Josia Pietsch
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71 lines
2.2 KiB
TeX
71 lines
2.2 KiB
TeX
\tutorial{04}{2023-11-14}{}
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\subsection{Sheet 4}
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% 14 / 20
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\subsubsection{Exercise 1}
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\begin{enumerate}[(a)]
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\item $\langle X_\alpha : \alpha\rangle$
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is a descending chain of closed sets (transfinite induction).
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Since $X$ is second countable, there cannot exist
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uncountable strictly decreasing chains of closed sets:
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Suppose $\langle X_{\alpha}, \alpha < \omega_1\rangle$
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was such a sequence,
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then $X \setminus X_{\alpha}$ is open for every $\alpha$,
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Let $\{U_n : n < \omega\}$ be a countable basis.
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Then $N(\alpha) = \{n | U_n \cap (X \setminus X_\alpha) \neq \emptyset\}$,
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is a strictly ascending chain in $\omega$.
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\item We need to show that $X_{\alpha_0}$ is perfect and closed.
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It is closed since all $X_{\alpha}$ are,
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and perfect, as a closed set $F$ is perfect
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iff it coincides $F'$.
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$X \setminus X_{\alpha_0}$
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is countable:
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$X_{\alpha} \setminus X_{\alpha + 1}$ is
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countable as for every $x$ there exists a basic open set $U$,
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such that $U \cap X_{\alpha} = \{x\}$,
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and the space is second countable.
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Hence $X \setminus X_{\alpha_0}$
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is countable as a countable union of countable sets.
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\end{enumerate}
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\subsection{Exercise 3}
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\begin{itemize}
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\item Let $Y \subseteq \R$ be $G_\delta$
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such that $Y$ and $\R \setminus Y$ are dense in $\R$.
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Then $Y \cong \cN$.
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$Y$ is Polish, since it is $G_\delta$.
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$Y$ is 0-dimensional,
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since the sets $(a,b) \cap Y$ for $a, b \in \R \setminus Y$
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form a clopen basis.
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Each compact subset of $Y$ has empty interior:
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Let $K \subseteq Y$ be compact
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and $U \subseteq K$ be open in $Y$.
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Then we can find cover of $U$ that has no finite subcover $\lightning$.
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\item Let $Y \subseteq \R$ be $G_\delta$ and dense
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such that $\R \setminus Y$ is dense as well.
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Define $Z \coloneqq \{x \in \R^2 | |x| \in Y\} \subseteq \R^2$.
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Then $Z$ is dense in $\R^2$
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and $\R^2 \setminus \Z$ is dense in $\R^2$.
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We have that for every $y \in Y$
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$\partial B_y(0) \subseteq Z$.
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Other example:
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Consider $\R^2 \setminus \Q^2$.
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\end{itemize}
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