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+\tutorial{04}{2023-11-14}{}
+
+\subsection{Sheet 4}
+
+% 14 / 20
+
+\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}