Josia Pietsch
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214 lines
7.4 KiB
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214 lines
7.4 KiB
TeX
\subsection{Sheet 2}
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\tutorial{03}{2023-10-31}{}
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% 15 / 16
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\begin{remark}
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$F_\sigma$ stands for \vocab{ferm\'e sum denumerable}.
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\end{remark}
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\nr 1
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Let $X$ be a Polish space.
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Then there exists an injection $f\colon X \to 2^\omega$
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such that for each $n < \omega$,
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the set $f^{-1}(\{(y_n) \in 2^\omega : y_n = 1\})$
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is open.
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Moreover if $V \subseteq 2^{ \omega}$ is closed,
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then $f^{-1}(V)$ is $G_\delta$.
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Let $(U_i)_{i < \omega}$ be a countable base of $X$.
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Define
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\begin{IEEEeqnarray*}{rCl}
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f\colon X &\longrightarrow & 2^{\omega} \\
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x &\longmapsto & (x_i)_{i < \omega}
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\end{IEEEeqnarray*}
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where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise.
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\gist{
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Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$
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is open.
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We have that $f$ is injective since $X$ is T1.
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Let $f\colon X \hookrightarrow 2^\omega$ be such that
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$f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\})$.
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Let $V \subseteq 2^{\omega}$ be closed.
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Then $2^{\omega} \setminus V$ is open, i.e.~has the form
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$\bigcup_{i \in I} ((\prod_{j<n_j} X_{i,j}) \times 2^{\omega})$
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for some $X_{i,j} \subseteq 2$.
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As $2^{\omega}$ is second countable, we may assume $I$ to be countable.
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Then $V = \bigcap_{i \in I} \left(2^{\omega} \setminus ((\prod_{i <n_j} X_{i,j}) \times 2^{\omega})\right)$.
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Since $f$ is injective, we have $f^{-1}(\bigcap_{a \in A} a) = \bigcap_{a \in A} f^{-1}(a)$.
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Thus it suffices to show that $f^{-1}(2^{\omega} \setminus ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
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is $G_{\delta}$, as a countable intersection of $G_{\delta}$-sets
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is $G_{\delta}$.
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We have that $U_k \coloneqq f^{-1}(\{y = (y_i) \in 2^{\omega} : y_k = 1\})$
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is open. Since $f$ is injective
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$f^{-1}(\{y = (y_i) \in 2^{\omega} : y_k = 0\}) = X \setminus U_k$
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is closed, in particular it is $G_\delta$.
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Let $x = (x_1,\ldots, x_n) \in 2^{n} \setminus (\prod_{i < n} X_i)$.
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Then $f^{-1}(\{x\} \times 2^\omega) = \bigcap_{i < n}\bigcap U'_n$
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is $G_{\delta}$, were $U'_i = U_i$ if $x_k = i$
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and $U'_i = X \setminus U_i$ otherwise.
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Since $2^{n} \setminus \left( \prod_{i < n} X_i \right)$
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is finite, we get that
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$f^{-1}(2^{\omega} \setminus ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
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is $G_\delta$ as a finite union of $G_{\delta}$ sets.
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}{}
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\nr 2
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Let $X$ be a Polish space. Then $X$ is homeomorphic to a closed subspace of $\R^{ \omega}$ :
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\todo{handwritten solution}
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% \begin{itemize}
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% \item
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% Let $f(x^{(i)})$ be a sequence in $f(X)$.
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% Suppose that $f(x^{(i)}) \to y$.
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% We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
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% Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
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% Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
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% \end{itemize}
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\nr 3
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\begin{example}
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Consider
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\begin{IEEEeqnarray*}{rCl}
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f\colon \R &\longrightarrow & [0,1] \\
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\frac{p}{q} &\longmapsto & \frac{1}{q}\\
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\R \setminus \Q \ni x &\longmapsto & 0
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\end{IEEEeqnarray*}
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Then $\osc_f(\frac{p}{q}) = \frac{1}{q}$
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and $\osc_f(x) = 0$ for $x \not\in \Q$.
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\end{example}
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\begin{definition}
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We say that $f\colon X \to Y$ is continuous
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at $a \in X$,
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if for $N$ a neighbourhood of $f(a)$ (i.e.~there exists
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$f(a) \in U \overset{\text{open}}{\subseteq} N$,
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then $f^{-1}(N)$ is a neighbourhood of $a$.
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\end{definition}
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\begin{theorem}[Kuratowski]
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Let $X$ be metrizable, $Y$ completely metrizable,
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$S \subseteq X$ and $f\colon S \to Y$ continuous.
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Then $f$ can be extended to a continuous function $\tilde{f}$
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on a $G_\delta$ set $G$ with $S \subseteq G \subseteq \overline{S}$.
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\end{theorem}
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\begin{proof}
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Let $G \coloneqq \overline{S} \cap \{x \in X | \osc_f(x) = 0\}$.
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Clearly $S \subseteq G$ as $f$ is continouos on $f$.
