\subsection{Sheet 4}
\tutorial{05}{2023-11-14}{}

% 14 / 20

\nr 1


\begin{enumerate}[(a)]
    \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.
\end{enumerate}

\nr 2
\todo{handwritten}


\nr 3
\begin{itemize}
    \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 $\R^2 \setminus \Q^2$.
\end{itemize}

\nr 4

\begin{enumerate}[(a)]
    \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 $s \in \omega^\omega$.
        Then
        \[
        \bigcap_{n < \omega} \overline{U_{s\defon{n}}}
        \]
        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.
\end{enumerate}