diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex index 289847a..444922f 100644 --- a/inputs/lecture_05.tex +++ b/inputs/lecture_05.tex @@ -164,8 +164,10 @@ \] \end{notation} +\gist{% The following similar to Fubini, but for meager sets: +}{} \begin{theorem}[Kuratowski-Ulam] \yalabel{Kuratowski-Ulam}{Kuratowski-Ulam}{thm:kuratowskiulam} @@ -193,6 +195,7 @@ but for meager sets: \end{enumerate} \end{theorem} \begin{refproof}{thm:kuratowskiulam} +\gist{ (ii) and (iii) are equivalent by passing to the complement. \begin{claim}%[1a] @@ -286,16 +289,11 @@ but for meager sets: $M_x$ is comeager as a countable intersection of comeager sets. \end{refproof} +}{} + % \phantom\qedhere % \end{refproof} % TODO fix claim numbers -\gist{% -\begin{remark} - Suppose that $A$ has the BP. - Then there is an open $U$ such that - $A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager. - Then $A = U \symdif M$. -\end{remark} -}{} + diff --git a/inputs/lecture_06.tex b/inputs/lecture_06.tex index f3bb7e0..63709a9 100644 --- a/inputs/lecture_06.tex +++ b/inputs/lecture_06.tex @@ -1,8 +1,8 @@ \lecture{06}{2023-11-03}{} - +\gist{% % \begin{refproof}{thm:kuratowskiulam} \begin{enumerate}[(i)] - \item Let $A$ be a set with the Baire Property. + \item Let $A$ be a set with the Baire property. Write $A = U \symdif M$ for $U$ open and $M$ meager. Then for all $x$, @@ -51,8 +51,8 @@ Towards a contradiction suppose that $A$ is not meager. Then $U$ is not meager. Since $X \times Y$ is second countable, - we have that $A$ is a countable union of open rectangles. - At least one of them, say $G \times H \subseteq A$, + we have that $U$ is a countable union of open rectangles. + At least one of them, say $G \times H \subseteq U$, is not meager. By \yaref{thm:kuratowskiulam:c2}, both $G$ and $H$ are not meager. @@ -71,7 +71,59 @@ ``$\implies$'' This is \yaref{thm:kuratowskiulam:c1b}. \end{enumerate} +}{% + \begin{itemize} + \item (ii) $\iff$ (iii): pass to complement. + \item $F \overset{\text{closed}}{\subseteq} X \times Y$ nwd. + $\implies \{x \in X : F_x \text{ nwd}\} $ comeager: + \begin{itemize} + \item $W = F^c$ is open and dense, show that $\{x : W_x \text{ dense}\}$ + is comeager. + \item $(V_n)$ enumeration of basis. Show that $U_n \coloneqq \{x : V_n \cap W_x \neq \emptyset\}$ + is comeager for all $n$. + \item $U_n$ is open (projection of open) and dense ($W$ is dense, hence $W \cap ( U \times V_n) \neq \emptyset$ for $U$ open). + \end{itemize} + \item $F \subseteq X \times Y$ is nwd $\implies \{x \in X: F_x \text{ nwd}\}$ comeager. + (consider $\overline{F}$). + \item (ii) $\implies$: + $M \subseteq X \times Y$ meager $\implies \{x \in X: M_x \text{ meager}\}$ comeager + (write $M$ as ctbl. union of nwd.) + \item (i): If $A$ has the Baire Property, + then $A = U \symdif M$, $A_x = U_x \symdif M_x$, + $U_x$ open and $\{x : M_x \text{ meager}\}$ comeager + $\implies$ (i). + \item $P \subseteq X$, $Q \subseteq Y$ BP, + then $P \times Q$ meager $\iff$ $P$ or $Q$ meager. + \begin{itemize} + \item $\impliedby$ easy + \item $\implies$ Suppose $P \times Q$ meager, $P$ not meager. + $\emptyset\neq P \cap \underbrace{\{x : (P \times Q)_x \text{ meager} \}}_{\text{comeager}} \ni x$. + $(P \times Q)_x = Q$ is meager. + \end{itemize} + \item (ii) $\impliedby$: + \begin{itemize} + \item $A$ BP, $\{x : A_x \text{ meager}\}$ comeager. + \item $A = U \symdif M$. + \item Suppose $A$ not meager $\leadsto$ $U$ not meager + $\leadsto \exists G \times H \subseteq U$ not meager. + \item $G$ and $H$ are not meager. + \item $\exists x_0 \in G \cap \underbrace{\{x: A_x \text{ meager } \land M_x \text{ meager}\}}_\text{comeager}$. + \item $H$ meager, as + \[ + H \subseteq U_{x_0} \subseteq A_{x_0} \cup M_{x_0}. + \] + \end{itemize} + \end{itemize} +} \end{refproof} +\gist{% +\begin{remark} + Suppose that $A$ has the BP. + Then there is an open $U$ such that + $A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager. + Then $A = U \symdif M$. +\end{remark} +}{} \section{Borel sets} % TODO: fix chapters diff --git a/inputs/lecture_20.tex b/inputs/lecture_20.tex index 2c6797b..afe5663 100644 --- a/inputs/lecture_20.tex +++ b/inputs/lecture_20.tex @@ -20,9 +20,6 @@ \end{remark} }{} - -% TODO ANKI-MARKER - We will be studying projections to the first $d$ coordinates, i.e. \[ @@ -49,6 +46,9 @@ Let $\pi_n\colon X \to (S^1)^n$ be the projection to the first $n$ coordinates. +% TODO ANKI-MARKER + + \begin{lemma} \label{lem:lec20:1} Let $x,x' \in X$ with $\pi_n(x) = \pi_n(x')$ diff --git a/inputs/lecture_22.tex b/inputs/lecture_22.tex index ea4af63..6032904 100644 --- a/inputs/lecture_22.tex +++ b/inputs/lecture_22.tex @@ -146,9 +146,7 @@ For this we define % 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:% - \gist{% - \footnote{This is not relevant for the exam.} - + \notexaminable{% Let $\langle E_n : n < \omega \rangle$ be an enumeration of a countable basis for $\mathbb{K}^I$. @@ -165,11 +163,10 @@ For this we define 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 The order of the flow is $\eta$:% - \gist{% - \footnote{This is not relevant for the exam.} + \notexaminable{% Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$. Consider the flows we get from $(f_i)_{i < j}$ resp.~$(f_i)_{i \le j}$ @@ -193,6 +190,6 @@ For this we define \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_24.tex b/inputs/lecture_24.tex index 77ddcae..366e135 100644 --- a/inputs/lecture_24.tex +++ b/inputs/lecture_24.tex @@ -70,7 +70,8 @@ \begin{notation} In this case we write $x = \ulim{\cU}_n x_n$. \end{notation} -\begin{refproof}{lem:ultrafilterlimit}[sketch] +\begin{refproof}{lem:ultrafilterlimit}\footnote{The proof from the lecture only works + for metric spaces.} Whenever we write $X = Y \cup Z$ we have $(\cU n) x_n \in Y$ or $(\cU n) x_n \in Z$. @@ -120,15 +121,14 @@ This gives $+ \colon \beta\N \times \beta\N \to \beta\N$. This is not commutative, but associative and $a \mapsto a + b$ is continuous -for a fixed $b$. -This is called a left compact topological semigroup. - +for a fixed $b$, +i.e.~it is a left compact topological semigroup. Let $X$ be a compact Hausdorff space and let $T \colon X \to X$ be continuous.% -\footnote{Note that this need not be a homeomorphism, i.e.~we only get a $\N$-action +\footnote{Note that this may not be a homeomorphism, i.e.~we only get a $\N$-action but not a $\Z$-action.} For any $\cU \in \beta\N$, we define $T^{\cU}$ by diff --git a/inputs/lecture_25.tex b/inputs/lecture_25.tex index e92aff3..b2e1ff0 100644 --- a/inputs/lecture_25.tex +++ b/inputs/lecture_25.tex @@ -132,6 +132,12 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. % TODO general fact: continuous functions agreeing on a dense set % agree everywhere (fact section) \end{proof} +\begin{trivial}+ + More generally, + $\beta$ is a functor from the category of topological + spaces to the category of compact Hausdorff spaces. + It is left adjoint to the inclusion functor. +\end{trivial} % RECAP \gist{% diff --git a/inputs/tutorial_14.tex b/inputs/tutorial_14.tex index b2f93de..ada3004 100644 --- a/inputs/tutorial_14.tex +++ b/inputs/tutorial_14.tex @@ -84,11 +84,12 @@ with parameter $\alpha \in \R$, $1 \cdot x \coloneqq x + \alpha$. \nr 4 % Examinable! +% TODO THINK! +\gist{% % RECAP -Let $X$ be a metrizable topological space. - -Let $K(X) \coloneqq \{ K \subseteq X : \text{ compact}\}$. +Let $X$ be a metrizable topological space +and let $K(X) \coloneqq \{ K \subseteq X : K \text{ compact}\}$. The Vietoris topology has a basis given by $\{K \subseteq U\}$, $U$ open (type 1) @@ -103,19 +104,21 @@ $\max_{a \in A} d(a,B)$. On previous sheets, we checked that $d_H$ is a metric. If $X$ is separable, then so is $K(X)$. % END RECAP +}{} \begin{fact} + \label{fact:s12e4} Let $(X,d)$ be a complete metric space. Then so is $(K(X), d_H)$. \end{fact} -\begin{proof} +\begin{refproof}{fact:s12e4} We need to show that $(K(X), d_H)$ is complete. Let $(K_n)_{ n< \omega}$ be Cauchy in $(K(X), d_H)$. Wlog.~$K_n \neq \emptyset$ for all $n$. Let $K = \{ x \in X : \forall x \in U \overset{\text{open}}{\subseteq} X.~ - \text{ $X$ intersects $K_n$ for infinitely many $n$}\}$. + \text{ $U \cap K_n \neq \emptyset$ for infinitely many $n$}\}$. Equivalently, $K = \{x : x \text{ is a cluster point of some subsequence $(x_n)$ with $x_n \in K_n$ for all $K_n$}\}$. @@ -123,12 +126,12 @@ Then so is $(K(X), d_H)$. (A cluster point is a limit of some subsequence). \begin{claim} + \label{fact:s12e4:c1} $K_n \to K$. \end{claim} - \begin{subproof} + \begin{refproof}{fact:s12e4:c1} Note that $K$ is closed (the complement is open). - \begin{claim} $K \neq \emptyset$. \end{claim} @@ -159,7 +162,7 @@ Then so is $(K(X), d_H)$. space, it is complete. So it suffices to show that $K$ is totally bounded. - Let $\epsilon > 0$ + Let $\epsilon > 0$. Take $N$ such that $d_H(K_i,K_j) < \epsilon$ for all $i,j \ge N$. @@ -200,9 +203,8 @@ Then so is $(K(X), d_H)$. To do this, construct a sequence of $y_{n_i} \in K_{n_i}$ starting with $y$ such that $d(y_{n_i}, y_{n_{i+1}}) < \frac{\epsilon}{2^{i+2}}$. (same trick as before). - \end{subproof} - -\end{proof} + \end{refproof} +\end{refproof} \begin{fact} If $X$ is compact metrisable, @@ -223,9 +225,3 @@ Then so is $(K(X), d_H)$. % TODO complete and totally bounded Sutherland metric and topological spaces - - - - - - diff --git a/inputs/tutorial_15.tex b/inputs/tutorial_15.tex index b3b0051..33c18a4 100644 --- a/inputs/tutorial_15.tex +++ b/inputs/tutorial_15.tex @@ -2,7 +2,7 @@ \tutorial{15}{2024-01-31}{Additions} The following is not relevant for the exam, -but gives a more general picture. +but aims to give a more general picture. Let $X$ be a topological space and let $\cF$ be a filter on $ X$. @@ -21,6 +21,7 @@ is contained in $\cF$. \end{proof} \begin{fact} + \label{fact:compactiffufconv} $X$ is (quasi-) compact iff every ultrafilter converges. \end{fact} @@ -29,7 +30,7 @@ is contained in $\cF$. Let $\cU$ be an ultrafilter. Consider the family $\cV = \{\overline{A} : A \in \cU\}$ of closed sets. - By the FIP we geht that there exist + By the FIP we get that there exist $c \in X$ such that $c \in \overline{A}$ for all $A \in \cU$. Let $N$ be an open neighbourhood of $c$. If $N^c \in \cU$, then $c \in N^c \lightning$ @@ -69,17 +70,19 @@ so is $f(\cB)$. \end{fact} \begin{proof} Consider $(f,g)^{-1}(\Delta) \supseteq A$. + The RHS is a dense closed set, i.e.~the entire space. \end{proof} -We can uniquely extend $f\colon X \to Y$ continuous +We can uniquely extend a continuous $f\colon X \to Y$ to a continuous $\overline{f}\colon \beta X \to Y$ by setting $\overline{f}(\cU) \coloneqq \lim_\cU f$. -Let $V$ be an open neighbourhood of $Y$ in $\overline{f}\left( U) \right) $. -Consider $f^{-1}(V)$. -Consider the basic open set -\[ -\{\cF \in \beta\N : \cF \ni f^{-1}(V)\}. -\] +% Let $V$ be an open neighbourhood of $y \in \overline{f}\left( U \right)$. +% Consider $f^{-1}(V)$. +% Then +% \[ +% \{\cF \in \beta\N : \cF \ni f^{-1}(V)\} +% \] +% is a basic open set. \todo{I missed the last 5 minutes}