diff --git a/.gitea/workflows/build.yaml b/.gitea/workflows/build.yaml index bcacea7..44106ae 100644 --- a/.gitea/workflows/build.yaml +++ b/.gitea/workflows/build.yaml @@ -12,7 +12,7 @@ jobs: - name: Prepare pages run: | mkdir public - mv build/logic3.pdf build/logic3.log README.md public + mv build/*.pdf build/*.log README.md public - name: Deploy to pages uses: actions/pages@v1 with: diff --git a/inputs/lecture_01.tex b/inputs/lecture_01.tex index c3ebec3..7aebc22 100644 --- a/inputs/lecture_01.tex +++ b/inputs/lecture_01.tex @@ -198,18 +198,19 @@ suffices to show that open balls in one metric are unions of open balls in the o such that $f: X \to f(X)$ is a homeomorphism. \end{proposition} \begin{proof} - $X$ is separable, so it has some countable dense subset, - which we order as a sequence $(x_n)_{n \in \omega}$. + \gist{$X$ is separable, so it has some countable dense subset, + which we order as a sequence $(x_n)_{n \in \omega}$.}% + {Let $(x_n)_{n \in \omega}$ be a countable dense subset.} - Let $d$ be a metric on $X$ which is compatible with the topology. - W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}). - - Let $d$ be the metric of $X$ and define + \gist{Let $d$ be a metric on $X$ which is compatible with the topology. + W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}).}% + {Let $d \le 1$ be a metric of $X$.} + Define \begin{IEEEeqnarray*}{rCl} f\colon X &\longrightarrow & [0,1]^{\omega} \\ x&\longmapsto & (d(x,x_n))_{n < \omega} \end{IEEEeqnarray*} - +\gist{ \begin{claim} $f$ is injective. \end{claim} @@ -237,17 +238,21 @@ suffices to show that open balls in one metric are unions of open balls in the o Then $f(U) = f(X) \cap [0,1]^{n} \times [0,\epsilon) \times [0,1]^{\omega}$ is open\footnote{as a subset of $f(X)$!}. \end{subproof} +}{} \end{proof} \begin{proposition} Countable disjoint unions of Polish spaces are Polish. \end{proposition} +\gist{ \begin{proof} Define a metric in the obvious way. \end{proof} +}{} \begin{proposition} Closed subspaces of Polish spaces are Polish. \end{proposition} +\gist{}{ \begin{proof} Let $X$ be Polish and $V \subseteq X$ closed. Let $d$ be a complete metric on $X$. @@ -255,15 +260,14 @@ suffices to show that open balls in one metric are unions of open balls in the o Subspaces of second countable spaces are second countable. \end{proof} +} \begin{definition} Let $X$ be a topological space. A subspace $A \subseteq X$ is called $G_\delta$\footnote{Gebietdurchschnitt}, if it is a countable intersection of open sets. \end{definition} - +\gist{ Next time: Closed sets are $G_\delta$. A subspace of a Polish space is Polish iff it is $G_{\delta}$ - - - +}{} diff --git a/inputs/lecture_02.tex b/inputs/lecture_02.tex index 3c6a526..6faa1f3 100644 --- a/inputs/lecture_02.tex +++ b/inputs/lecture_02.tex @@ -1,148 +1,158 @@ \lecture{02}{2023-10-13}{Subsets of Polish spaces} - \begin{theorem} - \label{subspacegdelta} - A subspace of a Polish space is Polish - iff it is $G_{\delta}$. - \end{theorem} +\begin{theorem} + \label{subspacegdelta} + A subspace of a Polish space is Polish + iff it is $G_{\delta}$. +\end{theorem} - \begin{remark} - Closed subsets of a metric space $(X, d )$ - are $G_{\delta}$. - \end{remark} - \begin{proof} - Let $C \subseteq X$ be closed. - Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$. - Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$. - Let $x \in \bigcap U_{\frac{1}{n}}$. - Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$. - The $x_n$ converge to $x$ and since $C$ is closed, - we get $x \in C$. - Hence $C = \bigcap U_{\frac{1}{n}}$ - is $G_{\delta}$. - \end{proof} +\begin{remark} + Closed subsets of a metric space $(X, d )$ + are $G_{\delta}$. +\end{remark} +\begin{proof} +\gist{ + Let $C \subseteq X$ be closed. + Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$. + Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$. + Let $x \in \bigcap U_{\frac{1}{n}}$. + Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$. + The $x_n$ converge to $x$ and since $C$ is closed, + we get $x \in C$. + Hence $C = \bigcap U_{\frac{1}{n}}$ + is $G_{\delta}$. +}{% + For $C \overset{\text{closed}}{\subseteq} X$, + we have $C = \bigcap_{n \in \N} \{x | d(x,C) < \frac{1}{n}\}$. +} +\end{proof} - \begin{example} - Let $ X$ be Polish. - Let $d$ be a complete metric on $X$. - \begin{enumerate}[a)] - \item If $Y \subseteq X$ is closed, - then $(Y,d\defon{Y})$ is complete. - \item $Y = (0,1) \subseteq \R$ - with the usual metric $d(x,y) = |x-y|$. - Then $x_n \to 0$ is Cauchy in $((0,1), d)$. +\gist{ +\begin{example} + Let $ X$ be Polish. + Let $d$ be a complete metric on $X$. + \begin{enumerate}[a)] + \item If $Y \subseteq X$ is closed, + then $(Y,d\defon{Y})$ is complete. + \item $Y = (0,1) \subseteq \R$ + with the usual metric $d(x,y) = |x-y|$. + Then $x_n \to 0$ is Cauchy in $((0,1), d)$. - But - \[ - d_1(x,y) \coloneqq | x -y| - + \left|\frac{1}{\min(x, 1- x)} - - \frac{1}{\min(y, 1-y)} - \right| - \] - also is a complete metric on $(0,1)$ - which is compatible with $d$. + But + \[ + d_1(x,y) \coloneqq | x -y| + + \left|\frac{1}{\min(x, 1- x)} + - \frac{1}{\min(y, 1-y)} + \right| + \] + also is a complete metric on $(0,1)$ + which is compatible with $d$. - We want to generalize this idea. - \end{enumerate} - \end{example} + We want to generalize this idea. + \end{enumerate} +\end{example} +}{} - \begin{refproof}{subspacegdelta} - \begin{claim} - \label{psubspacegdelta:c1} - If $Y \subseteq (X,d)$ is $G_{\delta}$, - then there exists a complete metric on $Y$. - \end{claim} - \begin{refproof}{psubspacegdelta:c1} - Let $Y = U$ be open in $X$. - Consider the map - \begin{IEEEeqnarray*}{rCl} - f_U\colon U &\longrightarrow & - \underbrace{X}_{d} \times \underbrace{\R}_{|\cdot |} \\ - x &\longmapsto & \left( x, \frac{1}{d(x, U^c)} \right). - \end{IEEEeqnarray*} +\begin{refproof}{subspacegdelta} + \begin{claim} + \label{psubspacegdelta:c1} + If $Y \subseteq (X,d)$ is $G_{\delta}$, + then there exists a complete metric on $Y$. + \end{claim} + \begin{refproof}{psubspacegdelta:c1} + Let $Y = U$ be open in $X$. + Consider the map + \begin{IEEEeqnarray*}{rCl} + f_U\colon U &\longrightarrow & + \underbrace{X}_{d} \times \underbrace{\R}_{|\cdot |} \\ + x &\longmapsto & \left( x, \frac{1}{d(x, U^c)} \right). + \end{IEEEeqnarray*} - Note that $X \times \R$ with the - \[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\] - metric is complete. + Note that $X \times \R$ with the + \[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\] + metric is complete. - $f_U$ is an embedding of $U$ into $X \times \R$: - \begin{itemize} - \item It is injective because of the first coordinate. - \item It is continuous since $d(x, U^c)$ is continuous - and only takes strictly positive values. % TODO - \item The inverse is continuous because projections - are continuous. - \end{itemize} + $f_U$ is an embedding of $U$ into $X \times \R$\gist{: + \begin{itemize} + \item It is injective because of the first coordinate. + \item It is continuous since $d(x, U^c)$ is continuous + and only takes strictly positive values. % TODO + \item The inverse is continuous because projections + are continuous. + \end{itemize} + }{.} - So we have shown that $U$ is homeomorphic to % TODO with ? - the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$. - The graph is closed in $U \times \R$, - because $\tilde{f_U}$ is continuous. - It is closed in $X \times \R$ because - $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$. - \todo{Make this precise} + So we have shown that $U$ and + the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$ + are homeomorphic. + The graph is closed \gist{in $U \times \R$, + because $\tilde{f_U}$ is continuous. + It is closed}{} in $X \times \R$ \gist{because + $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}. + \todo{Make this precise} - Therefore we identified $U$ with a closed subspace of - the Polish space $(X \times \R, d_1)$. - \end{refproof} + Therefore we identified $U$ with a closed subspace of + the Polish space $(X \times \R, d_1)$. + \end{refproof} - Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$. - Take - \begin{IEEEeqnarray*}{rCl} - f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\ - x &\longmapsto & - \left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right) - \end{IEEEeqnarray*} + Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$. + Take + \begin{IEEEeqnarray*}{rCl} + f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\ + x &\longmapsto & + \left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right) + \end{IEEEeqnarray*} - As for an open $U$, $f_Y$ is an embedding. - Since $X \times \R^{\N}$ - is completely metrizable, - so is the closed set $f_Y(Y) \subseteq X \times \R^\N$. + As for an open $U$, $f_Y$ is