diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 53092a8..decf836 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -233,6 +233,7 @@ Recall: correspond to metrics witnessing that the flow is isometric. \end{remark} \begin{proposition} + \label{prop:isomextdistal} An isometric extension of a distal flow is distal. \end{proposition} \begin{proof} @@ -263,11 +264,12 @@ Recall: % TODO THE inverse limit is A limit of $\Sigma$ iff \[ - \forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2). + \forall x_1 \neq x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2). \] \end{definition} \begin{proposition} + \label{prop:limitdistal} A limit of distal flows is distal. \end{proposition} \begin{proof} diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex index c0a5161..457bb11 100644 --- a/inputs/lecture_16.tex +++ b/inputs/lecture_16.tex @@ -1,5 +1,4 @@ \lecture{16}{2023-12-08}{} -% TODO ANKI-MARKER $X$ is always compact metrizable. @@ -18,16 +17,19 @@ $X$ is always compact metrizable. % and $h\colon x \mapsto x + \alpha$.\footnote{Again $x + \alpha$ denotes $x \cdot \alpha$ in $\C$.} % \end{example} \begin{proof} + % TODO TODO TODO Think! The action of $1$ determines $h$. Consider \[ - \{h^n : n \in \Z\} \subseteq \cC(X,X) = \{f\colon X \to X : f \text{ continuous}\}, + \{h^n : n \in \Z\} \subseteq \cC(X,X)\gist{ = \{f\colon X \to X : f \text{ continuous}\}}{}, \] where the topology is the uniform convergence topology. % TODO REF EXERCISE Let $G = \overline{ \{h^n : n \in \Z\} } \subseteq \cC(X,X)$. - Since + Since the family $\{h^n : n \in \Z\}$ is uniformly equicontinuous, + i.e.~ \[ - \forall \epsilon > 0.~\exists \delta > 0.~d(x,y) < \delta \implies \forall n.~d(h^n(x), h^n(y)) < \epsilon + \forall \epsilon > 0.~\exists \delta > 0.~d(x,y) < \delta \implies \forall n.~d(h^n(x), h^n(y)) < \epsilon, + % Here we use isometric \] we have by the Arzel\`a-Ascoli-Theorem % TODO REF that $G$ is compact. @@ -126,8 +128,8 @@ $X$ is always compact metrizable. Every quasi-isometric flow is distal. \end{corollary} \begin{proof} - \todo{TODO} - % The trivial flow is distal. + The trivial flow is distal. + Apply \yaref{prop:isomextdistal} and \yaref{prop:limitdistal}. \end{proof} \begin{theorem}[Furstenberg] @@ -199,7 +201,8 @@ The Hilbert cube $\bH = [0,1]^{\N}$ embeds all compact metric spaces. Thus we can consider $K(\bH)$, the space of compact subsets of $\bH$. -$K(\bH)$ is a Polish space.\todo{Exercise} +$K(\bH)$ is a Polish space.\footnote{cf.~\yaref{s9e2}, \yaref{s12e4}} +% TODO LEARN EXERCISES Consider $K(\bH^2)$. A flow $\Z \acts X$ corresponds to the graph of \begin{IEEEeqnarray*}{rCl} diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index b9e7020..9ce36cd 100644 --- a/inputs/lecture_17.tex +++ b/inputs/lecture_17.tex @@ -1,4 +1,5 @@ \subsection{The Ellis semigroup} +% TODO ANKI-MARKER \lecture{17}{2023-12-12}{The Ellis semigroup} Let $(X, d)$ be a compact metric space diff --git a/inputs/lecture_18.tex b/inputs/lecture_18.tex index 3587f49..d801fc4 100644 --- a/inputs/lecture_18.tex +++ b/inputs/lecture_18.tex @@ -68,61 +68,62 @@ This will follow from the following lemma: we get $U_{\epsilon}(x') \subseteq U_a(x_1) \cap U_b(x_2)$. \end{refproof} \begin{refproof}{lem:ftophelper}% - \footnote{This was not covered in class.} + \notexaminable{\footnote{This was not covered in class.