tutorial 09

inputs/lecture_12.tex

@@ -12,6 +12,7 @@
 \end{definition}
 
 \begin{proposition}
+    \label{prop:ifs11}
     $\IF \in  \Sigma^1_1(\Tr)$.
 \end{proposition}
 \begin{proof}

inputs/lecture_17.tex

@@ -197,16 +197,9 @@ However if we pick $y \in Y$, $Ty$ might not be dense.
 %     so $Gy \subseteq Gx$.
 %     Since $y = g_0 x \implies x = g_0^{-1}y$, we also have $x \in Gy$,
 %     hence $Gx \subseteq  Gy$
+%     TODO: WHY?
 \end{proof}
 \begin{corollary}
     If $(X,T)$ is distal and minimal,
     then $E(X,T) \acts X$ is transitive.
 \end{corollary}
-
-
-
-
-
-
-
-

inputs/tutorial_03.tex

@@ -125,7 +125,7 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
     % Then $f$ can be extended to a $G_{\delta}$ set $G \subseteq X$
     % with $Y \subseteq G \subseteq  \overline{Y}$.
     % $\tilde{f}$ and $\id_G$ agree on $Y$.
-    % $Y$ is dense in $G$ and the codomain of $f$ is ltd.\todo{????}
+    % $Y$ is dense in $G$ and the codomain of $f$ is Hausdorff.\todo{????}
     % So $f = \id_G$, i.e.~$G = Y$.
 \end{proof}
 

inputs/tutorial_08.tex

@@ -109,8 +109,6 @@ for some $B_i \in \cB(Y_i)$.
 \end{itemize}
 
 \nr 3
-\todo{Wait for mail}
-
 \begin{lemma}
     Let $X$ be a second-countable topological space.
     Then every base of $X$ contains a countable subset which
@@ -134,6 +132,9 @@ for some $B_i \in \cB(Y_i)$.
     such that $m \in M_{f(m)}$.
     Then $\bigcup_{m \in M} B_{f(m)} = C_n$.
 \end{proof}
+\begin{remark}
+    We don't actually need this.
+\end{remark}
 
 \begin{itemize}
     \item We use the same construction as in exercise 2 (a)
@@ -188,31 +189,71 @@ for some $B_i \in \cB(Y_i)$.
         \]
         is countable as a countable union of countable sets $\lightning$.
 
-    \item Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
-        as in the first part.
-        Clearly $f$ is also continuous with respect to the new topology,
-        so we may assume that $X$ is zero dimensional.
 
-        Let $W \subseteq X$ be such that $f\defon{W}$ is injective
+
-        and $f(W) = f(X)$ (this exists by the axiom of choice).
+        Other proof (without using the existence of a countable clopen
-        Since $f(X)$ is uncountable, so is $W$.
+         basis):
-        By the second point, there exist disjoint clopen sets
+
-        $U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
+        We can cover $X$ by countably many clopen sets of diameter
-        are uncountable.
+        $< \frac{1}{n}$:
-        Inductively construct $U_s$ for $s \in 2^{<\omega}$
+        Cover $X$ with open balls of diameter $< \frac{1}{n}$.
-        as follows:
+        Write each open ball as a union of clopen sets.
-        Suppose that $U_{s}$ has already been chosen.
+        That gives us a cover by clopen sets of diameter $< \frac{1}{n}$.
-        Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
+        As $X$ is Lindelöf, there exists a countable subcover.
-        be disjoint clopen such that $U_{s\concat 1} \cap W$
+        Then continue as in the first proof.
-        and $U_{s\concat  0} \cap W$ are uncountable.
+
-        Such sets exist, since $ U_s \cap W$ is uncountable
+
-        and $U_s$ is a zero dimensional space with the subspace topology.
+    \item Note that this step does not help us to prove the statement.
-        And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
+        It was an error on the exercise sheet.
-        in $U_s$ iff it is clopen in $X$.
+
+%         Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
+%         as in the first part.
+%         Clearly $f$ is also continuous with respect to the new topology,
+%         so we may assume that $X$ is zero dimensional.
+% 
+%         Let $W \subseteq X$ be such that $f\defon{W}$ is injective
+%         and $f(W) = f(X)$ (this exists by the axiom of choice).
+%         Since $f(X)$ is uncountable, so is $W$.
+%         By the second point, there exist disjoint clopen sets
+%         $U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
+%         are uncountable.
+%         Inductively construct $U_s$ for $s \in 2^{<\omega}$
+%         as follows:
+%         Suppose that $U_{s}$ has already been chosen.
+%         Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
+%         be disjoint clopen such that $U_{s\concat 1} \cap W$
+%         and $U_{s\concat  0} \cap W$ are uncountable.
+%         Such sets exist, since $ U_s \cap W$ is uncountable
+%         and $U_s$ is a zero dimensional space with the subspace topology.
+%         And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
+%         in $U_s$ iff it is clopen in $X$.
 
