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