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8
bibliography/references.bib
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@ -44,4 +44,12 @@ title = {Classical Descriptive Set Theory},
volume = {156},
year = {2012},
}
@MISC{3722713,
TITLE = {Embedding of countable linear orders into $\Bbb Q$ as topological spaces},
AUTHOR = {Eric Wofsey},
HOWPUBLISHED = {Mathematics Stack Exchange},
NOTE = {URL:https://math.stackexchange.com/q/3722713 (version: 2020-06-16)},
EPRINT = {https://math.stackexchange.com/q/3722713},
URL = {https://math.stackexchange.com/q/3722713}
}

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

16
inputs/lecture_13.tex
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@ -18,20 +18,10 @@ by associating a function $f\colon \Q \to \{0,1\}$
with $(f^{-1}(\{1\}), <)$.
\begin{lemma}
Any countable ordinal embeds into $(\Q,<)$.
Any countable wellorder embeds into $(\Q,<)$.
\end{lemma}
\begin{proof}[sketch]
Use transfinite induction.
Suppose we already have $\alpha \hookrightarrow (\Q, <)$,
we need to show that $\alpha +1 \hookrightarrow (\Q, <)$.
Since $(0,1) \cap \Q \cong \Q$,
we may assume $\alpha \hookrightarrow ((0,1), <)$
and can just set $\alpha \mapsto 2$.
For a limit $\alpha$
take a countable cofinal subsequence $\alpha_1 < \alpha_2 < \ldots \to \alpha$.
Then map $[0,\alpha_1)$ to $(0,1)$
and $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$.
\begin{proof}\footnote{In the lecture this was only done for countable \emph{ordinals}.}
Cf.~\cite{3722713}.
\end{proof}
% TODO $\WF \subseteq 2^\Q$ is $\Sigma^1_1$-complete.

4
inputs/lecture_15.tex
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@ -273,6 +273,7 @@ Recall:
A limit of distal flows is distal.
\end{proposition}
\begin{proof}
\gist{%
Let $(X,T)$ be a limit of $\Sigma = \{(X_i, T) : i \in I\}$.
Suppose that each $(X_i, T)$ is distal.
If $(X,T)$ was not distal,
@ -283,4 +284,7 @@ Recall:
But then $g_n \pi_i(x_1) \to \pi_i(z)$
and $g_n \pi_i(x_2) \to \pi_i(z)$,
which is a contradiction since $(X_i, T)$ is distal.
}{Suppose there is a proximal pair $x_1,x_2$.
Take $i$ such that $\pi_i(x_1) \neq \pi_i(x_2) \lightning$.
}
\end{proof}

27
inputs/lecture_16.tex
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@ -17,13 +17,13 @@ $X$ is always compact metrizable.
% and $h\colon x \mapsto x + \alpha$.\footnote{Again $x + \alpha$ denotes $x \cdot \alpha$ in $\C$.}
% \end{example}
\begin{proof}
% TODO TODO TODO Think!
\gist{%
The action of $1$ determines $h$.
Consider
\[
\{h^n : n \in \Z\} \subseteq \cC(X,X)\gist{ = \{f\colon X \to X : f \text{ continuous}\}}{},
\]
where the topology is the uniform convergence topology. % TODO REF EXERCISE
where the topology is the uniform convergence topology.
Let $G = \overline{ \{h^n : n \in \Z\} } \subseteq \cC(X,X)$.
Since the family $\{h^n : n \in \Z\}$ is uniformly equicontinuous,
i.e.~
@ -82,6 +82,15 @@ $X$ is always compact metrizable.
For $\alpha = h$ we get that
a flow $\Z \acts X$ corresponds to $\Z \acts K$
with $(1,x) \mapsto x + \alpha$.
}{
\begin{itemize}
\item $G \coloneqq \overline{\{h^n : n \in \Z\} } \subseteq \cC(X,X)$.
\item $G$ is compact (Arzela-Ascoli), abelian topological group (closure of ab. top. group)
\item Take any $x \in G$.
\item $Gx$ is compact (since $g \mapsto gx$ is continuous and $G$ is compact)
\item Stabilizer $G_x$ is closed. $K \coloneqq \faktor{G}{G_x}$, $K \to X, fG_x \mapsto f(x)$.
\end{itemize}
}
\end{proof}
\begin{definition}
Let $(X,T)$ be a flow
@ -155,12 +164,14 @@ By Zorn's lemma, this will follow from
\end{tikzcd}\]
\end{theorem}
\yaref{thm:furstenberg} allows us to talk about ranks of distal minimal flows:
\begin{definition}
Let $(X, \Z)$ be distal minimal.
Then $\rank((X,\Z)) \coloneqq \min \{\eta : (X, \Z) \cong (X_\eta, \Z)\}$
where $(X_{\eta}, \Z)$ is as from the definition of quasi-isometric flows,
i.e.~$\rank((X,\Z))$ is the minimal height such
that a tower as in the definition exists.
\begin{definition}[{\cite[{}13.1]{Furstenberg}}]
\label{def:floworder}
Let $(X,T)$ be a quasi-isometric flow,
and let $\eta$ be the smallest ordinal
such that there exists a quasi-isometric system $\{(X_\xi, T), \xi \le \eta\}$
with $(X,T) = (X_\xi, T)$.
Then $\eta$ is called the \vocab{rank} or \vocab{order} of the flow
and is denoted by $\rank((X,T))$.
\end{definition}
\begin{definition}+

