Josia Pietsch 3df55b6516
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\subsection{Sheet 4}
% 14 / 20
\nr 1
\item $\langle X_\alpha : \alpha\rangle$
is a descending chain of closed sets (transfinite induction).
Since $X$ is second countable, there cannot exist
uncountable strictly decreasing chains of closed sets:
Suppose $\langle X_{\alpha}, \alpha < \omega_1\rangle$
was such a sequence,
then $X \setminus X_{\alpha}$ is open for every $\alpha$,
Let $\{U_n : n < \omega\}$ be a countable basis.
Then $N(\alpha) = \{n | U_n \cap (X \setminus X_\alpha) \neq \emptyset\}$,
is a strictly ascending chain in $\omega$.
\item We need to show that $X_{\alpha_0}$ is perfect and closed.
It is closed since all $X_{\alpha}$ are,
and perfect, as a closed set $F$ is perfect
iff it coincides $F'$.
$X \setminus X_{\alpha_0}$
is countable:
$X_{\alpha} \setminus X_{\alpha + 1}$ is
countable as for every $x$ there exists a basic open set $U$,
such that $U \cap X_{\alpha} = \{x\}$,
and the space is second countable.
Hence $X \setminus X_{\alpha_0}$
is countable as a countable union of countable sets.
\nr 2
\nr 3
\item Let $Y \subseteq \R$ be $G_\delta$
such that $Y$ and $\R \setminus Y$ are dense in $\R$.
Then $Y \cong \cN$.
$Y$ is Polish, since it is $G_\delta$.
$Y$ is 0-dimensional,
since the sets $(a,b) \cap Y$ for $a, b \in \R \setminus Y$
form a clopen basis.
Each compact subset of $Y$ has empty interior:
Let $K \subseteq Y$ be compact
and $U \subseteq K$ be open in $Y$.
Then we can find cover of $U$ that has no finite subcover $\lightning$.
\item Let $Y \subseteq \R$ be $G_\delta$ and dense
such that $\R \setminus Y$ is dense as well.
Define $Z \coloneqq \{x \in \R^2 | |x| \in Y\} \subseteq \R^2$.
Then $Z$ is dense in $\R^2$
and $\R^2 \setminus \Z$ is dense in $\R^2$.
We have that for every $y \in Y$
$\partial B_y(0) \subseteq Z$.
Other example:
Consider $\R^2 \setminus \Q^2$.
\nr 4
\item Let $d$ be a compatible, complete metric on $X$, wlog.~$d \le 1$.
Set $ U_{\emptyset} \coloneqq X$.
Suppose that $U_{s}$ has already been chosen.
Then $D_s \coloneqq X \setminus U_s$ is closed.
Hence $U_s^{(n)} \coloneqq \{x \in X | \dist(x,D_s) > \frac{1}{n}\}$
is open.
Let $m$ be such that $D_s^{(m)} \neq \emptyset$.
Clearly $\overline{U_s^{(n)}} \subseteq U_s$.
Let $(B_k)_{k < \omega}$ be a countable cover of $X$
consisting of balls of diameter $2^{-|s|-2}$.
Take some bijection $\phi\colon \omega \to \omega \times (\omega \setminus m)$
and set $U_{s\concat i} \coloneqq U_s^{(\pi_1\left( \phi(i) \right))} \cap B_{\pi_2(\phi(i))}$,
where there $\pi_i$ denote the projections
(if this is empty, set $U_{s \concat i} \coloneqq U_s^{\pi_1(\phi(j))} \cap B_{\pi_2(\phi(j))}$
for some arbitarily chosen $j < \omega$ such that it is not empty).
Then $\overline{U_{s \concat i}} \subseteq \overline{U_s^{(n)}} \subseteq U_s$,
\bigcup_{i < \omega} U_{s \concat i} = \bigcup_{n < \omega} U_{s}^{\left( n \right) } = U_s
and $\diam(U_{s \concat i}) \le \diam(B_{\pi_2(\phi(i))})$.
\item Let $s \in \omega^\omega$.
\bigcap_{n < \omega} \overline{U_{s\defon{n}}}
contains exactly one point.
Let $f$ be the function that maps an $s \in \omega^{\omega}$
to the unique point in the intersection of the $\overline{U_{s\defon{n}}}$.
Let $x \in X$ be some point.
Then by induction we can construct a sequence $s \in \omega^{\omega}$
such that $x \in U_{s\defon{n}}$ for all $n$,
hence $x = f(s)$, i.e.~$f$ is surjective.
Let $B \overset{\text{open}}{\subseteq} X$.
Then $B = \bigcup_{i \in I} U_i$
for some $i \subseteq \omega^{<\omega}$,
as every basic open set can be recovered as a union of $U_i$
and $f^{-1}(B) = \bigcup_{i \in I} \left( \{i_0\} \times \ldots \{i_{|i|-1}\} \right) \times \omega^{\omega}$ is open,
hence $f$ is continuous.
On the other hand, consider an open ball $B \coloneqq \{\prod_{i < n} \{x_i\}\} \times \omega^{\omega} \subseteq \omega^{\omega}$.
Then $f(B) = U_{(x_0,\ldots,x_{n-1})}$ is open,
hence $f$ is open.