diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex index d305928..55c45e5 100644 --- a/inputs/lecture_09.tex +++ b/inputs/lecture_09.tex @@ -154,7 +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_16.tex b/inputs/lecture_16.tex index 457bb11..fdd1ed3 100644 --- a/inputs/lecture_16.tex +++ b/inputs/lecture_16.tex @@ -140,7 +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 $Y$. diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index fc81f10..1891dd0 100644 --- a/inputs/lecture_17.tex +++ b/inputs/lecture_17.tex @@ -15,7 +15,7 @@ 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. +$X^{X}$ 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. @@ -34,16 +34,20 @@ $X^{X}$ 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. -% TODO THINK ABOUT THIS \gist{ Properties of $(X,T)$ translate to properties of $E(X,T)$: @@ -58,15 +62,14 @@ since $X^X$ has these properties. 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 $th \in G$. - \gist{ - We have that $t^{-1}G$ is compact, - since $t^{-1}$ is continuous - and $G$ is compact. - }{$t^{-1}G$ is compact.} + We have that $t^{-1}G$ is compact, + since $t^{-1}$ is continuous + and $G$ is compact. It is $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$. @@ -93,6 +96,21 @@ since $X^X$ has these properties. 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} @@ -101,9 +119,11 @@ since $X^X$ has these properties. 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} @@ -112,6 +132,7 @@ since $X^X$ has these properties. 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 $s \in R$. @@ -128,25 +149,39 @@ since $X^X$ has these properties. 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 might 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 $\Gamma \coloneqq Gg$. @@ -164,19 +199,39 @@ But it is interesting for other semigroups. 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} -% TODO ANKI-MARKER - -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: @@ -189,6 +244,7 @@ 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 $G$. @@ -196,17 +252,18 @@ 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 d801fc4..4c9f35d 100644 --- a/inputs/lecture_18.tex +++ b/inputs/lecture_18.tex @@ -1,16 +1,11 @@ \subsection{Sketch of proof of \yaref{thm:l16:3}} \lecture{18}{2023-12-15}{Sketch of proof of \yaref{thm:l16:3}} +% TODO ANKI-MARKER + + 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. diff --git a/inputs/lecture_22.tex b/inputs/lecture_22.tex index 9e93df4..ea11b14 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: 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_12.tex b/inputs/tutorial_12.tex index 72d35dd..46ab2d5 100644 --- a/inputs/tutorial_12.tex +++ b/inputs/tutorial_12.tex @@ -102,7 +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 $d(gx_0,gx_1) = 0$