2
inputs/lecture_17.tex
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@ -127,7 +127,7 @@ since $X^X$ has these properties.
\begin{lemma}[Ellis–Numakura]
\yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura}
Every compact semigroup
Every non-empty compact semigroup
contains an \vocab{idempotent} element,
i.e.~$f$ such that $f^2 = f$.
\end{lemma}

22
inputs/lecture_19.tex
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@ -35,20 +35,13 @@ equicontinuity coincide.
By equicontinuity of $T$ we get that $\tilde{d}$ and $d$
induce the same topology on $X$.
\end{proof}
\gist{
Recall that we defined the order of a quasi-isometric flow
to be the minimal number of steps required when building the tower
to reach the flow with a quasi-isometric system (cf.~\yaref{thm:l16:3},
\yaref{def:floworder}).
}{}
\begin{question}
What is the minimal number of steps required
when building the tower to reach the flow
as in \yaref{thm:l16:3}?
\end{question}
\begin{definition}[{\cite[{}13.1]{Furstenberg}}]
Let $(X,T)$ be a quasi isometric flow,
and let $\eta$ be the smallest ordinal
such that there exists a quasi-isometric system $\{(X_\xi, T), \xi \le \eta\}$
with $(X,T) = (X_\xi, T)$.
Then $\eta$ is called the \vocab{order} of the flow.
\end{definition}
\begin{theorem}[Maximal isometric factor]
\label{thm:maxisomfactor}
For every flow $(X,T)$ there is a maximal factor $(Y,T)$, $\pi\colon X\to Y$,
@ -279,9 +272,6 @@ More generally we can show:
In particular,
$(X'_{\xi+1}, T)$ is a factor of $(X_{\xi+1}, T)$.
\end{proof}
% TODO ANKI-MARKER
\begin{example}[{\cite[p. 513]{Furstenberg}}]
\label{ex:19:inftorus}
Let $X$ be the infinite torus

7
inputs/lecture_20.tex
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@ -19,6 +19,7 @@
Here I'll try to only use multiplicative notation.
\end{remark}
}{}
We will be studying projections to the first $d$ coordinates,
i.e.
\[
@ -34,9 +35,9 @@ For $d = 1$ we get the circle rotation $x \mapsto e^{\i \alpha} x$.
\[
H_m \coloneqq \{x \in S^1 : x^m = 0\}
\]
for some $m \in \Z$.
for some $m \in \Z$.%
\footnote{cf.~\yaref{s12e2}}
\end{fact}
\todo{Homework!}
We will show that $\tau_d$ is minimal for all $d$,
i.e.~every orbit is dense.
From this it will follow that $\tau$ is minimal.
@ -45,6 +46,8 @@ 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}

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}

5
inputs/lecture_23.tex
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@ -37,10 +37,9 @@ Let $I$ be a linear order
S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
\end{IEEEeqnarray*}
\todo{Exercise sheet 12}
$S$ is Borel.
$S$ is Borel.\footnote{cf.~\yaref{s12e1}}
We will % TODO ?
We will
construct a reduction
\begin{IEEEeqnarray*}{rCl}
M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\

