Kuratowski-Ulam gist

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}
-}{}

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.
+            we have that $U$ is a countable union of open rectangles.
-            At least one of them, say $G \times H \subseteq A$,
+            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
 

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 $\pi_n(x) = \pi_n(x')$

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{%
+            \notexaminable{%
-                \footnote{This is not relevant for the exam.}
-
                 Let $\langle E_n : n < \omega \rangle$
                 be an enumeration of a countable basis for $\mathbb{K}^I$.
 
@@ -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{%
+            \notexaminable{%
-                \footnote{This is not relevant for the exam.}
                 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}

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  $b$.
+for a fixed  $b$,
-This is called a left compact topological semigroup.
+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

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{%

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 $X$ be a metrizable topological space
-
+and let $K(X) \coloneqq  \{ K \subseteq  X : K \text{ compact}\}$.
-Let $K(X) \coloneqq  \{ K \subseteq  X : \text{ compact}\}$.
 
 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{refproof}
-
+\end{refproof}
-\end{proof}
 
 \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
-
-
-
-
-
-

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) $.
+% Let $V$ be an open neighbourhood of $y \in \overline{f}\left( U \right)$.
-Consider $f^{-1}(V)$.
+% Consider $f^{-1}(V)$.
-Consider the basic open set
+% Then
-\[
+% \[
-\{\cF \in  \beta\N  : \cF \ni f^{-1}(V)\}.
+% \{\cF \in  \beta\N  : \cF \ni f^{-1}(V)\}
-\]
+% \]
+% is a basic open set.
 
 \todo{I missed the last 5 minutes}