diff --git a/inputs/lecture_01.tex b/inputs/lecture_01.tex index df7eb8c..acab3f3 100644 --- a/inputs/lecture_01.tex +++ b/inputs/lecture_01.tex @@ -185,11 +185,11 @@ suffices to show that open balls in one metric are unions of open balls in the o \begin{definition}[Our favourite Polish spaces] \leavevmode \begin{itemize} - \item $2^{\omega}$ is called the \vocab{Cantor set}. + \item $2^{\N}$ is called the \vocab{Cantor set}. (Consider $2$ with the discrete topology) - \item $\omega^{\omega}$ is called the \vocab{Baire space}. - ($\omega$ with descrete topology) - \item $[0,1]^{\omega}$ is called the \vocab{Hilbert cube}. + \item $\cN \coloneqq \N^{\N}$ is called the \vocab{Baire space}. + ($\N$ with descrete topology) + \item $\mathbb{H} \coloneqq [0,1]^{\N}$ is called the \vocab{Hilbert cube}. ($[0,1] \subseteq \R$ with the usual topology) \end{itemize} \end{definition} diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex index 2353482..88f85bc 100644 --- a/inputs/lecture_03.tex +++ b/inputs/lecture_03.tex @@ -205,8 +205,8 @@ \[ D \subseteq \N^\N \mathbin{\text{\reflectbox{$\coloneqq$}}} \cN \] - and a continuous bijection from - $D$ onto $X$ (the inverse does not need to be continuous). + and a continuous bijection $f\colon D \to X$ + (the inverse does not need to be continuous). Moreover there is a continuous surjection $g: \cN \to X$ extending $f$. diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex index 8267a0f..4b7ff7a 100644 --- a/inputs/lecture_04.tex +++ b/inputs/lecture_04.tex @@ -64,7 +64,7 @@ Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x\defon{n} = s\}$. Clearly $S$ is a pruned tree. - Moreover, since $D$ is closed, we have that (cf.~\yaref{s3e1}) + Moreover, since $D$ is closed, we have that\footnote{cf.~\yaref{s3e1}} \[ D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}. \] diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 97d494d..c58fca2 100644 --- a/inputs/lecture_08.tex +++ b/inputs/lecture_08.tex @@ -78,18 +78,17 @@ where $X$ is a metrizable, usually second countable space. we also have $A = \bigcup_n A_n'$ with $A'_n \in \Pi^0_{\xi_n}(X)$. - We construct a $(2^\omega)^\omega \cong 2^\omega$-universal set + We construct a $(2^{\omega \times \omega}) \cong 2^\omega$-universal set for $\Sigma^0_\xi(X)$. - For $(y_n) \in (2^\omega)^\omega$ + For $(y_{m,n}) \in (2^{\omega \times \omega})$ and $x \in X$ - we set $((y_n), x) \in \cU$ - iff $\exists n.~(y_n, x) \in U_{\xi_n}$, - i.e.~iff $\exists n.~x \in (U_{\xi_n})_{y_n}$. + we set $((y_{m,n}), x) \in \cU$ + iff $\exists n.~((y_{m,n})_{m < \omega}, x) \in U_{\xi_n}$, + i.e.~iff $\exists n.~x \in (U_{\xi_n})_{(y_{m,n})_{m < \omega}}$. Let $A \in \Sigma^0_\xi(X)$. Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$. - % TODO - Furthermore $\cU \in \Sigma^0_{\xi}((2^\omega)^\omega \times X)$. + Furthermore $\cU \in \Sigma^0_{\xi}((2^{\omega \times \omega} \times X)$. \end{proof} \begin{remark} Since $2^{\omega}$ embeds diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex index d3b2af2..55c45e5 100644 --- a/inputs/lecture_09.tex +++ b/inputs/lecture_09.tex @@ -45,9 +45,8 @@ We will see that not every analytic set is Borel. \begin{remark} In the definition we can replace the assertion that $f$ is continuous - by the weaker assertion of $f$ being Borel. - \todo{Copy exercise from sheet 5} - % TODO WHY? + by the weaker assertion of $f$ being Borel.% + \footnote{use \yaref{thm:clopenize}, cf.