28
inputs/lecture_24.tex
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@ -1,10 +1,10 @@
\lecture{24}{2024-01-23}{Combinatorics!}
% ANKI 2
\subsection{Applications to Combinatorics} % Ramsey Theory}
% TODO Define Ultrafilter
\begin{definition}
An \vocab{ultrafilter} on $\N$ (or any other set)
is a family $\cU \subseteq \cP(\N)$
@ -44,6 +44,7 @@
for $\{ n \in \N : \phi(n)\} \in \cU$.
We say that $\phi(n)$ holds for \vocab{$\cU$-almost all} $n$.
\end{notation}
\gist{%
\begin{observe}
Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas.
@ -53,6 +54,7 @@
\item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$.
\end{enumerate}
\end{observe}
}{}
\begin{lemma}
\label{lem:ultrafilterlimit}
Let $X $ be a compact Hausdorff space.
@ -70,7 +72,10 @@
\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.}
\gist{
For metric spaces:
Whenever we write $X = Y \cup Z$
we have $(\cU n) x_n \in Y$
or $(\cU n) x_n \in Z$.
@ -85,8 +90,13 @@
$C \in \cP_{n+1} \implies \exists C \subseteq D \in \cP_{n}$
and
$C_1 \supseteq C_2 \supseteq \ldots$, $C_i \in \cP_i $ $\implies | \bigcap_{i} C_i| = 1$.
It is clear that we can do this for metric spaces,
but such partition can be found for compact Hausdorff spaces as well.
It is clear that we can do this for metric spaces.
}{}
See \yaref{thm:uflimit} for the full proof.
See
\yaref{fact:compactiffufconv} and
\yaref{fact:hdifffilterlimit} for a more general statement.
\end{refproof}
Let $\beta \N$ be the Čech-Stone compactification of $\N$,
@ -120,15 +130,14 @@ 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
@ -157,7 +166,6 @@ is not necessarily continuous.
\[
\forall n.~\exists k < M.~ T^{n+k}(x) \in G.
\]
\end{definition}
\begin{fact}
Let $\cU, \cV \in \beta\N$

31
inputs/lecture_25.tex
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@ -7,15 +7,17 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
where a basis consist of sets
$V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$.
\gist{%
(For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$
and $\beta\N = V_\N$.)
}{}
\item Note also that for $A, B \subseteq \N$,
$V_{A \cup B} = V_A \cup V_B$,
$V_{A^c} = \beta\N \setminus V_A$.
\end{itemize}
\end{fact}
\gist{%
\begin{observe}
\label{ob:bNclopenbasis}
Note that the basis is clopen. In particular
@ -25,6 +27,7 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
If $F$ is closed, then $U = \beta\N \setminus F = \bigcup_{i\in I} V_{A_i}$,
so $F = \bigcap_{i \in I} V_{\N \setminus A_i}$.
\end{observe}
}{}
\begin{fact}
\label{fact:bNhd}
@ -54,12 +57,14 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
$\bigcap_{j=1}^k F_{i_j} \neq \emptyset$.
We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$.
\gist{%
Replacing each $F_i$ by $V_{A_j^i}$ such
that $F_i = \bigcap_{j \in J_i} V_{A_j^i}$
(cf.~\yaref{ob:bNclopenbasis})
we may assume that $F_i$ is of the form $V_{A_i}$.
We get $\{F_i = V_{A_i} : i \in I\}$
with the finite intersection property.
}{Wlog.~$F_i = V_{A_i}$.}
Hence
$\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$
has the finite intersection property.
@ -78,11 +83,10 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
\item $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
\item $\N \subseteq \beta\N$ is dense.
\end{itemize}
\todo{Easy exercise}
% TODO write down (exercise)
\end{fact}
\begin{theorem}
\label{thm:uflimit}
For every compact Hausdorff space $X$,
a sequence $(x_n)$ in $X$,
and $\cU \in \beta\N$,
@ -132,6 +136,11 @@ 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}+
$\beta$ 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{%
@ -216,13 +225,12 @@ to obtain
Take $x_2 > x_1$ that satisfies this.
\item Suppose we have chosen $\langle x_i : i < n \rangle$.
Since $\cU$ is idempotent, we have
\[
(\cU n)[
n \in P
\land (\cU_k) n + k \in P
\land \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)
\land (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
\]
\begin{IEEEeqnarray*}{rCl}
(\cU n)&& n \in P\\
&\land& (\cU_k) n + k \in P\\
&\land& \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)\\
&\land& (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
\end{IEEEeqnarray*}
Chose $x_n > x_{n-1}$ that satisfies this.
\end{itemize}
Set $H \coloneqq \{x_i : i < \omega\}$.
@ -231,6 +239,3 @@ to obtain
Next time we'll see another proof of this theorem.