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\begin{claim}
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$G$ is $G_\delta$
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\end{claim}
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\begin{subproof}
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$\overline{S}$ is closed
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and
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\[
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\bigcap_{n \ge 1} \{x : \osc_f(x) <\frac{1}{n}\}
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\]
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is an intersection of open sets.
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\end{subproof}
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is an intersection of open sets.
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For $x \in G$, as $x \in \overline{S}$,
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there exists $(x_n)_{x_n < \omega}$, $x_n \in S$
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such that $x_n \to x$.
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We have that $(f(x_n))_n$ is Cauchy,
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as $\osc_f(x) = 0$.
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\todo{Something is missing here}
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\end{proof}
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\begin{corollary}
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Let $X$ be Polish and $Y \subseteq X$ Polish.
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Then $Y$ is $G_{\delta}$.
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\end{corollary}
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\begin{proof}
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\todo{TODO}
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% Consider the identity $f\colon Y \hookrightarrow X$.
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% Then $f$ can be extended to a $G_{\delta}$ set $G \subseteq X$
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% with $Y \subseteq G \subseteq \overline{Y}$.
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% $\tilde{f}$ and $\id_G$ agree on $Y$.
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% $Y$ is dense in $G$ and the codomain of $f$ is Hausdorff.\todo{????}
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% So $f = \id_G$, i.e.~$G = Y$.
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\end{proof}
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\nr 4
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Define
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\begin{IEEEeqnarray*}{rCl}
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f\colon \omega^{\omega} &\longrightarrow & 2^\omega \\
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(x_n)&\longmapsto & 0^{x_0} 1 0^{x_1} 1 \ldots.
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\end{IEEEeqnarray*}
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\begin{enumerate}[(1)]
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\item $f$ is a topological embedding:
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Consider a basic open set
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$B = \prod_{i < n} X_i \times \omega^{\omega}$
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for some $X_i \subseteq \omega$.
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Then $f(B) = \left(\bigcup_{x \in \prod_{i<n} X_i} B_x \right) \cap f(\omega^{\omega})$
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is open in $f(\omega^{\omega})$,
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where $B_x \coloneqq \{0^{x_0}10^{x_1}1\ldots 10^{x_n-1}1\} \times 2^{\omega}$.
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On the other hand let $C = \{x_0x_1x_2x_3x_4 \ldots x_{n-1}\}\times 2^{\omega}$
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be some basic open set of $2^{\omega}$.
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W.l.o.g.~$x_0x_1\ldots x_{n-1}$
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has the form $0^{a_0}10^{a_1}1\ldots 10^{a_k}x_{n-1}$.
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If $x_{n-1} = 1$,
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we get
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\[
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f^{-1}(C \cap f(\omega^{\omega})) = \{(a_0,a_1,\ldots,a_k)\} \times \omega^{\omega}.
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\]
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In the case of $x_{n-1} = 0$,
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it is
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\[
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f^{-1}(C \cap f(\omega^\omega)) = \bigcup_{b > a_k} \{(a_0,a_1,\ldots,a_{k-1}, b)\} \times \omega^{\omega}.
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\]
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In both cases the preimage is open.
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\item $C \coloneqq 2^{\omega} \setminus f(\omega^\omega)$
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is countable and dense in $2^{\omega}$.
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We have $C = \{x \in 2^{\omega} | x_i = 0 \text{ for all but finitely many $i$}\} = \bigcup_{i < \omega} (2^{i} \times 1^{\omega})$.
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Clearly this is countable.
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For denseness take some $x \in 2^\omega$.
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Let $x^{(n)}$ be defined by $x^{(n)}_i = x_i$ for $i < n$
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and $x^{(n)}_i = 0$ for $i \ge n$.
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Then $x^{(n)} \in C$ for all $n$,
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and $x^{(n)}$ converges to $x$.
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\item $f(\omega^\omega)$ is $G_\delta$:
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We have
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\begin{IEEEeqnarray*}{rCl}
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f(\omega^\omega) &=& 2^\omega \setminus \left(\bigcup_{i < \omega} (2^{i} \times 1^{\omega})\right)\\
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&=& \bigcap_{i < \omega} \left(2^{\omega} \setminus(2^{i} \times 1^{\omega})\right).
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\end{IEEEeqnarray*}
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\item $C$ as in (2) is homeomorphic to $\Q$.
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Go to the right in the even digits, go to the left for the odd digits,
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i.e.~let $C = (1,-1,1,-1, \ldots)$
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and set $x < y$ iff $C \cdot x <_{\text{lex}} C \cdot y$,
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where $<_{\text{lex}}$ denotes the lexicographical ordering.
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Note that the order topology of $<$ on $C$
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agrees with the subspace topology from $2^\omega$.
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By Cantor's theorem for countable, unbounded, dense linear
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linear orders,
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we get an order isomorphism $C \leftrightarrow \Q$.
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This is also a homeomorphism, as the topologies
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on $C$ and $\Q$ are the respective order topologies.
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\end{enumerate}
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