diff --git a/inputs/tutorial_03.tex b/inputs/tutorial_03.tex new file mode 100644 index 0000000..93e170b --- /dev/null +++ b/inputs/tutorial_03.tex @@ -0,0 +1,100 @@ +\ctutorial{03}{2023-10-31}{} + +% 15 / 16 + +\begin{remark} + $F_\sigma$ + stands for ferm\'e sum denumerable. +\end{remark} + + +\subsection{Exercise 2} + +(b) +Let $f(x^{(i)})$ be a sequence in $f(X)$. +Suppose that $f(x^{(i)}) \to y$. +We have that $f^{-1} = \pi_{\text{odd}}$ is continuous. +Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$. +Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$. + + +\subsection{Exercise 3} + +\begin{example} + Consider + \begin{IEEEeqnarray*}{rCl} + f\colon \R &\longrightarrow & [0,1] \\ + \frac{p}{q} &\longmapsto & \frac{1}{q}\\ + \R \setminus \Q \ni x &\longmapsto & 0 + \end{IEEEeqnarray*} + Then $\osc_f(\frac{p}{q}) = \frac{1}{q}$ + and $\osc_f(x) = 0$ for $x \not\in \Q$. +\end{example} + + +\begin{definition} + We say that $f\colon X \to Y$ is continuous + at $a \in X$, + if for $N$ a neighbourhood of $f(a)$ (i.e.~there exists + $f(a) \in U \overset{\text{open}}{\subseteq} N$, + then $f^{-1}(N)$ is a neighbourhood of $a$. +\end{definition} + + + +\begin{theorem}[Kuratowski] + Let $X$ be metrizable, $Y$ completely metrizable, + $f\colon S \to Y$ continuous. + Then $f$ can be extended to a continuous fnuction $f_n$ + on a $G_\delta$ set $G$ with $S \subseteq G \subseteq \overline{G}$. +\end{theorem} +\begin{proof} + Let $G \coloneqq \overline{S} \cap \{x \in X | \osc_f(x) = 0\}$. + Clearly $S \subseteq G$ as $f$ is continouos on $f$. + \begin{claim} + $G$ is $G_\delta$ + \end{claim} + \begin{subproof} + $\overline{S}$ is closed + and + \[ + \bigcap_{n \ge 1} \{x : \osc_f(x) <\frac{1}{n}\} + \] + is an intersection of open sets. + \end{subproof} + is an intersection of open sets. + + For $x \in G$, as $x \in \overline{S}$, + there exists $(x_n)_{x_n < \omega}$, $x_n \in S_$ + such that $x_n \to x$. + We have that $(f(x_n))_n$ is Cauchy, + as $\osc_f(x) = 0$. + % TODO + + \todo{Something is missing here} + + +\end{proof} + +\begin{corollary} + Let $X$ be Polish and $Y \subseteq X$ Polish. + Then $Y$ is $G_{\delta}$. +\end{corollary} +\begin{proof} + Consider the identity $f\colon Y \hookrightarrow X$. + Then $f$ can be extended to a $G_{\delta}$ set. + $f$ and $\id_G$ agree on $Y$. + Hence $Y \subseteq G \subseteq \overline{Y}$. + $Y$ is dense in $G$ and $\cod(f)$ is ltd.\todo{????} + So $f = \id_G$, i.e.~$G = Y$. +\end{proof} + +\subsection{Exercise 4} + +Let $C$ be the subspace of $2^{\omega}$ consisting +of sequences with finitely many $1$s. +We want to show that $C \cong \Q$. + +Go to the right in the even digits, go to the left for the odd digits, +i.e.~let $C = (1,-1,1,-1, \ldots)$ +and set $x < y$ iff $C \cdot x <_{\text{lex}} C \cdot y$.