99
inputs/tutorial_02.tex
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@ -3,12 +3,36 @@
% Points: 15 / 16
\nr 1
\todo{handwritten solution}
Let $(X,d)$ be a metric space and $\emptyset \neq A \subseteq X$.
Let $d(x,A) \coloneqq \inf(d(x,a) : a \in A\}$.
\begin{itemize}
\item $d(-,A)$ is uniformly continuous:
Clearly $|d(x,A) - d(y,A)| \le d(x,y)$.
\todo{Add details}
\item $d(x,A) = 0 \iff x \in \overline{A}$.
$d(x,A) = 0$ iff there is a sequence in $A$
converging towards $x$ iff $x \in \overline{A}$.
\end{itemize}
\nr 2
Let $X$ be a discrete space.
For $f,g \in X^{\N}$ define
\[
d(f,g) \coloneqq \begin{cases}
(1 + \min \{n: f(n) \neq g(n)\})^{-1} &: f \neq g,\\
0 &: f= g.
\end{cases}
\]
\begin{enumerate}[(a)]
\item $d$ is an ultrametric:
\item $d$ is an \vocab{ultrametric},
i.e.~$d(f,g) \le \max \{d(f,h), d(g,h)\}$ for all $f,g,h \in X^{\N}$ :
Let $f,g,h \in X^{\N}$.
@ -70,10 +94,15 @@
\nr 3
Consider $\N$ as a discrete space and $\N^{\N}$ with the product topology.
Let
\[
S_{\infty} = \{f\colon \N \to \N \text{ bijective}\} \subseteq \N^{\N}.
\]
\begin{enumerate}[(a)]
\item $S_{\infty}$ is a Polish space:
From (2) we know that $\N^{\N}$ is Polish.
From \yaref{s1e2} we know that $\N^{\N}$ is Polish.
Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$
with respect to $\N^\N$.
@ -111,69 +140,9 @@
Clearly there cannot exist a finite subcover
as $B$ is the disjoint union of the $B_j$.
% TODO Think about this
\end{enumerate}
\nr 4
% (uniform metric)
%
% \begin{enumerate}[(a)]
% \item $d_u$ is a metric on $\cC(X,Y)$:
%
% It is clear that $d_u(f,f) = 0$.
%
% Let $f \neq g$. Then there exists $x \in X$ with
% $f(x) \neq g(x)$, hence $d_u(f,g) \ge d(f(x), g(x)) > 0$.
%
% Since $d$ is symmetric, so is $d_u$.
%
% Let $f,g,h \in \cC(X,Y)$.
% Take some $\epsilon > 0$
% choose $x_1, x_2 \in X$
% with $d_u(f,g) \le d(f(x_1), g(x_1)) + \epsilon$,
% $d_u(g,h) \le d(g(x_2), h(x_2)) + \epsilon$.
%
% Then for all $x \in X$
% \begin{IEEEeqnarray*}{rCl}
% d(f(x), h(x)) &\le &
% d(f(x), g(x)) + d(g(x), h(x))\\
% &\le & d(f(x_1), g(x_1)) + d(g(x_2), h(x_2))-2\epsilon\\
% &\le & d_u(f,g) + d_u(g,h) - 2\epsilon.
% \end{IEEEeqnarray*}
% Thus $d_u(f,g) \le d_u(f,g) + d_u(g,h) - 2\epsilon$.
% Taking $\epsilon \to 0$ yields the triangle inequality.
%
% \item $\cC(X,Y)$ is a Polish space:
% \todo{handwritten solution}
%
% \begin{itemize}
% \item $d_u$ is a complete metric:
%
% Let $(f_n)_n$ be a Cauchy series with respect to $d_u$.
%
% Then clearly $(f_n(x))_n$ is a Cauchy sequence with respect
% to $d$ for every $x$.
% Hence there exists a pointwise limit $f$ of the $f_n$.
% We need to show that $f$ is continuous.
%
% %\todo{something something uniform convergence theorem}
%
% \item $(\cC(X,Y), d_u)$ is separable:
%
% Since $Y$ is separable, there exists a countable
% dense subset $S \subseteq Y$.
%
% Consider $\cC(X,S) \subseteq \cC(X,Y)$.
% Take some $f \in \cC(X,Y)$.
% Since $X$ is compact,
%
%
% % TODO
%
% \end{itemize}
% \end{enumerate}
\begin{fact}
Let $X $ be a compact Hausdorff space.
Then the following are equivalent:
@ -205,7 +174,7 @@
Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish}
and $Y $ Polish.
Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$.
Consider the metric $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$.
Consider the \vocab{uniform metric} $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$.
Clearly $d_u$ is a metric.
\begin{claim}
@ -243,7 +212,7 @@ Clearly $d_u$ is a metric.
for each $y \in X_m$.
Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$:
Indeed if $f \in \cC(X,Y)$ and $\eta > 0$,
we finde $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$,
we find $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$,
since $f$ is uniformly continuous.
Let $g \in D_{m,n}$ be such that $\forall y \in X_m.~d(f(y), g(y)) < \frac{1}{n+1}$.
We have $d_u(f,g) \le \eta$,

