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Kuratowski-Ulam gist

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14
inputs/lecture_05.tex
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@ -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}
}{}

60
inputs/lecture_06.tex
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@ -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

6
inputs/lecture_20.tex
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@ -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 $\pi_n(x) = \pi_n(x')$

11
inputs/lecture_22.tex
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@ -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}

10
inputs/lecture_24.tex
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@ -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

6
inputs/lecture_25.tex
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@ -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{%

30
inputs/tutorial_14.tex
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@ -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

21
inputs/tutorial_15.tex
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@ -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}