} - Let $T = \bigcup_n T_n$,% TODO Why does this exist? - $T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and - let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$. - Take $b$ such that $F(x,x') < b < a$. - Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$ - is open in $G(x,x')$ - and since $F(x,x') < b$ we have $U \neq \emptyset$. - \begin{claim} - There exists $n$ such that - \[ - \forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset. - \] - \end{claim} - \begin{subproof} - Suppose not. - Then for all $n$, there is $(u_n, u_n') \in G(x,x')$ - with - \[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\] - Note that the RHS is closed. - For $m > n$ we have - $T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$ - since $T_n \subseteq T_m$. - By compactness of $X$, - there exists $v,v'$ and some subsequence - such that $(u_{n_k}, u'_{n_k}) \to (v,v')$. + Let $T = \bigcup_n T_n$,% TODO Why does this exist? + $T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and + let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$. + Take $b$ such that $F(x,x') < b < a$. + Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$ + is open in $G(x,x')$ + and since $F(x,x') < b$ we have $U \neq \emptyset$. + \begin{claim} + There exists $n$ such that + \[ + \forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset. + \] + \end{claim} + \begin{subproof} + Suppose not. + Then for all $n$, there is $(u_n, u_n') \in G(x,x')$ + with + \[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\] + Note that the RHS is closed. + For $m > n$ we have + $T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$ + since $T_n \subseteq T_m$. + By compactness of $X$, + there exists $v,v'$ and some subsequence + such that $(u_{n_k}, u'_{n_k}) \to (v,v')$. - So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$, - hence $T(v,v') \cap U = \emptyset$, - so $G(v,v') \cap U = \emptyset$. - But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$. - \end{subproof} - The map - \begin{IEEEeqnarray*}{rCl} - T\times X&\longrightarrow & X \\ - (t,x) &\longmapsto & tx - \end{IEEEeqnarray*} - is continuous. - Since $T_n$ is compact, - we have that $\{(x,t) \mapsto tx : t \in T_n\}$ - is equicontinuous.\todo{Sheet 11} - So there is $\epsilon > 0$ such that - $d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$ - for all $t \in T_n$. + So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$, + hence $T(v,v') \cap U = \emptyset$, + so $G(v,v') \cap U = \emptyset$. + But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$. + \end{subproof} + The map + \begin{IEEEeqnarray*}{rCl} + T\times X&\longrightarrow & X \\ + (t,x) &\longmapsto & tx + \end{IEEEeqnarray*} + is continuous. + Since $T_n$ is compact, + we have that $\{(x,t) \mapsto tx : t \in T_n\}$ + is equicontinuous.\todo{Sheet 11} + So there is $\epsilon > 0$ such that + $d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$ + for all $t \in T_n$. - Suppose now that $F(x', x'') < \epsilon$. - Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$, - hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in T_n$. - Since $(t_0x, t_0x') \in G(x,x')$, - there is $t_1 \in T_n$ - with $(t_1t_0x, t_1t_0x') \in U$, - i.e.~$d(t_1t_0x, t_1t_0x') < b$ - and therefore - $F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$. + Suppose now that $F(x', x'') < \epsilon$. + Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$, + hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in T_n$. + Since $(t_0x, t_0x') \in G(x,x')$, + there is $t_1 \in T_n$ + with $(t_1t_0x, t_1t_0x') \in U$, + i.e.