lecture 15

inputs/lecture_12.tex

@@ -119,7 +119,7 @@ For the proof we need some prerequisites:
 \end{refproof}
 
 \begin{corollary}
-\label{cor:ifsi11c}
+\label{cor:ifs11c}
     $\IF$ is $\Sigma^1_1$-complete.
 \end{corollary}
 \begin{proof}

inputs/lecture_13.tex

@@ -51,7 +51,7 @@ with $(f^{-1}(\{1\}), <)$.
 \begin{proposition}
     Suppose that $(A, <)$ is a countable well ordering.
     Then for a tree $T \subseteq A^{<\N}$ on $A$,
-    Then $T$ is well-fonuded iff
+    Then $T$ is well-founded iff
     $(T, <_{KB}\defon{T})$ is well ordered.
 \end{proposition}
 \begin{proof}
@@ -87,7 +87,7 @@ with $(f^{-1}(\{1\}), <)$.
     \]
     (equivalently $x \in \IF \iff f(x) \in \LO \setminus \WO$).
     This suffices, since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete
-    (see \yaref{cor:ifiss11c}).
+    (see \yaref{cor:ifs11c}).
 
     Fix a bijection $b\colon \N \to \N^{<\N}$.
 

inputs/lecture_14.tex

@@ -134,6 +134,7 @@ Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
     \todo{TODO}
 \end{exercise}
 \begin{theorem}[Kunen-Martin]
+    \yalabel{Kunen-Martin}{Kunen-Martin}{thm:kunenmartin}
     If $(X, \prec)$ is wellfounded
     and $\prec \subseteq X^2$ is $\Sigma^1_1$
     then $\rho(\prec) < \omega_1$.

