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Josia Pietsch 2024-02-07 02:05:20 +01:00
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commit e887f46a5d
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9 changed files with 92 additions and 96 deletions

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@ -127,7 +127,7 @@ since $X^X$ has these properties.
\begin{lemma}[EllisNumakura]
\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}

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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. %\\

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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.
@ -72,6 +74,8 @@
\end{notation}
\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 $(\cU n) x_n \in Z$.
@ -86,8 +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 $\N$,
@ -157,7 +166,6 @@ is not necessarily continuous.
\[
\forall n.~\exists k < M.~ T^{n+k}(x) \in G.
\]
\end{definition}
\begin{fact}
Let $\cU, \cV \in \beta\N$

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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 $\cU \in \beta\N$,
@ -133,7 +137,6 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
% 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.
@ -222,13 +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 $H \coloneqq \{x_i : i < \omega\}$.
@ -237,6 +239,3 @@ to obtain
Next time we'll see another proof of this theorem.

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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 $d_u(f,g) \le \eta$,

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@ -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

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@ -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:

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@ -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 $Y = X$.

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@ -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}