Josia Pietsch e887f46a5d
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\subsection{Sheet 1}
% Points: 15 / 16
\nr 1
Let $(X,d)$ be a metric space and $\emptyset \neq A \subseteq X$.
Let $d(x,A) \coloneqq \inf(d(x,a) : a \in A\}$.
\item $d(-,A)$ is uniformly continuous:
Clearly $|d(x,A) - d(y,A)| \le d(x,y)$.
\todo{Add details}
\item $d(x,A) = 0 \iff x \in \overline{A}$.
$d(x,A) = 0$ iff there is a sequence in $A$
converging towards $x$ iff $x \in \overline{A}$.
\nr 2
Let $X$ be a discrete space.
For $f,g \in X^{\N}$ define
d(f,g) \coloneqq \begin{cases}
(1 + \min \{n: f(n) \neq g(n)\})^{-1} &: f \neq g,\\
0 &: f= g.
\item $d$ is an \vocab{ultrametric},
i.e.~$d(f,g) \le \max \{d(f,h), d(g,h)\}$ for all $f,g,h \in X^{\N}$ :
Let $f,g,h \in X^{\N}$.
We need to show that $d(f,g) \le \max(d(f,h), d(g,h))$.
If $f = g$ this is trivial.
Otherwise let $n$ be minimal such that $f(n) \neq g(n)$.
Then $h(n) \neq f(n)$ or $ h(n) \neq g(n)$
must be the case.
W.l.o.g.~$h(n) \neq f(n)$.
Then $d(f,g) = \frac{1}{1+n} \le d(f,h)$.
\item $d$ induces the product topology on $X^{\N}$:
It suffices to show that the $\epsilon$-balls with respect to $d$
are exactly the basic open set of the product topology,
i.e.~the sets of the form
\{x_1\} \times \ldots \times \{x_n\} \times X^{\N}
for some $n \in \N$, $x_1,\ldots,x _n \in X$.
Let $\epsilon > 0$. Let $n$ be minimal such that $\frac{1}{1+n} \ge \epsilon$.
Then $B_{\epsilon}((x_i)_{i \in \N}) = \{x_1\} \times \{x_n\} \times X^{\N}$.
Since $\N \ni n \mapsto \frac{1}{1+n}$
is injective, every basic open set of the product topology
can be written in this way.
\item $d$ is complete:
Let $(f_n)_{n \in \N}$ be a Cauchy sequence with respect to $d$.
For $n \in \N$ take $N_n \in \N$
such that $d(f_i, f_j) < \frac{1}{1 + n}$.
Clearly $f_i(n) = f_j(n)$ for all $n > N_n$.
Define $f \in X^\N$ by $f(n) \coloneqq f_{N_n}(n)$.
Then $ (f_n)_{n \in \N}$
converges to $f$,
since for all $n > N_n$ $f_n$
\item If $X$ is countable, then $X^{\N}$ with the product topology
is a Polish space:
(We assume that $X$ is non-empty, as otherwise the claim is wrong)
We need to show that there exists a countable dense subset.
To this end, pick some $x_0 \in X$ and
consider the set $D \coloneqq \bigcup_{n\in \N} (X^n \times \{x_{0}\}^{\N})$.
Since $X$ is countable, so is $D$.
Take some $(a_n)_{n \in \N} \in X^{\N}$
and consider $B \coloneqq B_{\epsilon}((a_n)_{n \in \N})$.
Let $m$ be such that $\frac{1}{1+m} < \epsilon$.
Then $(b_{n})_{n \in \N} \in B \cap D$,
where $b_n \coloneqq a_n$ for $n \le m$ and $b_n \coloneqq x_0$
Hence $D$ is dense.
\nr 3
Consider $\N$ as a discrete space and $\N^{\N}$ with the product topology.
S_{\infty} = \{f\colon \N \to \N \text{ bijective}\} \subseteq \N^{\N}.
\item $S_{\infty}$ is a Polish space:
From \yaref{s1e2} we know that $\N^{\N}$ is Polish.
Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$
with respect to $\N^\N$.
Consider the sets $I \coloneqq \bigcap_{(i,j) \in \N^2, i < j} \{f \in \N^{\N} | f(i) \neq f(j)\}$
and $S \coloneqq \bigcap_{n \in \N} \{f \in \N^\N | n \in \im f\}$.