32
inputs/tutorial_03.tex
View file

@ -12,6 +12,14 @@
\nr 1
Let $X$ be a Polish space.
Then there exists an injection $f\colon X \to 2^\omega$
such that for each $n < \omega$,
the set $f^{-1}(\{(y_n) \in 2^\omega : y_n = 1\})$
is open.
Moreover if $V \subseteq 2^{ \omega}$ is closed,
then $f^{-1}(V)$ is $G_\delta$.
Let $(U_i)_{i < \omega}$ be a countable base of $X$.
Define
\begin{IEEEeqnarray*}{rCl}
@ -19,6 +27,7 @@ Define
x &\longmapsto & (x_i)_{i < \omega}
\end{IEEEeqnarray*}
where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise.
\gist{
Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$
is open.
We have that $f$ is injective since $X$ is T1.
@ -51,17 +60,21 @@ Since $2^{n} \setminus \left( \prod_{i < n} X_i \right)$
is finite, we get that
$f^{-1}(2^{\omega} \setminus ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
is $G_\delta$ as a finite union of $G_{\delta}$ sets.
}{}
\nr 2
Let $X$ be a Polish space. Then $X$ is homeomorphic to a closed subspace of $\R^{ \omega}$ :
\todo{handwritten solution}
(b)
Let $f(x^{(i)})$ be a sequence in $f(X)$.
Suppose that $f(x^{(i)}) \to y$.
We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
% \begin{itemize}
% \item
% Let $f(x^{(i)})$ be a sequence in $f(X)$.
% Suppose that $f(x^{(i)}) \to y$.
% We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
% Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
% Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
% \end{itemize}
\nr 3
@ -130,6 +143,13 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
\end{proof}
\nr 4
Define
\begin{IEEEeqnarray*}{rCl}
f\colon \omega^{\omega} &\longrightarrow & 2^\omega \\
(x_n)&\longmapsto & 0^{x_0} 1 0^{x_1} 1 \ldots.
\end{IEEEeqnarray*}
\begin{enumerate}[(1)]
\item $f$ is a topological embedding:
Consider a basic open set

2
inputs/tutorial_04.tex
View file

@ -8,7 +8,7 @@ Let $A \neq \emptyset$ be discrete.
For $D \subseteq A^{\omega}$,
let
\[
T_D \coloneqq \{x\defon{n} \in A^{<\omega} | x \in D, n \in \N\}..
T_D \coloneqq \{x\defon{n} \in A^{<\omega} | x \in D, n \in \N\}.
\]
\begin{enumerate}[(a)]
\item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree:

4
inputs/tutorial_08.tex
View file

@ -267,14 +267,14 @@ $X_\omega \coloneqq \bigcap X_i$ and $Y_\omega \coloneqq \bigcap Y_i$.
% https://q.uiver.app/#q=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
\adjustbox{scale=0.7,center}{%
\[\begin{tikzcd}
\begin{tikzcd}
{X \setminus X_\omega =} & {(X_0 \setminus X_1)} & \cup & {(X_1 \setminus X_2)} & \cup & {(X_2 \setminus X_3)} & \cdots & {} \\
{Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_1 \setminus Y_2)} & \cup & {(Y_2 \setminus Y_3)} & \cdots & {}
\arrow["f"'{pos=0.7}, from=1-2, to=2-4]
\arrow["g"{pos=0.1}, from=2-2, to=1-4]
\arrow["f"{pos=0.8}, from=1-6, to=2-8]
\arrow["g"{pos=0.1}, from=2-6, to=1-8]
\end{tikzcd}\]
\end{tikzcd}
}
By \autoref{thm:lusinsouslin}

2
inputs/tutorial_11.tex
View file

@ -63,7 +63,7 @@ Flows are always on non-empty spaces $X$.
\begin{proof}
(i) $\implies$ (ii):
Let $(Y,T)$ be a subflow of $(X,T)$.
take $y \in Y$. Then $Ty$ is dense in mKX.
take $y \in Y$. Then $Ty$ is dense in $X$.
But $Ty \subseteq Y$, so $Y$ is dense in $X$.
Since $Y$ is closed, we get $Y = X$.

