lecture 21

inputs/lecture_04.tex

@@ -102,14 +102,13 @@
     Equivalently
     \begin{itemize}
         \item $\overline{A}$ is nwd,
-        \item $X \setminus A$ is dense in $X$,
+        \item $X \setminus \overline{A}$ is dense in $X$,%
         \item $\forall  \emptyset \neq  U \overset{\text{open}}{\subseteq} X.~
             \exists  \emptyset \neq V \overset{\text{open}}{\subseteq} U.~
             V\cap A = \emptyset$.
             (If we intersect $A$ with an open $U$,
             then $A \cap U$ is not dense in $U$).
     \end{itemize}
-    %\todo{Think about this}
 
     A set $B \subseteq  X$ is \vocab{meager} 
     (or \vocab{first category}),

inputs/lecture_05.tex

@@ -1,14 +1,22 @@
 \lecture{05}{2023-10-31}{}
 
 \begin{fact}
-    A set $A$ is nwd iff $\overline{A}$ is nwd.
+    \begin{itemize}
+        \item A set $A$ is nwd iff $\overline{A}$ is nwd.
+        \item If $F$ is closed then $F$ is nwd iff $X \setminus F$ is open and dense.
+        \item Any meager set $B$ is contained in a meager $F_{\sigma}$-set.
+    \end{itemize}
 
-    If $F$ is closed then
-    $F$ is nwd iff $X \setminus F$ is open and dense.
 
-    Any meager set $B$ is contained in
-    a meager $F_{\sigma}$-set.
 \end{fact}
+\begin{proof} % remove?
+    \begin{itemize}
+        \item This follows from the definition as $\overline{\overline{A}} = \overline{A}$.
+        \item Trivial.
+        \item Let $B = \bigcup_{n < \omega} B_n$ be a union of nwd sets.
+            Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$.
+    \end{itemize}
+\end{proof}
 
 \begin{definition}
     A  \vocab{$\sigma$-algebra} on a set $X$
@@ -46,7 +54,7 @@
     \[
         \left( \bigcup_{n < \omega} A_n \right) \symdif \left( \bigcup_{n < \omega} U_n \right)
     \]
-    is meager,\todo{small exercise}
+    is meager,
     hence $\bigcup_{n < \omega} A_n \in \cA$.
 
     Let $A \in \cA$.
@@ -61,7 +69,9 @@
         In particular, $F \symdif \inter(F)$ is nwd.
     \end{claim}
     \begin{refproof}{thm:bairesigma:c1}
-        \todo{TODO}
+        $F \setminus \inter(F)$ is closed,
+        hence $\overline{F \setminus \inter(F)} = F \setminus \inter(F)$.
+        Clearly $\inter(F\setminus\inter(F)) = \emptyset$.
     \end{refproof}
 
     From the claim we get that

inputs/lecture_06.tex

@@ -113,7 +113,7 @@ and $\Pi^0_2 = G_\delta$.
 
 Furthermore define
 \[
-\Delta^0_\alpha(X(X)) \coloneqq \Sigma^0_\alpha(X) \cap \Pi^0_\alpha(X),
+\Delta^0_\alpha(X) \coloneqq \Sigma^0_\alpha(X) \cap \Pi^0_\alpha(X),
 \]
 i.e.~$\Delta^0_1$ is the set of clopen sets.
 

inputs/lecture_20.tex

@@ -1,4 +1,4 @@
-\lecture{20}{2024-01-09}{The infinite Torus}
+\lecture{20}{2024-01-09}{The Infinite Torus}
 
 \begin{example}
     \footnote{This is the same as \yaref{ex:19:inftorus},
@@ -132,6 +132,7 @@ coordinates.
 \end{definition}
 
 \begin{remark}
+    \label{rem:l20:sigma}
     Note that for
     \begin{IEEEeqnarray*}{rCl}
         \sigma\colon (S^1)^d &\longrightarrow & S^1 \\
@@ -167,14 +168,15 @@ coordinates.
 \end{proof}
 
 \begin{refproof}{thm:taudminimal:help}
-    Let $s \coloneqq  \tau_d$ and $Y \coloneqq (S^1)^d$.
+    Let $S \coloneqq  \tau_d$, $T \coloneqq \tau_{d+1}$ and $Y \coloneqq (S^1)^d$.
     Consider
     \begin{IEEEeqnarray*}{rCl}
         \gamma\colon S^1 &\longrightarrow & Y \\
-        x &\longmapsto & (x,x,\ldots,x)
+        x &\longmapsto & (x,x,\ldots,x).
     \end{IEEEeqnarray*}
+    Note that
     \begin{enumerate}[(a)]
-        \item $\gamma$ and $s \circ \gamma$ are homotopic
+        \item $\gamma$ and $S \circ \gamma$ are homotopic
             via
             \begin{IEEEeqnarray*}{rCl}
                 H\colon S^1 \times [0,1] &\longrightarrow & (S^1)^d \\
@@ -185,9 +187,7 @@ coordinates.
             since $\sigma(\gamma(x)) = \sigma((x,\ldots,x)) = x$.
     \end{enumerate}
 
