test gist

.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:

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,
+    \gist{$X$ is separable, so it has some countable dense subset,
-    which we order as a sequence $(x_n)_{n \in \omega}$.
+    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.
+    \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}).
+    W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}).}%
-
+    {Let $d \le 1$ be a metric of $X$.}
-    Let $d$ be the metric of $X$ and define
+    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 $G_{\delta}$
-
+}{}
-
-

inputs/lecture_02.tex

@@ -11,6 +11,7 @@
     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 $C \subseteq \bigcap U_{\frac{1}{n}}$.
@@ -20,8 +21,13 @@
     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}
 
+\gist{
 \begin{example}
     Let $ X$ be Polish.
     Let $d$ be a complete metric on $X$.
@@ -45,6 +51,7 @@
            We want to generalize this idea.
     \end{enumerate}
 \end{example}
+}{}
 
 \begin{refproof}{subspacegdelta}
     \begin{claim}
@@ -65,7 +72,7 @@
         \[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
         metric is complete.
 
-         $f_U$ is an embedding of $U$ into $X \times \R$:
+        $f_U$ 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
@@ -73,13 +80,15 @@
             \item The inverse is continuous because projections
                 are continuous.
         \end{itemize}
+    }{.}
 
-         So we have shown that $U$ is homeomorphic to % TODO with ?
+        So we have shown that $U$ and
-         the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$.
+        the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
-         The graph is closed in $U \times  \R$,
+        are homeomorphic.
+        The graph is closed \gist{in $U \times  \R$,
         because $\tilde{f_U}$ is continuous.
-         It is closed in $X \times \R$ because
+        It is closed}{} in $X \times \R$ \gist{because
-         $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$.
+        $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
         \todo{Make this precise}
 
         Therefore we identified $U$ with a closed subspace of
@@ -114,7 +123,7 @@
             \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
@@ -143,6 +152,7 @@
         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}
 

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 $(f_i)_{i < j}$