\tutorial{04}{2023-11-14}{}

\subsection{Sheet 4}

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\subsubsection{Exercise 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}


\subsection{Exercise 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}