-    [to be continued]
     \phantom\qedhere
-    
 \end{refproof}
 
 

inputs/lecture_21.tex

@@ -0,0 +1,87 @@
+\lecture{21}{2024-01-12}{Iterated Skew Shift}
+
+\begin{refproof}{thm:taudminimal:help}
+    Suppose towards a contradiction that
+    $Y \times S^1$ contains a proper minimal subflow $Z$.
+    Consider the projection $\pi\colon Y \times S^1 \to Y$.
+    By minimality of $Y$, we have $\pi(Z) = Y$.
+    Note that for every $\theta \in S^1$, $\theta \cdot  Z$ is minimal,
+    so either $\theta \cdot  Z = Z$ or $(\theta \cdot  Z)\cap Z = \emptyset$.
+
+    Let $H = \{\theta \in S^1 : \theta \cdot  Z = Z\}$.
+    $H$ is a closed subgroup of $S^1$.
+    % H is a rotation of Z containing 1 (?)
+    Therefore either $H = S^1$ (but in that case $Z = Y \times S^1$),
+    or there exists $m \in \Z$ such that $H = \{ \xi \in S^1 : \xi^m = 1 \}$
+    by \yaref{fact:tau1minimal}.
+
+    Note that if $(y, \beta) \in Z$ then for $t \in S^1$,
+    we have
+    \[
+        (y, \beta \cdot  t) \in  Z \iff t^m = 1.
+    \]
+    Therefore for every $y \in Y$, there are exactly $m$ many
+    $\xi \in S^1$
+    such that $(y, \xi) \in Z$.
+
+    Specifically for all $y$ there exists $\beta^{(y)} \in S^1$
+    such that $(y,\xi) \in Z$ iff
+    \[
+        \xi \in  \{\beta^{(y)} \cdot  t_1, \beta^{(y)} \cdot  t_2, \ldots,\beta^{(y)} \cdot  t_m\},
+    \]
+    where the $t_i \in S^1$
+    are such that
+    $t_i^m = 1$ for all $i$ and $i \neq j \implies t_i \neq t_j$,
+    i.e.~the $t_i$ are the $m$\textsuperscript{th} roots of unity.
+
+    Consider $f \colon  (y,\xi) \mapsto (y, \xi^m)$.
+    Since $(\beta^{(y)} \cdot  t_i)^m = (\beta^{(y)})^m$
+    we get a continuous\todo{Why is this continuous?}
+    function $\phi\colon Y \to S^1$
+    such that
+    \[
+    Z = \{(y,\xi) \in Y \times S^1 : \xi^m = \phi(y)\}.
+    \]
+    % namely
+    % \begin{IEEEeqnarray*}{rCl}
+    %     \phi\colon Y &\longrightarrow & S^1 \\
+    %     y &\longmapsto & \beta^{(y)}.
+    % \end{IEEEeqnarray*}
+
+    Note that $f(Z)$ is homeomorphic to $Y$.\todo{Why?}
+
+    \begin{claim}
+        $\phi(S(y)) = \phi(y) \cdot  (\sigma(y))^m$.
+    \end{claim}
+    \begin{subproof}
+        We have $T(y, \xi) = (S(y), \sigma(y) \cdot \xi)$
+        (cf.~\yaref{rem:l20:sigma}).
+        $Z$ is invariant under $T$.
+        So for $(y, \xi) \in Z$ we get $T(y, \xi) = ({\color{red}S(y)}, {\color{blue}\sigma(y) \cdot  \xi}) \in Z$.
+        Thus
+        \begin{IEEEeqnarray*}{rCl}
+            \phi({\color{red}S(y)}) &=& ({\color{blue}\sigma(y) \cdot \xi})^m\\
+                                    &=& (\sigma(y))^m \cdot \xi^m\\
+                                    &=& (\sigma(y))^m \cdot  \phi(y).
+        \end{IEEEeqnarray*}
+    \end{subproof}
+    Applying $\gamma$ we obtain
+    \[
+        [\phi \circ S \circ \gamma] = [\phi \circ \gamma] + [x \mapsto (\sigma(\gamma(x))^n].
+    \]
+    $S\circ \gamma$ is homotopic to $\gamma$,
+    so $[\phi \circ S \circ \gamma] = [\phi \circ \gamma]$.
+    Thus $[x \mapsto (\sigma(\gamma(x))^n] = 0$,
+    but that is a contradiction to (b) $\lightning$
+\end{refproof}
+
+Let $X_n \coloneqq  (S^1)^n$ and $X \coloneqq (S^1)^{\N}$.
+\begin{theorem}
+    $(X_n, \tau_n)$ is the maximal isometric extension of  $(X_{n-1}, \tau_{n-1})$
+    in $(X,\tau)$.
+\end{theorem}
+\begin{corollary}
+    The order of $(X,\tau)$ is $\omega$.
+\end{corollary}
+\todo{I could not attend lecture 21 as I was sick. The official notes on the lecture are very short.
+    Is something missing in the official notes?}