Josia Pietsch
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122 lines
4.6 KiB
TeX
122 lines
4.6 KiB
TeX
\subsection{Sheet 4}
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\tutorial{05}{2023-11-14}{}
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% 14 / 20
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\nr 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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\nr 2
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\todo{handwritten}
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\nr 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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\nr 4
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\begin{enumerate}[(a)]
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\item Let $d$ be a compatible, complete metric on $X$, wlog.~$d \le 1$.
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Set $ U_{\emptyset} \coloneqq X$.
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Suppose that $U_{s}$ has already been chosen.
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Then $D_s \coloneqq X \setminus U_s$ is closed.
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Hence $U_s^{(n)} \coloneqq \{x \in X | \dist(x,D_s) > \frac{1}{n}\}$
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is open.
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Let $m$ be such that $D_s^{(m)} \neq \emptyset$.
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Clearly $\overline{U_s^{(n)}} \subseteq U_s$.
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Let $(B_k)_{k < \omega}$ be a countable cover of $X$
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consisting of balls of diameter $2^{-|s|-2}$.
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Take some bijection $\phi\colon \omega \to \omega \times (\omega \setminus m)$
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and set $U_{s\concat i} \coloneqq U_s^{(\pi_1\left( \phi(i) \right))} \cap B_{\pi_2(\phi(i))}$,
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where there $\pi_i$ denote the projections
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(if this is empty, set $U_{s \concat i} \coloneqq U_s^{\pi_1(\phi(j))} \cap B_{\pi_2(\phi(j))}$
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for some arbitarily chosen $j < \omega$ such that it is not empty).
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Then $\overline{U_{s \concat i}} \subseteq \overline{U_s^{(n)}} \subseteq U_s$,
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\[
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\bigcup_{i < \omega} U_{s \concat i} = \bigcup_{n < \omega} U_{s}^{\left( n \right) } = U_s
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\]
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and $\diam(U_{s \concat i}) \le \diam(B_{\pi_2(\phi(i))})$.
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\item Let $s \in \omega^\omega$.
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Then
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\[
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\bigcap_{n < \omega} \overline{U_{s\defon{n}}}
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\]
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contains exactly one point.
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Let $f$ be the function that maps an $s \in \omega^{\omega}$
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to the unique point in the intersection of the $\overline{U_{s\defon{n}}}$.
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Let $x \in X$ be some point.
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Then by induction we can construct a sequence $s \in \omega^{\omega}$
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such that $x \in U_{s\defon{n}}$ for all $n$,
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hence $x = f(s)$, i.e.~$f$ is surjective.
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Let $B \overset{\text{open}}{\subseteq} X$.
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Then $B = \bigcup_{i \in I} U_i$
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for some $i \subseteq \omega^{<\omega}$,
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as every basic open set can be recovered as a union of $U_i$
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and $f^{-1}(B) = \bigcup_{i \in I} \left( \{i_0\} \times \ldots \{i_{|i|-1}\} \right) \times \omega^{\omega}$ is open,
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hence $f$ is continuous.
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On the other hand, consider an open ball $B \coloneqq \{\prod_{i < n} \{x_i\}\} \times \omega^{\omega} \subseteq \omega^{\omega}$.
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Then $f(B) = U_{(x_0,\ldots,x_{n-1})}$ is open,
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hence $f$ is open.
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\end{enumerate}
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