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Josia Pietsch 2024-02-07 02:05:20 +01:00
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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] \begin{lemma}[EllisNumakura]
\yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura} \yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura}
Every compact semigroup Every non-empty compact semigroup
contains an \vocab{idempotent} element, contains an \vocab{idempotent} element,
i.e.~$f$ such that $f^2 = f$. i.e.~$f$ such that $f^2 = f$.
\end{lemma} \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},\\ S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\} &&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
\todo{Exercise sheet 12} $S$ is Borel.\footnote{cf.~\yaref{s12e1}}
$S$ is Borel.
We will % TODO ? We will
construct a reduction construct a reduction
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\ 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!} \lecture{24}{2024-01-23}{Combinatorics!}
% ANKI 2
\subsection{Applications to Combinatorics} % Ramsey Theory} \subsection{Applications to Combinatorics} % Ramsey Theory}
% TODO Define Ultrafilter
\begin{definition} \begin{definition}
An \vocab{ultrafilter} on $\N$ (or any other set) An \vocab{ultrafilter} on $\N$ (or any other set)
is a family $\cU \subseteq \cP(\N)$ is a family $\cU \subseteq \cP(\N)$
@ -44,6 +44,7 @@
for $\{ n \in \N : \phi(n)\} \in \cU$. for $\{ n \in \N : \phi(n)\} \in \cU$.
We say that $\phi(n)$ holds for \vocab{$\cU$-almost all} $n$. We say that $\phi(n)$ holds for \vocab{$\cU$-almost all} $n$.
\end{notation} \end{notation}
\gist{%
\begin{observe} \begin{observe}
Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas. Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas.
@ -53,6 +54,7 @@
\item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$. \item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$.
\end{enumerate} \end{enumerate}
\end{observe} \end{observe}
}{}
\begin{lemma} \begin{lemma}
\label{lem:ultrafilterlimit} \label{lem:ultrafilterlimit}
Let $X $ be a compact Hausdorff space. Let $X $ be a compact Hausdorff space.
@ -72,6 +74,8 @@
\end{notation} \end{notation}
\begin{refproof}{lem:ultrafilterlimit}\footnote{The proof from the lecture only works \begin{refproof}{lem:ultrafilterlimit}\footnote{The proof from the lecture only works
for metric spaces.} for metric spaces.}
\gist{
For metric spaces:
Whenever we write $X = Y \cup Z$ Whenever we write $X = Y \cup Z$
we have $(\cU n) x_n \in Y$ we have $(\cU n) x_n \in Y$
or $(\cU n) x_n \in Z$. or $(\cU n) x_n \in Z$.
@ -86,8 +90,13 @@
$C \in \cP_{n+1} \implies \exists C \subseteq D \in \cP_{n}$ $C \in \cP_{n+1} \implies \exists C \subseteq D \in \cP_{n}$
and and
$C_1 \supseteq C_2 \supseteq \ldots$, $C_i \in \cP_i $ $\implies | \bigcap_{i} C_i| = 1$. $C_1 \supseteq C_2 \supseteq \ldots$, $C_i \in \cP_i $ $\implies | \bigcap_{i} C_i| = 1$.
It is clear that we can do this for metric spaces, It is clear that we can do this for metric spaces.
but such partition can be found for compact Hausdorff spaces as well.
}{}
See \yaref{thm:uflimit} for the full proof.
See
\yaref{fact:compactiffufconv} and
\yaref{fact:hdifffilterlimit} for a more general statement.
\end{refproof} \end{refproof}
Let $\beta \N$ be the Čech-Stone compactification of $\N$, Let $\beta \N$ be the Čech-Stone compactification of $\N$,
@ -157,7 +166,6 @@ is not necessarily continuous.
\[ \[
\forall n.~\exists k < M.~ T^{n+k}(x) \in G. \forall n.~\exists k < M.~ T^{n+k}(x) \in G.
\] \]
\end{definition} \end{definition}
\begin{fact} \begin{fact}
Let $\cU, \cV \in \beta\N$ 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 where a basis consist of sets
$V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$. $V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$.
\gist{%
(For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$ (For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$
and $\beta\N = V_\N$.) and $\beta\N = V_\N$.)