inputs/lecture_15.tex

@@ -1,3 +1,161 @@
 \lecture{15}{2023-12-05}{}
 
-\todo{Lecture 15 is missing}
+\begin{theorem}[The Boundedness Theorem]
+    \yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}
+
+    Let $X$ be Polish, $C \subseteq X$ coanalytic,
+    $\phi\colon  C \to \omega_1$ a coanalytic rank on $C$,
+    $A \subseteq C$ analytic, i.e.~$A \in \Sigma^1_1(X)$.
+    Then $\sup \{\phi(x) : x \in A\} < \omega_1$.
+
+    Moreover for all $\xi < \omega_1$,
+    \[
+    D_\xi \coloneqq  \{x \in C : \phi(x) < \xi\}
+    \]
+    and
+    \[
+    E_\xi \coloneqq  \{x \in C : \phi(x) \le \xi\} 
+    \]
+    are Borel subsets of $X$.
+\end{theorem}
+\begin{proof}
+%     Let
+%     \begin{IEEEeqnarray*}{rCl}
+%         x \prec y&:\iff& x,y \in A \land \phi(x) < \phi(y)\\
+%                  &\iff& x,y \in A \land y \not\le_\phi^\ast x.
+%     \end{IEEEeqnarray*}
+%     Since $A$ is analytic,
+%     this relation is analytic and wellfounded on $X$.
+%     By \yaref{thm:kunenmartin}
+%     we get $\rho(\prec) < \omega_1$.
+%     Thus $\sup \{\phi(x) : x \in A\} < \omega_1$.
+% 
+%     Since $D_\xi = \bigcup_{\eta < \xi} E_\xi$,
+%     it suffices to check $E_\xi \in \Sigma_1^1(X)$.
+%     If $\alpha = \sup \{\phi(x) : x \in C\}$,
+%     we have $E_\xi = E_\alpha$ for all $\alpha < \xi < \omega_1$.
+
+    \todo{TODO: Copy from official notes}
+\end{proof}
+
+\section{Abstract Topological Dynamics}
+
+\begin{definition}
+    Let $T$ be a topological group\footnote{usually $T = \Z$ with the discrete topology}
+    and let $X$ be a compact metrizable space.
+
+    A \vocab{flow}  $(X, T)$, sometimes denoted $T \acts X$
+    is a continuous action
+    \begin{IEEEeqnarray*}{rCl}
+        T \times X&\longrightarrow & X \\
+        (t,x) &\longmapsto & tx.
+    \end{IEEEeqnarray*}
+
+    A flow is \vocab{minimal}  iff every orbit is dense.
+
+    $(Y,T)$ is a \vocab{subflow}  of $(X,T)$ if $Y \subseteq X$
+    and $Y$ is invariant under $T$,
+    i.e.~$\forall t \in T,y \in Y.~ty \in Y$.
+
+    A flow $(X,T)$ is \vocab{isometric}
+    iff there is a metric $d$ on $X$ such that
+    for all $t \in T$ the map
+    \begin{IEEEeqnarray*}{rCl}
+        a_t\colon X &\longrightarrow & X \\
+        x &\longmapsto & tx
+    \end{IEEEeqnarray*}
+    is an \vocab{isometry},
+    i.e.~$\forall t \in T.~\forall x,y \in X.~d(a_t(x),a_t(y)) = d(x,y)$.
+
+    If $(X,T)$ is a flow, then a pair $(x,y)$, $x \neq y$
+    is \vocab{proximal} iff
+    \[
+    \exists  z \in X.~\exists (t_n)_{n < } \in T^{\omega}.~t_n x \xrightarrow{n \to \infty}  z \land t_n y \xrightarrow{n \to \infty} z.
+    \]
+
+    A flow is \vocab{distal} iff
+    it has no proximal pair.
+\end{definition}
+\begin{remark}
+    Note that a flow is minimal iff it has no proper subflows.
+\end{remark}
+% \begin{example}
+%     Recall that $S_1 = \{z \in \C : |z| = 1\}$.
+%     Let $X = S_1$, $T = S_1$
+%     $(\alpha,\beta) \mapsto \alpha + \beta$ is isometric.
+%     % TODO: In the official notes it says \alpha + \beta, but this is no group action.
+%     % Maybe \alpha * \beta ?
+% \end{example}
+
+\begin{definition}
+    Let $X,Y$ be compact metric spaces
+    and $\pi\colon (X,T) \to (Y,T)$ a factor map.
+    Then $(X,T)$ is an \vocab{isometric extension} 
+    of $(Y,T)$ if there is a real valued $\rho:$
+    defined on the pullback, $\{(x_1,x_2) \in X^2 : \pi(x_1) = \pi(x_2)\}$, % TODO nice notation?
+    such that
+    \begin{enumerate}[(a)]
+        \item $\rho$ is continuous.
+        \item For each $y \in Y$, $\rho$ is a metric on the fibre
+            $X_y \coloneqq \{x \in X: \pi(x) = y\}$.
+        \item $\forall t \in T.~\rho(tx_1,tx_2) = \rho(x_1,x_2)$.
+        \item $\forall y,y' \in Y.~$
+            the metric spaces $(X_y, \rho)$ and  $(X_{y'}, \rho)$
+            are isometric.
+    \end{enumerate}
+\end{definition}
+\begin{remark}
+    A flow is isometric iff it is an isometric extension
+    of the trivial flow,
+    i.e.~the flow acting on a singleton.
+    % TODO THINK ABOUT THIS!
+\end{remark}
+\begin{proposition}
+    An isometric extension of a distal flow is distal.
+\end{proposition}
+\begin{proof}
+    Let $\pi\colon  X\to Y$ be an isometric extension.
+    Towards a contradiction,
+    suppose that $x_1,x_2 \in X$ are proximal.
+    Take $z \in X$ and $(g_n) \in T^{\omega}$
+    such that $g_n x_1 \to z$ and $g_n x_2 \to z$.
+
+    Then $g_n \pi(x_1) \to  \pi(z)$
+    and $g_n \pi(x_2) \to \pi(z)$,
+    so by distality of $Y$
+    we have $\pi(x_1) = \pi(x_{2})$.
+    Then $\rho(g_n x_1, g_n x_2)$
+    is defined and equal to $\rho(x_1,x_2)$.
+    By the continuity of $\rho$,
+    we get $\rho(g_n x_1, g_n x_2) \to \rho(z,z) = 0$.
+    Therefore $\rho(x_1,x_2) = 0$.
+    Hence $x_1 = x_2$ $\lightning$.
+\end{proof}
+
+\begin{definition}
+    Let $\Sigma = \{(X_i, T) : i \in I\} $
+    be a collection of factors of $(X,T)$. % TODO State precise definition of a factor
+    Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map.
+    Then $(X, T)$ is the \vocab{limit}  of $\Sigma$
+    iff
+    \[
+    \forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
+    \]
+    % TODO think about abstract nonsense
+\end{definition}
+
+\begin{proposition}
+    A limit of distal flows is distal.
+\end{proposition}
+\begin{proof}
+    Let $(X,T)$ be a limit of $\Sigma = \{(X_i, T) : i \in I\}$.
+    Suppose that each $(X_i, T)$ is distal.
+    If $(X,T)$ was not distal,
+    then there were $x_1, x_2, z \in X$
+    and a sequence $(g_n)$ in $T$
+    with $g_n x_1 \to z$ and $g_n x_2 \to z$.
+    Take $i \in I$ such that $\pi_i(x_1) \neq \pi_i(x_2)$.
+    But then $g_n \pi_i(x_1) \to \pi_i(z)$
+    and $g_n \pi_i(x_2) \to \pi_i(z)$,
+    which is a contradiction since $(X_i, T)$ is distal.
+\end{proof}

logic.sty

@@ -145,6 +145,7 @@
 \DeclareMathOperator{\symdif}{\triangle}
 \DeclareSimpleMathOperator{proj}
 \newcommand{\fc}{\mathfrak{c}}
+\DeclareMathOperator{\acts}{\curvearrowright}
 
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
 \newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}