~\yaref{s6e2}} \end{remark} \begin{theorem} @@ -155,7 +154,7 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$. \begin{proof} Take $\cU \subseteq Y \times X \times \cN$ which is $Y$-universal for $\Pi^0_1(X \times \cN)$. - Let $\cV \coloneqq \proj_{Y \times Y}(\cU)$. + Let $\cV \coloneqq \proj_{Y \times X}(\cU)$. Then $\cV$ is $Y$-universal for $\Sigma^1_1(X)$: \begin{itemize} \item $\cV \in \Sigma^1_1(Y \times X)$ diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex index a8fb5ea..514ee1f 100644 --- a/inputs/lecture_10.tex +++ b/inputs/lecture_10.tex @@ -34,7 +34,8 @@ we need the following definition: \begin{lemma} \label{lem:lusinsephelp} If $P = \bigcup_{m < \omega} P_m$, $Q = \bigcup_{n < \omega} Q_n$ are such that - for any $m, n$ the sets $P_m$ and $Q_n$ are Borel separable. + for any $m, n$ the sets $P_m$ and $Q_n$ are Borel separable, + then $P$ and $Q$ are Borel separable. \end{lemma} \begin{proof} For all $m, n$ pick $R_{m,n}$ Borel, @@ -42,7 +43,7 @@ we need the following definition: and $Q_n \cap R_{m,n} = \emptyset$. Then $R = \bigcup_m \bigcap_n R_{m,n}$ has the desired property - that $R \subseteq R$ and $R \cap Q = \emptyset$. + that $P \subseteq R$ and $R \cap Q = \emptyset$. \end{proof} \begin{notation} @@ -60,7 +61,7 @@ we need the following definition: Write $A_s \coloneqq f(\cN_s)$ and $B_s \coloneqq g(\cN_s)$. Note that $A_s = \bigcup_m A_{s\concat m}$ - and $B_ns = \bigcup_{n < \omega} B_{s\concat n}$. + and $B_s = \bigcup_{n < \omega} B_{s\concat n}$. In particular $A = \bigcup_{m < \omega} A_{\underbrace{\langle m \rangle}_{\in \omega^1}}$ diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index a495a93..019002f 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -183,7 +183,7 @@ i.e.}{} Let $X$ be Polish and $C \subseteq X$ coanalytic. Then $\phi\colon C \to \Ord$ is a \vocab[Rank!$\Pi^1_1$-rank]{$\Pi^1_1$-rank} - provided that $\le^\ast$ and $<^\ast$ are coanalytic subsets of $X \times X$, + provided that $\le^\ast_\phi$ and $<^\ast_\phi$ are coanalytic subsets of $X \times X$, where $x \le^\ast_{\phi} y$ iff diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 7b5989a..e7d2ce2 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -104,7 +104,7 @@ \forall x \in X.~(\exists n.~(x,n) \in R \iff \exists! n.~(x,n)\in R^\ast). \] We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.% - \footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}} + \footnote{Wikimedia has a nice \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{picture}.} \end{theorem} \begin{proof} Let $\phi\colon R \to \Ord$ diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 53092a8..fa9c486 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -55,7 +55,7 @@ then $\xi = \alpha$ and \[ - E_\xi = E_\alpha = \bigcup_{\eta < \alpha} + E_\xi = E_\alpha = \bigcup_{\eta < \alpha} E_\eta \] is a countable union of Borel sets by the previous case. \end{itemize} @@ -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..fdd1ed3 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] @@ -138,7 +140,7 @@ By Zorn's lemma, this will follow from \begin{theorem}[Furstenberg] \label{thm:l16:3} Let $(X, T)$ be a minimal distal flow - and let $(Y, T)$ be a proper factor. + and let $(Y, T)$ be a proper factor.% \footnote{i.e.~$(X,T)$ and $(Y,T)$ are not isomorphic} Then there is another factor $(Z,T)$ of $(X,T)$ which is a proper isometric extension of $Y$. @@ -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..c9b265d 100644 --- a/inputs/lecture_17.tex +++ b/inputs/lecture_17.tex @@ -15,16 +15,16 @@ U_{\epsilon}(x,y) \coloneqq \{f \in X^X : d(x,f(y)) < \epsilon\}. \] for all $x,y \in X$, $\epsilon > 0$. -$X^{X}$ is a compact Hausdorff space. -\begin{remark} - \todo{Copy from exercise sheet 10} +$X^{X}$ is a compact Hausdorff space.\footnote{cf.~\href{https://en.wikipedia.org/wiki/Tychonoff's_theorem}{Tychonoff's theorem}} +\begin{remark}% + \footnote{cf.~\yaref{s11e1}} Let $f_0 \in X^X$ be fixed. \begin{itemize} \item $X^X \ni f \mapsto f \circ f_0$ is continuous: Consider $\{f : f \circ f_0 \in U_{\epsilon}(x,y)\}$. - We have $ff_0 \in U_{\epsilon}(x,y)$ + We have $f \circ f_0 \in U_{\epsilon}(x,y)$ iff $f \in U_\epsilon(x,f_0(y))$. \item Fix $x_0 \in X$. Then $f \mapsto f(x_0)$ is continuous. @@ -34,27 +34,35 @@ $X^{X}$ is a compact Hausdorff space. \end{remark} \begin{definition} -Let $(X,T)$ be a flow. -Then the \vocab{Ellis semigroup} -is defined by -$E(X,T) \coloneqq \overline{T} \subseteq X^X$, -i.e.