Josia Pietsch
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100 lines
2.7 KiB
TeX
100 lines
2.7 KiB
TeX
\ctutorial{03}{2023-10-31}{}
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% 15 / 16
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\begin{remark}
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$F_\sigma$
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stands for ferm\'e sum denumerable.
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\end{remark}
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\subsection{Exercise 2}
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(b)
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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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\subsection{Exercise 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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$f\colon S \to Y$ continuous.
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Then $f$ can be extended to a continuous fnuction $f_n$
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on a $G_\delta$ set $G$ with $S \subseteq G \subseteq \overline{G}$.
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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
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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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Consider the identity $f\colon Y \hookrightarrow X$.
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Then $f$ can be extended to a $G_{\delta}$ set.
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$f$ and $\id_G$ agree on $Y$.
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Hence $Y \subseteq G \subseteq \overline{Y}$.
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$Y$ is dense in $G$ and $\cod(f)$ is ltd.\todo{????}
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So $f = \id_G$, i.e.~$G = Y$.
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\end{proof}
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\subsection{Exercise 4}
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Let $C$ be the subspace of $2^{\omega}$ consisting
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of sequences with finitely many $1$s.
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We want to show that $C \cong \Q$.
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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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