neq \emptyset$.
Suppose that the orbit of $x_0 \in X$ is dense.
Then there exist $k,l \in \Z$
such that $f^k(x_0)\in U$ and $f^l(x_0) \in V$,
so $f^{l-k} U \cap V \neq \emptyset$.
For basic open sets $V$
A_V \coloneqq \{ x \in X: \exists n.~ f^n(x) \in V\}.
By assumption, all the $A_V$ are dense.
Hence $\bigcap_{V}A_V$ is dense by the \yaref{thm:bct}.
$A_V = \bigcup_{n \in \Z} f^n(V)$ is open.
The condition can be written as a $G_\delta$ set.
For $f_0(U) \cap V \neq \emptyset$
take $u \in U$ such that $f_0(u) \in V$.
Then there exists $\epsilon > 0$ such that $B_{\epsilon}(f_0(u)) \subseteq U$,
hence $B_{\epsilon}(f_0)$ is an open neighbourhood contained
in $\{f : f(U) \cap V \neq \emptyset \} $.
For $n = 2$ note that
$d(f^2(u), f^2_0(u) \le d(f(f(u)), f_0(f(u))) + d(f_0(f(u)), f_0(f_0(u)))$.
The first part can be bounded by $d(f,f_0)$.
For the second part,
note that there exists $\delta$ such that
\[d(a,b) < \delta \implies d(f_0(a), f_0(b)) < \frac{\epsilon}{2}.\]
Let $\eta \coloneqq \min \{\delta, \frac{\epsilon}{2}\}$
and consider $d_\infty(f,f_0) < \epsilon$.
For other $n$ it is some more work, which is left as an exercise.