~identify $t \in T$ with $x \mapsto tx$ -and take the closure in $X^X$. +\gist{% + Let $(X,T)$ be a flow. + Then the \vocab{Ellis semigroup} + is defined by + $E(X,T) \coloneqq \overline{T} \subseteq X^X$, + i.e.~identify $t \in T$ with $x \mapsto tx$ + and take the closure in $X^X$. +}{% + The \vocab{Ellis semigroup} of a flow $(X,T)$ + is $E(X,T) \coloneqq \overline{T} \subseteq X^X$. +} \end{definition} $E(X,T)$ is compact and Hausdorff, since $X^X$ has these properties. -Properties of $(X,T)$ translate to properties of $E(X,T)$: -\begin{goal} - We want to show that if $(X,T)$ is distal, - then $E(X,T)$ is a group. -\end{goal} +\gist{ + Properties of $(X,T)$ translate to properties of $E(X,T)$: + \begin{goal} + We want to show that if $(X,T)$ is distal, + then $E(X,T)$ is a group. + \end{goal} +}{} \begin{proposition} $E(X,T)$ is a semigroup, i.e.~closed under composition. \end{proposition} \begin{proof} +\gist{ Let $G \coloneqq E(X,T)$. Take $t \in T$. We want to show that $tG \subseteq G$, i.e.~for all $h \in G$ we have $th \in G$. @@ -63,7 +71,8 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$: since $t^{-1}$ is continuous and $G$ is compact. - Then $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$. + It is $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$. + So $G = \overline{T} \subseteq t^{-1}G$. Hence $tG \subseteq G$. @@ -74,7 +83,7 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$: \] \end{claim} \begin{subproof} - \todo{Homework} + Cf.~\yaref{s11e1} \end{subproof} Let $g \in G$. @@ -87,17 +96,34 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$: Since $G$ compact, and $Tg \subseteq G$, we have $ \overline{Tg} \subseteq G$. +}{ + $G \coloneqq E(X,T)$. + \begin{itemize} + \item $\forall t \in T. ~ tG \subseteq G$: + \begin{itemize} + \item $t^{-1}G$ is compact. + \item $T \subseteq t^{-1}G$, + \item $\leadsto G = \overline{T} \subseteq t^{-1}G$, + i.e.~$tG \subseteq G$. + \end{itemize} + \item $\forall g \in G.~\overline{T}g = \overline{Tg}$ (cts.~map from compact to Hausdorff) + \item $\forall g \in G.~Gg \subseteq G$ : + $Gg = \overline{T}g = \overline{Tg} \overset{G \text{ compact}, Tg \subseteq G}{\subseteq} G$. + \end{itemize} +} \end{proof} \begin{definition} A \vocab{compact semigroup} $S$ - is a nonempty semigroup with a compact - Hausdorff topology, + is a nonempty semigroup\footnote{may not contain inverses or the identity} + with a compact Hausdorff topology, such that $S \ni x \mapsto xs$ is continuous for all $s$. \end{definition} +\gist{ \begin{example} The Ellis semigroup is a compact semigroup. \end{example} +}{} \begin{lemma}[Ellis–Numakura] \yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura} @@ -106,6 +132,7 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$: i.e.~$f$ such that $f^2 = f$. \end{lemma} \begin{proof} +\gist{ Using Zorn's lemma, take a $\subseteq$-minimal compact subsemigroup $R$ of $S$ and let $s \in R$. @@ -122,25 +149,39 @@ Properties of $(X,T)$ translate to properties of $E(X,T)$: Thus $P = R$ by minimality, so $s \in P$, i.e.~$s^2 = s$. +}{ + \begin{itemize} + \item Take $R \subseteq S$ minimal compact subsemigroup (Zorn), + $s \in R$. + \item $Rs \subseteq R \implies Rs = R$. + \item $P \coloneqq \{x \in R : xs = s\}$: + \begin{itemize} + \item $P \neq \emptyset$, since $s \in Rs$ + \item $P$ compact, since $P = \alpha^{-1}(s) \cap R$, + $\alpha: x \mapsto xs$ cts. + \item $P = R \implies s^2 = s$. + \end{itemize} + \end{itemize}% +} \end{proof} -The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$, -since we already know that it has an identity, -in fact we have chosen $R = \{1\}$ in the proof. -But it is interesting for other semigroups. +\gist{ + The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$, + since we already know that it has an identity. + %in fact we might have chosen $R = \{1\}$ in the proof. + But it is interesting for other semigroups. +}{} \begin{theorem}[Ellis] $(X,T)$ is distal iff $E(X,T)$ is a group. \end{theorem} \begin{proof} - - Let $G \coloneqq E(X,T)$. - Let $d$ be a metric on $X$. - + Let $G \coloneqq E(X,T)$ and let $d$ be a metric on $X$. +\gist{ For all $g \in G$ we need to show that $x \mapsto gx$ is bijective. If we had $gx = gy$, then $d(gx,gy) = 0$. - Then $\inf d(tx,ty) = 0$, but the flow is distal, + Then $\inf_{t \in T} d(tx,ty) = 0$, but the flow is distal, hence $x = y$. Let $g \in G$. Consider the compact semigroup $\Gamma \coloneqq Gg$. @@ -150,25 +191,46 @@ But it is interesting for other semigroups. Since $f$ is injective, we get that $x = f(x)$, i.e.~$f = \id$. - Take $g' \in G$ such that $f = g' \circ g$.% - %\footnote{This exists since $f \in Gg$.} + Since $f \in Gg$, there exists $g' \in G$ such that $f = g' \circ g$. It is $g' = g'gg'$, so $\forall x .~g'(x) = g'(g g'(x))$. Hence $g'$ is bijective and $x = gg'(x)$, i.e.~$g g' = \id$. +}{ + \begin{itemize} + \item $x \mapsto gx$ injective for all $g \in G$: + \[gx = gy + \implies d(gx,gy) = 0 + \implies \inf_{t \in T} d(tx, ty) = 0 + \overset{\text{distal}}{\implies} x = y. + \] + \item Fix $g \in G$. + \begin{itemize} + \item $\Gamma \coloneqq Gg$ is a compact semigroup. + \item $\exists f\in \Gamma.~f^2 = f$ (\yaref{lem:ellisnumakura}) + \item $f$ is injective, hence $f = \id$. + \item Take $g'$ such that $g'g = f$. $g'gg' = g' \implies gg' = \id$. + \end{itemize} + \end{itemize} +} - \todo{The other direction is left as an easy exercise.} +\gist{ + On the other hand if $(x_0,x_1)$ is proximal, + then there exists $g \in G$ such that $gx_0 = gx_1$.% + \footnote{cf.~\yaref{s11e1} (e)} + It follows that an inverse to $g$ can not exist. +}{If $(x_0,x_1)$ is proximal, there is $g \in G$ with $gx_0 = gx_1$, i.e.~no inverse to $g$.} \end{proof} -Let $(X,T)$ be a flow. -Then by Zorn's lemma, there exists $X_0 \subseteq X$ -such that $(X_0, T)$ is minimal. -In particular, -for $x \in X$ and $\overline{Tx} = Y$ -we have that $(Y,T)$ is a flow. -However if we pick $y \in Y$, $Ty$ might not be dense. +% Let $(X,T)$ be a flow. +% Then by Zorn's lemma, there exists $X_0 \subseteq X$ +% such that $(X_0, T)$ is minimal. +% In particular, +% for $x \in X$ and $\overline{Tx} = Y$ +% we have that $(Y,T)$ is a flow. +% However if we pick $y \in Y$, $Ty$ might not be dense. % TODO: question! % TODO: think about this! % We want to a minimal subflow in a nice way: @@ -181,6 +243,7 @@ However if we pick $y \in Y$, $Ty$ might not be dense. \end{theorem} \begin{proof} Let $G = E(X,T)$. +\gist{ Note that for all $x \in X$, we have that $Gx \subseteq X$ is compact and invariant under the action of $G$. @@ -188,17 +251,18 @@ However if we pick $y \in Y$, $Ty$ might not be dense. Since $G$ is a group, the orbits partition $X$.% \footnote{Note that in general this does not hold for semigroups.} - % Clearly the sets $Gx$ cover $X$. We want to show that they - % partition $X$. - % It suffices to show that $y \in Gx \implies Gy = Gx$. - -% Take some $y \in Gx$. -% Recall that $\overline{Ty} = \overline{T} y = Gy$. -% We have $\overline{Ty} \subseteq Gx$, -% 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? + We need to show that $(Gx, T)$ is minimal. + Suppose that $y \in Gx$, i.e.~$Gx = Gy$. + Since $g \mapsto gy$ is continuous, + we have $Gx = Gy = \overline{T}y = \overline{Ty}$, + so $Ty$ is dense in $Gx$. +}{ + \begin{itemize} + \item $G$ is a group, so the $G$-orbits partition $X$. + \item Let $y \in Gx$, then $Gx \overset{y \in Gx}{=} Gy \overset{\text{def}}{=} \overline{T}y \overset{g \mapsto gy \text{ cts.}}{=} \overline{Ty}$, + i.e.~$(Gx,T)$ is minimal. + \end{itemize} +} \end{proof} \begin{corollary} If $(X,T)$ is distal and minimal, diff --git a/inputs/lecture_18.tex b/inputs/lecture_18.tex index 3587f49..ab5a461 100644 --- a/inputs/lecture_18.tex +++ b/inputs/lecture_18.tex @@ -3,14 +3,6 @@ The goal for this lecture is to give a very rough sketch of \yaref{thm:l16:3} in the case of $|Z| = 1$. -% \begin{theorem}[Furstenberg] -% Let $(X, T)$ be a minimal distal flow -% and let $(Z,T)$ be a proper factor of $X$% -% \footnote{i.e.