~$d(t_1t_0x, t_1t_0x') < b$ + and therefore + $F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$. + } \end{refproof} Now assume $Z = \{\star\}$. diff --git a/inputs/lecture_22.tex b/inputs/lecture_22.tex index 0cd13ac..9e93df4 100644 --- a/inputs/lecture_22.tex +++ b/inputs/lecture_22.tex @@ -145,48 +145,54 @@ For this we define This is the same as for iterated skew shifts. % TODO since for $\overline{x}, \overline{y} \in \mathbb{K}^I$, % $d(x_\alpha, y_\alpha) = d((f(\overline{x})_\alpha, (f(\overline{y})_\alpha))$. - \item Minimality: + \item Minimality:% + \gist{% + \footnote{This is not relevant for the exam.} - Let $\langle E_n : n < \omega \rangle$ - be an enumeration of a countable basis for $\mathbb{K}^I$. + Let $\langle E_n : n < \omega \rangle$ + be an enumeration of a countable basis for $\mathbb{K}^I$. - For all $n$ let - \[ - U_n \coloneqq \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\} - \] - where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$. + For all $n$ let + \[ + U_n \coloneqq \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\} + \] + where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$. - Beleznay and Foreman showed that $U_n$ is open - and dense for all $n$. + Beleznay and Foreman showed that $U_n$ is open + and dense for all $n$. - So if $\overline{f} \in \bigcap_{n} U_n$, then $\overline{1}$ - 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}). + So if $\overline{f} \in \bigcap_{n} U_n$, then $\overline{1}$ + 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}). + }{ Not relevant for the exam.} - \item Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$. - Consider the flows we get from $(f_i)_{i < j}$ - resp.~$(f_i)_{i \le j}$ - denoted by $X_{ j+1.~z_k = 1,\\ - &&d(f^{k_m}(\overline{x_m}), f^{k_m}(\overline{z})) < \epsilon \text{ and }\\ - &&d(f^{k_n}(\overline{x_n}), f^{k_n}(\overline{z})) < \epsilon\\ - &&\} - \end{IEEEeqnarray*} - Beleznay and Foreman show that this is open and dense.% - \footnote{This is not relevant for the exam.} - % TODO similarities to the lemma used today + Fix a countable dense set $(\overline{x_n})$ in $\mathbb{K}^I$. + For $\epsilon \in \Q$ let + \begin{IEEEeqnarray*}{rCl} + V_{j,m,n,\epsilon} &\coloneqq \{\overline{f} \in \mathbb{K}_I :& \\ + &&\text{if } \Pi_{j+1}(\overline{x}_n) = \Pi_{j+1}(\overline{x_m}),\\ + &&\text{then there are $k_m$, $k_n$, $\overline{z}$ such that}\\ + &&\pi_j(\overline{x_n}) = \pi_j(\overline{z}), \forall k> j+1.~z_k = 1,\\ + &&d(f^{k_m}(\overline{x_m}), f^{k_m}(\overline{z})) < \epsilon \text{ and }\\ + &&d(f^{k_n}(\overline{x_n}), f^{k_n}(\overline{z})) < \epsilon\\ + &&\} + \end{IEEEeqnarray*} + Beleznay and Foreman show that this is open and dense.% + % TODO similarities to the lemma used today + }{ Not relevant for the exam.} \end{itemize} \end{proof} diff --git a/inputs/lecture_23.tex b/inputs/lecture_23.tex index c751b74..c26f5b4 100644 --- a/inputs/lecture_23.tex +++ b/inputs/lecture_23.tex @@ -24,47 +24,49 @@ Let $I$ be a linear order \end{theorem} \begin{proof}[sketch] - Consider $\WO(\N) \subset \LO(\N)$. - We know that this is $\Pi_1^1$-complete. % TODO ref + \notexaminable{% + Consider $\WO(\N) \subset \LO(\N)$. + We know that this is $\Pi_1^1$-complete. % TODO ref - Let - \begin{IEEEeqnarray*}{rCll} - S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\ - &&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\} - \end{IEEEeqnarray*} - \todo{Exercise sheet 12} - $S$ is Borel. + Let + \begin{IEEEeqnarray*}{rCll} + S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\ + &&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\} + \end{IEEEeqnarray*} + \todo{Exercise sheet 12} + $S$ is Borel. - We will % TODO ? - construct a reduction - \begin{IEEEeqnarray*}{rCl} - M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\ - % \alpha &\longmapsto & M(\alpha) - \end{IEEEeqnarray*} - We want that $\alpha \in \WO(\N) \iff M(\alpha)$ - codes a distal minimal flow of rank $\alpha$. + We will % TODO ? + construct a reduction + \begin{IEEEeqnarray*}{rCl} + M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\ + % \alpha &\longmapsto & M(\alpha) + \end{IEEEeqnarray*} + We want that $\alpha \in \WO(\N) \iff M(\alpha)$ + codes a distal minimal flow of rank $\alpha$. - \begin{enumerate}[1.] - \item For any $\alpha \in S$, $M(\alpha)$ is a code for - a flow which is coded by a generic $(f_i)_{i \in I}$. - Specifically we will take a flow - corresponding to some $(f_i)_{i \in I}$ - which is in the intersection of all - $U_n$, $V_{j,m,n,\frac{p}{q}}$ - (cf.~proof of \yaref{thm:distalminimalofallranks}). + \begin{enumerate}[1.] + \item For any $\alpha \in S$, $M(\alpha)$ is a code for + a flow which is coded by a generic $(f_i)_{i \in I}$. + Specifically we will take a flow + corresponding to some $(f_i)_{i \in I}$ + which is in the intersection of all + $U_n$, $V_{j,m,n,\frac{p}{q}}$ + (cf.~proof of \yaref{thm:distalminimalofallranks}). - \item If $\alpha \in \WO(\N)$, - then additionally $(f_i)_{i \in I}$ will code - a distal minimal flow of ordertype $\alpha$. - \end{enumerate} - - One can get a Borel map $S \ni \alpha \mapsto \{T_n^{\alpha} : n \in \N\}$, - such that $T^{\alpha}_n$ is closed, - $T^{\alpha}_n \neq \emptyset$, $\diam(T^\alpha_n) \xrightarrow{n \to \infty} 0$, - $T^\alpha_{n+1} \subseteq T^\alpha_n$, - $T^{\alpha}_n \subseteq W^{\alpha}_n$, - where $W^{\alpha}_n$ is an enumeration of $U_m^\alpha$,$V^\alpha_{j,m,n,\frac{p}{q}}$. - Then $(f_i)_{i \in I} \in \bigcap_{n} T_{n}^\alpha$. + \item If $\alpha \in \WO(\N)$, + then additionally $(f_i)_{i \in I}$ will code + a distal minimal flow of ordertype $\alpha$. + \end{enumerate} + + One can get a Borel map $S \ni \alpha \mapsto \{T_n^{\alpha} : n \in \N\}$, + such that $T^{\alpha}_n$ is closed, + $T^{\alpha}_n \neq \emptyset$, $\diam(T^\alpha_n) \xrightarrow{n \to \infty} 0$, + $T^\alpha_{n+1} \subseteq T^\alpha_n$, + $T^{\alpha}_n \subseteq W^{\alpha}_n$, + where $W^{\alpha}_n$ is an enumeration of $U_m^\alpha$,$V^\alpha_{j,m,n,\frac{p}{q}}$. + Then $(f_i)_{i \in I} \in \bigcap_{n} T_{n}^\alpha$. + } \end{proof} \begin{lemma} Let $\{(X_\xi, T) : \xi \le \eta\}$ be a normal diff --git a/logic.sty b/logic.sty index 057fb5c..e458dac 100644 --- a/logic.sty +++ b/logic.sty @@ -156,5 +156,6 @@ \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}} \newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}} \newcommand\nr[1]{\subsubsection{Exercise #1}\yalabel{Sheet \arabic{subsection}, Exercise #1}{E \arabic{subsection}.#1}{s\arabic{subsection}e#1}} +\newcommand\notexaminable[1]{\gist{\footnote{Not relevant for the exam.}#1}{Not relevant for the exam.}} \usepackage[bibfile=bibliography/references.bib, imagefile=bibliography/images.bib]{mkessler-bibliography}