We have that $\{f \in \N^\N | f(i) \neq f(j)\} = \bigcup_{n \in \N} \N^{i-1} \times \{n\} \times \N^{i - j -1 } \times (\N \setminus \{n\} ) \times \N^\N$
is open.
Hence $I$ is $G_{\delta}$.
Furthermore $\{f \in \N^{\N} | n \in \im f\} = \bigcup_{k \in \N} \N^k \times \{n\} \times \N^\N$j
is open,
thus $S$ is $G_\delta$ as well.
In particular $S \cap G$ is $G_\delta$.
Since $I$ is the subset of injective functions
and $S$ is the subset of surjective functions,
we have that $S_{\infty} = I \cap S$.
\item $S_{\infty}$ is not locally compact:
Consider the point $x = (i)_{i \in \N} \in S_{\infty}$.
Let $x \in B$ be open. We need to show that there
is no closed compact set $C \supseteq B$
W.l.o.g.~let $B = (\{0\} \times \ldots \times \{n\} \times \N^\N) \cap S_\infty$
for some $n \in \N$.
Let $C \supseteq B$ be some closed set.
Consider the open covering
\{S_{\infty} \setminus B\} \cup \{ B_j | j > n\}.
B_j \coloneqq (\{0\} \times \ldots \times \{n\} \times \{j\} \times \N^{\N}) \cap S_\infty.
Clearly there cannot exist a finite subcover
as $B$ is the disjoint union of the $B_j$.
\nr 4
Let $X $ be a compact Hausdorff space.
Then the following are equivalent:
\item $X$ is Polish,
\item $X$ is metrisable,
\item $X$ is second countable.
(i) $\implies$ (ii) clear
(i) $\implies$ (iii) clear
(ii) $\implies$ (i) Consider the cover $\{B_{\epsilon}(x) | x \in X\}$
for every $\epsilon \in \Q$
and chose a finite subcover.
Then the midpoints of the balls from the cover
form a countable dense subset.
The metric is complete as $X$ is compact.
(For metric spaces: compact $\iff$ seq.~compact $\iff$ complete and totally bounded)
(iii) $\implies$ (ii)
Use Urysohn's metrisation theorem and the fact that compact
Hausdorff spaces are normal
Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish}
and $Y $ Polish.
Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$.
Consider the \vocab{uniform metric} $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$.
Clearly $d_u$ is a metric.
$d_u$ is complete.
Let $(f_n)$ be a Cauchy sequence in $\cC(X,Y)$.
A $Y$ is complete,
there exists a pointwise limit $f$.
$f_n$ converges uniformly to $f$:
d(f_n(x), f(x)) \le \overbrace{d(f_n(x), f_m(x))}^{\mathclap{\text{$(f_n)$ is Cauchy}}}
+ \underbrace{d(f_m(x), f(x))}_{\mathclap{\text{small for appropriate $m$}}}.
$f$ is continuous by the uniform convergence theorem.
There exists a countable dense subset.
Fix a metric $d_X$ on $X$ defining its topology.
C_{m,n} \coloneqq \{f \in \cC(X,Y) : \forall x,y \in X.~\left( d_X(x,y) < \frac{1}{m+1} \implies d(f(x), f(y)) <\frac{1}{n+1}\right) \}.
Choose $X_m \subseteq X$ finite with $X \subseteq \bigcup_{x \in X_m} B_{\frac{1}{m+1}}(x)$.
Let $D_{m,n} \subseteq C_{m,n}$ be countable,
such that for every $f \in C_{m,n}$ and every $\eta > 0$,
there is $g \in D_{m,n}$ with $d(f(y), g(y)) < \frac{\eta}{3}$
for each $y \in X_m$.
Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$:
Indeed if $f \in \cC(X,Y)$ and $\eta > 0$,
we find $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$,
since $f$ is uniformly continuous.
Let $g \in D_{m,n}$ be such that $\forall y \in X_m.~d(f(y), g(y)) < \frac{1}{n+1}$.
We have $d_u(f,g) \le \eta$,
since for every $x \in X$, we find $y \in X_m$ with $d_X(x,y) < \frac{1}{m+1}$,
d_Y(f(x), g(x)) &\le& d_Y(f(x), f(y)) + d_Y(f(y), g(y)) + d_Y(g(y), g(x))\\
&\le& \frac{1}{n+1} + \frac{1}{n+1} + \frac{1}{n+1} \le \eta.