~$(X,T)$ and $(Z,T)$ are not isomorphic.} -% Then three is another factor $(Y,T)$ of $(X,T)$ -% which is a proper isometric extension of $Z$. -% \end{theorem} - Let $(X,T)$ be a distal flow. Then $G \coloneqq E(X,T)$ is a group. @@ -26,7 +18,9 @@ F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}. \item $F(x,x') = F(x', x)$, \item $F(x,x') \ge 0$ and $F(x,x') = 0$ iff $x = x'$. \item $F(gx, gx') = F(x,x')$ since $G$ is a group. - \item $F$ is an upper semi-continuous function on $X^2$, + \item $F$ is an \vocab{upper semi-continuous}\footnote{% + Wikimedia has a nice \href{https://upload.wikimedia.org/wikipedia/commons/c/c0/Upper_semi.svg}{picture}.} + function on $X^2$, i.e.~$\forall a \in R.~\{(x,x') \in X^2 : F(x,x') < a\} \overset{\text{open}}{\subseteq} X^2$. This holds because $F$ is the infimum of continuous functions @@ -52,77 +46,83 @@ This will follow from the following lemma: \begin{lemma} \label{lem:ftophelper} Let $F(x,x') < a$. + \gist{% Then there exists $\epsilon > 0$ such that whenever $F(x',x'') < \epsilon$, then $F(x,x'') < a$. + }{Then $\exists \epsilon > 0.~\forall x''.~F(x',x'') < \epsilon \implies F(x,x'') < a$.} \end{lemma} \begin{refproof}{def:ftop} +\gist{% We have to show that if $U_a(x_1) \cap U_b(x_2) \neq \emptyset$, then this intersection is the union of sets of this kind. - Let $x' \in U_a(x_1)$. +}{} + Let $x' \in U_a(x_1) \cap U_b(x_2)$. Then by \yaref{lem:ftophelper}, there exists $\epsilon_1 > 0$ with $U_{\epsilon_1}(x') \subseteq U_a(x_1)$. - Similarly there exists $\epsilon_2 > 0$ - such that $U_{\epsilon_2}(x') \subseteq U_b(x_2)$. + Similarly there exists $\epsilon_2 > 0$\gist{ + such that $U_{\epsilon_2}(x') \subseteq U_b(x_2)$.}{.} So for $\epsilon \le \epsilon_1, \epsilon_2$, 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.} + % TODO: maybe learn? - 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 $Z = \{\star\}$. @@ -168,8 +168,7 @@ i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow \item One can show that $H$ is a topological group and $(M,H)$ is a flow.\footnote{This is non-trivial.} \item Since $H$ is compact, - $(M,H)$ is equicontinuous, %\todo{We didn't define this} - i.e.~it is isometric. + $(M,H)$ is equicontinuous, i.e.~it is isometric. In particular, $(M,T)$ is isometric. \end{enumerate} \item $M \neq \{\star\}$, i.e.~$(M,T)$ is non-trivial: @@ -213,9 +212,9 @@ i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow Let $X$ be a metric space and $\Gamma\colon X \to \R$ be upper semicontinuous. Then the set of continuity points of $\Gamma$ is comeager. - \todo{Missing figure: upper semicontinuous function} \end{theorem} \begin{proof} +\notexaminable{ Take $x$ such that $\Gamma$ is not continuous at $x$. Then there is an $\epsilon > 0$ and $x_n \to x$ such that @@ -227,10 +226,11 @@ i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow \] $X \setminus B_q = \{a \in X : \Gamma(a) < q\}$ is open, i.e.~$B_q$ is closed. - Note that $x \in F_q \coloneqq B_q \setminus B_q^\circ$ - and $B_q \setminus B_q^\circ$ is nwd + Note that $x \in F_q \coloneqq B_q \setminus \inter(B_q)$ + and $B_q \setminus \inter(B_q)$ is nwd as it is closed and has empty interior, so $\bigcup_{q \in \Q} F_q$ is meager. +} \end{proof} diff --git a/inputs/lecture_19.tex b/inputs/lecture_19.tex index 159746e..fdd9135 100644 --- a/inputs/lecture_19.tex +++ b/inputs/lecture_19.tex @@ -1,5 +1,5 @@ -\subsection{The order of a flow} -\lecture{19}{2023-12-19}{Orders of flows} +\subsection{The Order of a Flow} +\lecture{19}{2023-12-19}{Orders of Flows} See also \cite[\href{https://terrytao.wordpress.com/2008/01/24/254a-lecture-6-isometric-systems-and-isometric-extensions/}{Lecture 6}]{tao}. @@ -71,7 +71,6 @@ equicontinuity coincide. i.e.~if $(Y_\alpha, f_{\alpha,\beta})$, $\beta < \alpha \le \Theta$ are isometric, then the inverse limit $Y$ is isometric.