81
inputs/tutorial_14.tex
View file

@ -4,9 +4,58 @@
\nr 1
% Examinable
% TODO (there is a more direct way to do it, not using analytic / coanalytic)
Let $\LO(\N) \overset{\text{closed}}{\subseteq} 2^{\N\times \N}$ denote the set of linear orders on $\N$.
Let $S \subseteq \LO(\N)$ be the set of orders having a least
element and such that every element has an immediate successor.
\begin{itemize}
\item $S$ is Borel in $\LO(\N)$:
Let $M_n \subseteq \LO(\N)$ be the set of orders with minimal element $n$.
Let $I_{n,m} \subseteq \LO(\N)$ be the set of orders such
that $m$ is the immediate successor of $n$.
Clearly $S = \left(\bigcap_n \bigcup_{m\neq n} I_{n,m}\right) \cap \bigcup_n M_n$,
so it suffices to show that $M_n$ and $I_{n,m}$ are Borel.
It is $M_n = \bigcap_{m\neq n} \{\prec : m \not\prec n\}$
and $I_{n,m} = \{\prec: n \prec m\} \cap \bigcap_{i} \{\prec : n \preceq i \preceq m \implies n = i \lor n = m \}$.
\item Give an example of an element of $S$ which is not well-ordered:
Consider $\{1 - \frac{1}{n} : n \in \N^+\} \cup \{1 + \frac{1}{n} : n \in \N^{+}\} \subseteq \R$
with the order $<_\R$.
This is an element of $S$,
but $\{x \in S: x \ge 1\}$ has no minimal element,
hence it is not well-ordered.
\end{itemize}
\nr 2
% Examinable
Recall the definition of the circle shift flow $(\R / \Z, \Z)$
with parameter $\alpha \in \R$, $1 \cdot x \coloneqq x + \alpha$.
\begin{itemize}
\item If $\alpha \not\in \Q$, then $(\R / \Z, \Z)$ is minimal:
This is known as \href{https://en.wikipedia.org/wiki/Dirichlet's_approximation_theorem}{Dirichlet's Approximation Theorem}.
\item Consider $\R/\Z$ as a topological group.
Any subgroup $H$ of $\R / \Z$ is dense in $\R / \Z$
or of the form $H = \{ x \in \R / \Z | mx = 0\}$
for some $m \in \Z$.
If $H$ contains an irrational element $\alpha$, then
it is dense by the previous point.
Suppose that $H \subseteq \Q / \Z$.
Let $D$ be the set of denominators of elements of $H$
written as irreducible fractions.
If $D$ is finite,
then $H = \{x \in \R / \Z : \mathop{lcm}(D)x = 0\}$.
Otherwise $H$ is dense, as it contains
elements of arbitrarily large denominator.
\end{itemize}
\nr 3
@ -35,11 +84,12 @@
\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)
@ -54,19 +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 $K_n$}\}$.
@ -74,12 +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}
@ -110,7 +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 $i,j \ge N$.
@ -151,9 +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,
@ -174,9 +225,3 @@ Then so is $(K(X), d_H)$.
% TODO complete and totally bounded Sutherland metric and topological spaces

22
inputs/tutorial_15.tex
View file

@ -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$.
@ -12,6 +12,7 @@ all sets containing an open neighbourhood of $x$,
is contained in $\cF$.
\begin{fact}
\label{fact:hdifffilterlimit}
$X$ is Hausdorff iff every filter has at most one limit point.
\end{fact}
\begin{proof}
@ -21,6 +22,7 @@ is contained in $\cF$.
\end{proof}
\begin{fact}
\label{fact:compactiffufconv}
$X$ is (quasi-) compact
iff every ultrafilter converges.
\end{fact}
@ -29,7 +31,7 @@ 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 +71,19 @@ 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}