         Clearly this defines a Cantor scheme.
-    \item \todo{TODO}
+    \item Let $Y$ be a Polish space and $A \subseteq Y$ analytic
+        and uncountable.
+        Expand the topology on $Y$ so that $Y$ is zero dimensional
+        and $A$ is still analytic.
 
+        Then there exists a Polish space $X$
+        and a continuous function $f\colon X \to Y$
+        such that $f(X) = A$.
+
+        $A$ is uncountable, so by (2)
+        there exists non-empty disjoint clopen $V_0$, $V_1$
+        such that $V_0 \cap A$ and $V_1 \cap A$
+        are uncountable.
+
+        Let $W_0 = f^{-1}(V_0 \cap A)$ and $W_1 = f^{-1}(V_1 \cap A)$.
+        $ W_0$ and $W_1$ are clopen and disjoint.
+        We can cover $W_0$ with countably many open sets
+        of diameter $\le \frac{1}{n}$
+        and similarly for $W_1$.
+        Then pick open sets such that there image is uncountable.
+
+        Repeating this construction we get a Cantor scheme on $X$.
+        So we get $2^{\N} \overset{s}{\hookrightarrow} X$
+        and by construction of the cantor scheme,
+        we get that $f \circ s$ is injective and continuous.
 \end{itemize}
 