% - \todo{Why does an inverse limit exist?} % https://q.uiver.app/#q=WzAsNCxbMSwwLCJZX1xcYWxwaGEiXSxbMSwxLCJZX1xcYmV0YSJdLFswLDAsIlkiXSxbMiwwLCJYIl0sWzAsMSwiZl97XFxhbHBoYSwgXFxiZXRhfSJdLFsyLDAsImZfXFxhbHBoYSJdLFsyLDEsImZfXFxiZXRhIiwyXSxbMywwLCJcXHBpX1xcYWxwaGEiLDJdLFszLDEsIlxccGlfXFxiZXRhIl1d \[\begin{tikzcd} Y & {Y_\alpha} & X \\ @@ -193,8 +192,8 @@ More generally we can show: on the fibers of $Y$ over $Z_2$ and invariant under $T$. - $\sigma$ is a metric, since if - if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$, + $\sigma$ is a metric, + since if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$, then $\pi_1(y) = \pi_1(y')$ or $y = y'$. \end{proof} @@ -221,6 +220,7 @@ More generally we can show: For this, we show that for all $\xi < \eta$, $(X_\xi', T)$ is a factor of $(X_\xi ,T)$ using transfinite induction. + % https://q.uiver.app/#q=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 \[\begin{tikzcd} {X'_{\eta'}} & \dots & {X'_3} & {X'_2} & {X_1'} \\ @@ -242,7 +242,8 @@ More generally we can show: \arrow["{\pi'_2}", curve={height=-18pt}, from=1-1, to=1-4] \arrow["{\pi'_1}", curve={height=-30pt}, from=1-1, to=1-5] \end{tikzcd}\] - % TODO: induction start? + + We'll only show the successor step: Suppose we have $(X'_\xi, T) = \theta((X_\xi, T)$. @@ -251,18 +252,21 @@ More generally we can show: \[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times X'_{\xi+ 1}: x \in X\}.\] Then - % https://q.uiver.app/#q=WzAsNSxbMCwwLCIoWF97XFx4aSsxfSxUKSJdLFsyLDAsIihZLFQpIl0sWzMsMSwiKFgnX3tcXHhpKzF9LFQpIl0sWzIsMiwiKFgnX1xceGksVCkiXSxbMSwxLCIoWF9cXHhpLFQpIl0sWzAsNCwiXFx0ZXh0e21heC5+aXNvfSIsMV0sWzQsMywiXFx0aGV0YSIsMV0sWzIsMywie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsNCwie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsMl0sWzAsMSwiIiwwLHsiY3VydmUiOi0xLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0= -\[\begin{tikzcd} + % https://q.uiver.app/#q=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 +\begin{tikzcd} {(X_{\xi+1},T)} && {(Y,T)} \\ & {(X_\xi,T)} && {(X'_{\xi+1},T)} \\ && {(X'_\xi,T)} \arrow["{\text{max.~iso}}"{description}, from=1-1, to=2-2] \arrow["\theta"{description}, from=2-2, to=3-3] - \arrow["{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3] - \arrow["{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2] + \arrow[""{name=0, anchor=center, inner sep=0}, "{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3] + \arrow[""{name=1, anchor=center, inner sep=0}, "{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2] \arrow[from=1-3, to=2-4] \arrow[curve={height=-6pt}, dashed, from=1-1, to=1-3] -\end{tikzcd}\] + \arrow["{\pi'}", draw=none, from=1-3, to=2-4] + \arrow["{\theta'}", draw=none, from=0, to=3-3] + \arrow["\pi"', draw=none, from=1-3, to=1] +\end{tikzcd} The diagram commutes, since all maps are the induced maps. By definition of $Y$ is clear that $\pi$ and $\pi'$ separate points in $Y$. @@ -275,6 +279,9 @@ More generally we can show: In particular, $(X'_{\xi+1}, T)$ is a factor of $(X_{\xi+1}, T)$. \end{proof} + +% TODO ANKI-MARKER + \begin{example}[{\cite[p. 513]{Furstenberg}}] \label{ex:19:inftorus} Let $X$ be the infinite torus diff --git a/inputs/lecture_20.tex b/inputs/lecture_20.tex index bd5690f..27259a1 100644 --- a/inputs/lecture_20.tex +++ b/inputs/lecture_20.tex @@ -1,5 +1,6 @@ \lecture{20}{2024-01-09}{The Infinite Torus} +\gist{ \begin{example} \footnote{This is the same as \yaref{ex:19:inftorus}, but with new notation.} @@ -14,9 +15,10 @@ \begin{remark}+ Note that we can identify $S^1$ with a subset of $\C$ (and use multiplication) or with $\faktor{\R}{\Z}$ (and use addition). - In the lecture both notations were used.