From 5de0726c3c97b4f483fa60c2e26b4b45d38672a4 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 12 Dec 2023 13:31:59 +0100 Subject: [PATCH] tutorial 09 --- inputs/lecture_12.tex | 1 + inputs/lecture_17.tex | 9 +-- inputs/tutorial_03.tex | 2 +- inputs/tutorial_08.tex | 87 ++++++++++++++++------ inputs/tutorial_09.tex | 159 +++++++++++++++++++++++++++++++++++++++++ logic.sty | 3 + logic3.tex | 1 + 7 files changed, 230 insertions(+), 32 deletions(-) create mode 100644 inputs/tutorial_09.tex diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index ec91c05..5b9d551 100644 --- a/inputs/lecture_12.tex +++ b/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} diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index 7c7f253..05cd27d 100644 --- a/inputs/lecture_17.tex +++ b/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} - - - - - - - - diff --git a/inputs/tutorial_03.tex b/inputs/tutorial_03.tex index d5e685f..5aeb648 100644 --- a/inputs/tutorial_03.tex +++ b/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} diff --git a/inputs/tutorial_08.tex b/inputs/tutorial_08.tex index 91be24e..ad112a8 100644 --- a/inputs/tutorial_08.tex +++ b/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). - 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 $X$. + + Other proof (without using the existence of a countable clopen + basis): + + We can cover $X$ by countably many clopen sets of diameter + $< \frac{1}{n}$: + Cover $X$ with open balls of diameter $< \frac{1}{n}$. + Write each open ball as a union of clopen sets. + That gives us a cover by clopen sets of diameter $< \frac{1}{n}$. + As $X$ is Lindelöf, there exists a countable subcover. + Then continue as in the first proof. + + + \item Note that this step does not help us to prove the statement. + It was an error on the exercise sheet. + +% 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 diff --git a/inputs/tutorial_09.tex b/inputs/tutorial_09.tex new file mode 100644 index 0000000..9e50816 --- /dev/null +++ b/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} + + diff --git a/logic.sty b/logic.sty index eb78b62..22104a7 100644 --- a/logic.sty +++ b/logic.sty @@ -134,6 +134,9 @@ \DeclareSimpleMathOperator{WO} % well orderings +\DeclareSimpleMathOperator{Homeo} + + \DeclareSimpleMathOperator{osc} % oscillation \newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}} diff --git a/logic3.tex b/logic3.tex index ba544f9..f0d961a 100644 --- a/logic3.tex +++ b/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}