an embedding. + Since $X \times \R^{\N}$ + is completely metrizable, + so is the closed set $f_Y(Y) \subseteq X \times \R^\N$. - \begin{claim} - \label{psubspacegdelta:c2} - If $Y \subseteq (X,d)$ is completely metrizable, - then $Y$ is a $G_{\delta}$ subspace. - \end{claim} - \begin{refproof}{psubspacegdelta:c2} - There exists a complete metric $d_Y$ on $Y$. - For every $n$, - let $V_n \subseteq X$ be the union - of all open sets $U \subseteq X$ such that - \begin{enumerate}[(i)] - \item $U \cap Y \neq \emptyset$, - \item $\diam_d(U) \le \frac{1}{n}$, - \item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$. - \end{enumerate} + \begin{claim} + \label{psubspacegdelta:c2} + If $Y \subseteq (X,d)$ is completely metrizable, + then $Y$ is a $G_{\delta}$ subspace. + \end{claim} + \begin{refproof}{psubspacegdelta:c2} + There exists a complete metric $d_Y$ on $Y$. + For every $n$, + let $V_n \subseteq X$ be the union + of all open sets $U \subseteq X$ such that + \begin{enumerate}[(i)] + \item $U \cap Y \neq \emptyset$, + \item $\diam_d(U) \le \frac{1}{n}$, + \item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$. + \end{enumerate} + \gist{ + We want to show that $Y = \bigcap_{n \in \N} V_n$. + For $x \in Y$, $n \in \N$ we have $x \in V_n$, + as we can choose two neighbourhoods + $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$, + such that $\diam_{d_Y}(U) < \frac{1}{n}$ + and $U_2 \cap Y = U_1$. + Additionally choose $x \in U_3$ open in $X$ + with $\diam_{d}(U_3) < \frac{1}{n}$. + Then consider $U_2 \cap U_3 \subseteq V_n$. + Hence $Y \subseteq \bigcap_{n \in \N} V_n$. - We want to show that $Y = \bigcap_{n \in \N} V_n$. - For $x \in Y$, $n \in \N$ we have $x \in V_n$, - as we can choose two neighbourhoods - $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$, - such that $\diam_{d_Y}(U) < \frac{1}{n}$ - and $U_2 \cap Y = U_1$. - Additionally choose $x \in U_3$ open in $X$ - with $\diam_{d}(U_3) < \frac{1}{n}$. - Then consider $U_2 \cap U_3 \subseteq V_n$. - Hence $Y \subseteq \bigcap_{n \in \N} V_n$. - - Now let $x \in \bigcap_{n \in \N} V_n$. - For each $n$ pick $x \in U_n \subseteq X$ open - satisfying (i), (ii), (iii). - From (i) and (ii) it follows that $x \in \overline{Y}$, - since we can consider a sequence of points $y_n \in U_n \cap Y$ - and get $y_n \xrightarrow{d} x$. - For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$ - is an open set containing $x$, - hence $U_n' \cap Y \neq \emptyset$. - Thus we may assume that the $U_i$ form a decreasing sequence. - We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$. - If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$, - since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$ - and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$. - The sequence $y_n$ converges to the unique point in - $\bigcap_{n} \overline{U_n \cap Y}$. - Since the topologies agree, this point is $x$. - \end{refproof} - \end{refproof} + Now let $x \in \bigcap_{n \in \N} V_n$. + For each $n$ pick $x \in U_n \subseteq X$ open + satisfying (i), (ii), (iii). + From (i) and (ii) it follows that $x \in \overline{Y}$, + since we can consider a sequence of points $y_n \in U_n \cap Y$ + and get $y_n \xrightarrow{d} x$. + For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$ + is an open set containing $x$, + hence $U_n' \cap Y \neq \emptyset$. + Thus we may assume that the $U_i$ form a decreasing sequence. + We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$. + If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$, + since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$ + and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$. + The sequence $y_n$ converges to the unique point in + $\bigcap_{n} \overline{U_n \cap Y}$. + Since the topologies agree, this point is $x$. + }{Then $Y = \bigcap_n U_n$.} + \end{refproof} +\end{refproof} diff --git a/inputs/lecture_22.tex b/inputs/lecture_22.tex index 44943fb..0cd13ac 100644 --- a/inputs/lecture_22.tex +++ b/inputs/lecture_22.tex @@ -163,7 +163,6 @@ For this we define is dense in $\overline{x} \mapsto f(\overline{x})$. Since the flow is distal, it suffices to show that one orbit is dense (cf.~\yaref{thm:distalflowpartition}). - % TODO REF Distal flow can be decomposed into disjoint minimal flows \item Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$. Consider the flows we get from $(f_i)_{i < j}$