% to make things extra confusing. + In the lecture both notations were used. % to make things extra confusing. Here I'll try to only use multiplicative notation. \end{remark} +}{} We will be studying projections to the first $d$ coordinates, i.e. \[ diff --git a/inputs/lecture_22.tex b/inputs/lecture_22.tex index 0cd13ac..ea4af63 100644 --- a/inputs/lecture_22.tex +++ b/inputs/lecture_22.tex @@ -78,7 +78,7 @@ is Borel for all $\alpha < \omega_1$. \end{enumerate} \end{theorem} -\todo{This was already stated as \yaref{thm:beleznayforeman} in lecture 16 +\todo{This was already stated as \yaref{thm:beleznay-foreman} in lecture 16 and should not have two numbers.} A few words on the proof: @@ -86,7 +86,7 @@ Let $\mathbb{K} = S^1$ and $I$ a countable linear order. Let $\mathbb{K}^I$ be the product of $|I|$ many $\mathbb{K}$, $\mathbb{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.% - \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..aedf7f5 100644 --- a/inputs/lecture_23.tex +++ b/inputs/lecture_23.tex @@ -1,5 +1,9 @@ \lecture{23}{2024-01-19}{More sketches of ideas of Beleznay and Foreman} +% TODO read notes +% TODO def. almost distal +% From Lecture 23, you need to know the proposition on page 7 (with the proof), but I won't ask you for other proofs from that lecture + \begin{notation} Let $X$ be a Polish space and $\cP$ a property of elements of $X$, then we say that $x_0 \in X$ is \vocab{generic} @@ -24,47 +28,49 @@ 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 @@ -102,28 +108,31 @@ Let $I$ be a linear order The order %TODO (Furstenberg rank) is a $\Pi^1_1$-rank. \end{theorem} -For the proof one shows that $\le^\ast$ and $<^\ast$ -are $\Pi^1_1$, where -\begin{enumerate}[(1)] - \item $p_1 \le^\ast p_2$ iff $p_1$ codes - a distal minimal flow and if - $p_2$ also codes a distal minimal flow, - then $\mathop{order}(p_1) \le \mathop{order}(p_2)$. - \item $p_1 <^\ast p_2$ iff $p_1$ codes - a distal minimal flow and if - $p_2$ also codes a distal minimal flow, - then $\mathop{order}(p_1) < \mathop{order}(p_2)$. -\end{enumerate} - -One uses that $(Y_{i+1}, T)$ is a maximal -isometric extension of $(Y_i,T)$ -ind $(X,T)$ -iff for all $x_1,x_2$ from a fixed countable dense set -in $X$, -for all $i$ with $\pi_{i\oplus 1}(x_1) = \pi_{i \oplus 1}(x_2)$, -there is a sequence $(z_k)$ such that $\pi_i(z_k) = \pi_i(x_1)$, -$F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$. +\begin{proof}[sketch] + \notexaminable{ + For the proof one shows that $\le^\ast$ and $<^\ast$ + are $\Pi^1_1$, where + \begin{enumerate}[(1)] + \item $p_1 \le^\ast p_2$ iff $p_1$ codes + a distal minimal flow and if + $p_2$ also codes a distal minimal flow, + then $\mathop{order}(p_1) \le \mathop{order}(p_2)$. + \item $p_1 <^\ast p_2$ iff $p_1$ codes + a distal minimal flow and if + $p_2$ also codes a distal minimal flow, + then $\mathop{order}(p_1) < \mathop{order}(p_2)$. + \end{enumerate} + One uses that $(Y_{i+1}, T)$ is a maximal + isometric extension of $(Y_i,T)$ + ind $(X,T)$ + iff for all $x_1,x_2$ from a fixed countable dense set + in $X$, + for all $i$ with $\pi_{i\oplus 1}(x_1) = \pi_{i \oplus 1}(x_2)$, + there is a sequence $(z_k)$ such that $\pi_i(z_k) = \pi_i(x_1)$, + $F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$. + } +\end{proof} \begin{proposition} The order of a minimal distal flow on a separable, metric space is countable. @@ -169,4 +178,3 @@ $F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$. Then $\alpha \mapsto U_\alpha$ is an injection. \end{proof} - diff --git a/inputs/lecture_24.tex b/inputs/lecture_24.tex index 09b6a77..77ddcae 100644 --- a/inputs/lecture_24.tex +++ b/inputs/lecture_24.tex @@ -17,17 +17,19 @@ we have $X \in \cU \lor \N \setminus X \in \cU$. \end{enumerate} \end{definition} +\gist{ \begin{remark} \begin{itemize} - \item If $X \cup Y \in \cU$ then $X \in \cU \lor Y$ or $Y \in \cU$: + \item If $X \cup Y \in \cU$ then $X \in \cU$ or $Y \in \cU$: Consider $((\N \setminus X) \cap (\N \setminus Y) = \N \setminus (X \cup Y)$. \item Every filter can be extended to an ultrafilter. (Zorn's lemma) \end{itemize} \end{remark} +}{} \begin{definition} - An ultrafilter is called \vocab[Ultrafilter!principal]{principal} or \vocab[Ultrafilter!trivial]{trivial} - if it is of the form + An ultrafilter is called \vocab[Ultrafilter!principal]{principal} or \vocab[Ultrafilter!trivial]{trivial} + iff it is of the form \[ \hat{n} = \{X \subseteq \N : n \in X\}. \] @@ -46,9 +48,9 @@ Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas. \begin{enumerate}[(1)] - \item $(\cU n) (\phi(n) \land \psi(m)) \iff (\cU n) \phi(n) \land (\cU n) \psi(n)$. - \item $(\cU n) (\phi(n) \lor \psi(m)) \iff (\cU n) \phi(n) \lor (\cU n) \psi(n)$. - \item $(\cU n) \lnot \phi(n) \iff \lnot (\cU n) \phi(n)$. + \item $(\cU n) ~ (\phi(n) \land \psi(m)) \iff (\cU n) \phi(n) \land (\cU n) \psi(n)$. + \item $(\cU n) ~ (\phi(n) \lor \psi(m)) \iff (\cU n) ~ \phi(n) \lor (\cU n) ~ \psi(n)$. + \item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$. \end{enumerate} \end{observe} \begin{lemma} @@ -59,7 +61,7 @@ there is a unique $x \in X$, such that \[ - (\cU n) (x_n \in G) + (\cU n)~(x_n \in G) \] for every neighbourhood% \footnote{$G \subseteq X$ is a neighbourhood iff $x \in \inter G$.} diff --git a/inputs/tutorial_06.tex b/inputs/tutorial_06.tex index 7128041..963ecd3 100644 --- a/inputs/tutorial_06.tex +++ b/inputs/tutorial_06.tex @@ -108,13 +108,13 @@ This requires the use of the axiom of choice. \item Let $B = \{x \in \R | (F_i)_x \text{ is meager}\}$. Then $B$ is comeager in $\R$ and $|B| = \fc$. - We have $|B| = \fc$, since $B$ contains a comeager - $G_\delta$ set, $B'$: + We have $|B| = \fc$: + $B$ contains a comeager $G_\delta$ set, say $B'$. $B'$ is Polish, hence $B' = P \cup C$ for $P$ perfect and $C$ countable, and $|P| \in \{\fc, 0\}$. - But $B'$ can't contain isolated point. + But $B'$ can't contain an isolated point. \item We use $B$ to find a suitable point $a_i$: To ensure that (i) holds, it suffices to chose diff --git a/inputs/tutorial_07.tex b/inputs/tutorial_07.tex index 820928b..17beb1a 100644 --- a/inputs/tutorial_07.tex +++ b/inputs/tutorial_07.tex @@ -57,7 +57,7 @@ form a $\sigma$-algebra). Let $(U_n)_{n < \omega}$ be a countable base of $(X,\tau)$. Each $U_n$ is open, hence Borel, - so by a theorem from the lecture$^{\text{tm}}$ + so by \hyperref[thm:clopenize]{a theorem from the lecture™} there exists a Polish topology $\tau_n$ such that $U_n$ is clopen, preserving Borel sets. diff --git a/inputs/tutorial_09.tex b/inputs/tutorial_09.tex index 49682d1..801e178 100644 --- a/inputs/tutorial_09.tex +++ b/inputs/tutorial_09.tex @@ -116,7 +116,7 @@ amounts to a finite number of conditions on the preimage. \] is closed as an intersection of clopen sets. - Clearly $\pr_{LO(\N)}(\cF)$ is the complement + Clearly $\proj_{LO(\N)}(\cF)$ is the complement of $WO(\N)$, hence $WO(\N)$ is coanalytic. \end{itemize} \nr 4 diff --git a/inputs/tutorial_12.tex b/inputs/tutorial_12.tex index abb2752..46ab2d5 100644 --- a/inputs/tutorial_12.tex +++ b/inputs/tutorial_12.tex @@ -59,6 +59,8 @@ and since $B$ is Hausdorff, compact subsets of $B$ are closed. \end{subproof} +\nr 1 + Let $(X,T)$ be a flow and $G = E(X,T)$ its Ellis semigroup. Let $d$ be a compatible metric on $X$. \begin{enumerate}[(a)] @@ -100,7 +102,7 @@ Let $d$ be a compatible metric on $X$. Consider $\ev_x \colon X^X \to X$. $X^X$ is compact and $X$ is Hausdorff. Hence we can apply - \label{fact:t12:2}. + \yaref{fact:t12:2}. \item Let $x_0 \neq x_1 \in X$. Then $(x_0,x_1)$ is a proximal pair iff $d(gx_0,gx_1) = 0$ diff --git a/inputs/tutorial_12b.tex b/inputs/tutorial_12b.tex index dca0440..bddca3d 100644 --- a/inputs/tutorial_12b.tex +++ b/inputs/tutorial_12b.tex @@ -17,18 +17,12 @@ Then $t_n y \to (0) = t_n x$. - - - - +% TODO this is redundant \begin{refproof}{fact:isometriciffequicontinuous}. - $d$ and $d'(x,y) \coloneqq \sup_{t \in T} d(tx,ty)$ induce the same topology. Let $\tau, \tau'$ be the corresponding topologies. $\tau \subseteq \tau'$ easy, $\tau' \subseteq \tau'$ : use equicontinuity. - - \end{refproof} 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}