 \nr 4

inputs/tutorial_09.tex

@@ -0,0 +1,159 @@
+\tutorial{09}{2023-12-12}{}
+
+\begin{fact}
+    Let $X,Y$ be topological spaces
+    $X$ (quasi-)compact and
+    $Y$ Hausdorff.
+    Let $f\colon X\to Y$ be a continuous bijection.
+    Then $f$ is a homeomorphism.
+\end{fact}
+\begin{proof}
+    Compact subsets of Hausdorff spaces are closed.
+\end{proof}
+
+
+\subsection{Sheet 8}
+
+Material on topological dynamic
+\begin{itemize}
+    \item Terence Tao's notes on ergodic theory 254 A.
+    \item Furstenberg (uses very different notation!).
+\end{itemize}
+
+
+
+\nr 1
+
+\begin{remark}
+    $\Sigma^1_1$-complete sets are in some sense the ``worst''
+    $\Sigma^1_1$-sets:
+    Deciding whether an element is contained in the
+    $\Sigma^1_1$-complete set
+    is at least as ``hard''  as as for any $\Sigma^1_1$ set.
+
+    In particular, $\Sigma^1_1$-complete sets
+    are not Borel.
+\end{remark}
+
+
+
+Similarly as in \yaref{prop:ifs11} it can be shown that $L \in \Sigma^1_1(X)$:
+Consider $\{(x, \beta) \in X \times \cN : \forall n.~x_{\beta_n} | x_{\beta_{n+1}}\}$.
+This is closed in $X \times \cN$, since it is a countable intersection of clopen
+sets and $L = \proj_X(D)$.
+
+Since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete,
+it suffices to find a Borel map $f\colon \Tr \to X$
+such that $x \in \IF \iff f(x) \in L$.
+Let $\phi\colon \omega^2 + \omega \to \omega$ be bijective
+and let $p_i$ denote the $i$-th prime.
+Define
+\begin{IEEEeqnarray*}{rCl}
+    \psi\colon \omega^{<\omega} &\longrightarrow & \omega \setminus \{0\}  \\
+      (s_0, s_1, \ldots, s_{n-1})&\longmapsto & \prod_{i < n} p_{\phi(\omega \cdot i + s_i)}.
+\end{IEEEeqnarray*}
+Note that $\psi$ is injective and that $s \in \omega^{<\omega}$ is an initial segment
+of $t \in \omega^{<\omega}$ iff $\psi(s) | \psi(t)$.
+Let
+\begin{IEEEeqnarray*}{rCl}
+    f' \colon \Tr &\longrightarrow & \cP(\omega \setminus \{0\}) \\
+    T &\longmapsto & \{\phi(s) : s \in T\}.
+\end{IEEEeqnarray*}
+We can turn this into a function
+$f\colon \Tr \to (\omega \setminus \{0\})^{\omega}$
+by mapping a subset of $\omega \setminus \{0\}$
+to the unique strictly increasing sequence whose elements are from
+that subset
+(appending $\phi(\omega^2 + n), n \in \omega$, if the subset was finite).
+Note that $T \in \IF \iff f(T) \in L$.
+Furthermore $f$ is Borel, since fixing
+a finite initial sequence (i.e.~a basic open set of $(\omega \setminus \{0\})^{\omega}$)
+amounts to a finite number of conditions on the preimage.
+
+\nr 2
+\todo{handwritten}
+
+% Aron
+% 
+% \begin{enumerate}[1.]
+%     \item This is trivial $\sup \sup$.
+%     \item Clearly there are trees of rank $n$ for all $n < \omega$.
+%         Glue them together.
+%         \[
+%         \{(0,0,i,\underbrace{0,\ldots,0}_{i \text{~times}}) | i < \omega\}.
+%         \]
+%     \item Map infinite branches. $\sup$ the $\le $.
+%     \item Induction on $\rho(S)$.
+%         Cofinal subsequences bla bla.
+% \end{enumerate}
+\nr 3
+\begin{itemize}
+    \item $LO(\N) \overset{\text{closed}}{\subseteq} 2^{\N\times \N}$:
+
+        We have $< \in LO(\N)$ iff
+        \begin{itemize}
+            \item $\forall x,y.~ (x \neq y \implies (x< y \lor x > y))$,
+            \item $\forall x.~(x \not < x)$,
+            \item $\forall x,y,x.~(x < y < z \implies x < z)$.
+        \end{itemize}
+
+        Write this with $\bigcap$,
+        i.e.
+        \begin{IEEEeqnarray*}{rCl}
+            LO(\N) &=& \bigcap_{n \in \N} \{R: (n,n) \not\in R\}\\
+                   && \cap \bigcap_{m < n \in \N} (\{R: (n,m) \in R\} \cup \{R: (m,n) \in R\})\\
+                   &&\cap \bigcap_{a,b,c \in \N} (\{R: (a,b) \in R \land (b,c) \in R \implies (a,c) \in R\}.
+        \end{IEEEeqnarray*}
+        This is closed as an intersection of clopen sets.
+    \item We apply \yaref{thm:borel} (iv).
+        Let $\cF \subseteq LO(\N) \times \cN$ be
+        such that the $\cN$-coordinate encodes a strictly
+        decreasing sequence, i.e.~
+        \[(R, s) \in \cF :\iff \forall n \in \N.~(s(n+1), s(n)) \in R.\]
+
+        We have that
+        \[
+        \cF = \bigcap_{n \in \N} \{(R,s) \in LO(\N)\times \cN : (s(n+1), s(n)) \in R\}
+        \]
+        is closed as an intersection of clopen sets.
+
+        Clearly $\pr_{LO(\N)}(\cF)$ is the complement
+        of $WO(\N)$, hence $WO(\N)$ is coanalytic.
+\end{itemize}
+\nr 4
+
+\begin{remark}
+    In the lecture we only look at metrizable flows,
+    so the definitions from the exercise sheet and from
+    the lecture don't agree.
+\end{remark}
+
+\begin{itemize}
+    \item Consider
+        \begin{IEEEeqnarray*}{rCl}
+            \Phi\colon \Z\text{-flows on } X &\longrightarrow & \Homeo(X) \\
+            (\alpha\colon  \Z\times X \to X) &\longmapsto & \alpha(1, \cdot)\\
+            \begin{pmatrix*}[l]
+                \Z\times X &\longrightarrow & X \\
+                (z,x) &\longmapsto & \beta^{z}(x)
+            \end{pmatrix*}&\longmapsfrom & \beta \in \Homeo(X).
+        \end{IEEEeqnarray*}
+        Clearly this has the desired properties.
+    \item We have
+        \begin{IEEEeqnarray*}{Cl}
+            & \Z \circlearrowright X \text{ has a dense orbit}\\
+            \iff& \exists  x \in X.~ \overline{\Z\cdot x} = X\\
+            \iff& \exists x \in X.~\forall U\overset{\text{open}}{\subseteq} X.~\exists z \in \Z.~
+            z \cdot x \in U\\
+            \iff&\exists x \in X.~\forall U \overset{\text{open}}{\subseteq} X.~
+            \exists z \in \Z.~f^z(x) \in U.
+        \end{IEEEeqnarray*}
+    \item Not solved.
+    \item It suffices to check the condition from part (b)
+        for open sets $U$ of a countable basis
+        and points  $x \in X$ belonging to a countable dense subset.
+        Replacing quantifiers by unions resp.~intersections
+        gives that $D$ is Borel.
+\end{itemize}
+
+

logic.sty

@@ -134,6 +134,9 @@
 \DeclareSimpleMathOperator{WO} % well orderings
 
 
+\DeclareSimpleMathOperator{Homeo}
+
+
 \DeclareSimpleMathOperator{osc} % oscillation
 
 \newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}}

logic3.tex

@@ -60,6 +60,7 @@
 \input{inputs/tutorial_06}
 \input{inputs/tutorial_07}
 \input{inputs/tutorial_08}
+\input{inputs/tutorial_09}
 
 \section{Facts}
 \input{inputs/facts}