}{}
\item Note also that for $A, B \subseteq \N$, \item Note also that for $A, B \subseteq \N$,
$V_{A \cup B} = V_A \cup V_B$, $V_{A \cup B} = V_A \cup V_B$,
$V_{A^c} = \beta\N \setminus V_A$. $V_{A^c} = \beta\N \setminus V_A$.
\end{itemize} \end{itemize}
\end{fact} \end{fact}
\gist{%
\begin{observe} \begin{observe}
\label{ob:bNclopenbasis} \label{ob:bNclopenbasis}
Note that the basis is clopen. In particular 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 $U = \beta\N \setminus F = \bigcup_{i\in I} V_{A_i}$, If $F$ is closed, then $U = \beta\N \setminus F = \bigcup_{i\in I} V_{A_i}$,
so $F = \bigcap_{i \in I} V_{\N \setminus A_i}$. so $F = \bigcap_{i \in I} V_{\N \setminus A_i}$.
\end{observe} \end{observe}
}{}
\begin{fact} \begin{fact}
\label{fact:bNhd} \label{fact:bNhd}
@ -54,12 +57,14 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
$\bigcap_{j=1}^k F_{i_j} \neq \emptyset$. $\bigcap_{j=1}^k F_{i_j} \neq \emptyset$.
We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$. We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$.
\gist{%
Replacing each $F_i$ by $V_{A_j^i}$ such Replacing each $F_i$ by $V_{A_j^i}$ such
that $F_i = \bigcap_{j \in J_i} V_{A_j^i}$ that $F_i = \bigcap_{j \in J_i} V_{A_j^i}$
(cf.~\yaref{ob:bNclopenbasis}) (cf.~\yaref{ob:bNclopenbasis})
we may assume that $F_i$ is of the form $V_{A_i}$. we may assume that $F_i$ is of the form $V_{A_i}$.
We get $\{F_i = V_{A_i} : i \in I\}$ We get $\{F_i = V_{A_i} : i \in I\}$
with the finite intersection property. with the finite intersection property.
}{Wlog.~$F_i = V_{A_i}$.}
Hence Hence
$\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$ $\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$
has the finite intersection property. 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 $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
\item $\N \subseteq \beta\N$ is dense. \item $\N \subseteq \beta\N$ is dense.
\end{itemize} \end{itemize}
\todo{Easy exercise}
% TODO write down (exercise)
\end{fact} \end{fact}
\begin{theorem} \begin{theorem}
\label{thm:uflimit}
For every compact Hausdorff space $X$, For every compact Hausdorff space $X$,
a sequence $(x_n)$ in $X$, a sequence $(x_n)$ in $X$,
and $\cU \in \beta\N$, and $\cU \in \beta\N$,
@ -133,7 +137,6 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
% agree everywhere (fact section) % agree everywhere (fact section)
\end{proof} \end{proof}
\begin{trivial}+ \begin{trivial}+
More generally,
$\beta$ is a functor from the category of topological $\beta$ is a functor from the category of topological
spaces to the category of compact Hausdorff spaces. spaces to the category of compact Hausdorff spaces.
It is left adjoint to the inclusion functor. It is left adjoint to the inclusion functor.
@ -222,13 +225,12 @@ to obtain
Take $x_2 > x_1$ that satisfies this. Take $x_2 > x_1$ that satisfies this.
\item Suppose we have chosen $\langle x_i : i < n \rangle$. \item Suppose we have chosen $\langle x_i : i < n \rangle$.
Since $\cU$ is idempotent, we have Since $\cU$ is idempotent, we have
\[ \begin{IEEEeqnarray*}{rCl}
(\cU n)[ (\cU n)&& n \in P\\
n \in P &\land& (\cU_k) n + k \in P\\
\land (\cU_k) n + k \in P &\land& \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)\\
\land \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P) &\land& (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
\land (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right). \end{IEEEeqnarray*}
\]
Chose $x_n > x_{n-1}$ that satisfies this. Chose $x_n > x_{n-1}$ that satisfies this.
\end{itemize} \end{itemize}
Set $H \coloneqq \{x_i : i < \omega\}$. Set $H \coloneqq \{x_i : i < \omega\}$.
@ -237,6 +239,3 @@ to obtain
Next time we'll see another proof of this theorem. Next time we'll see another proof of this theorem.

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@ -3,12 +3,36 @@
% Points: 15 / 16 % Points: 15 / 16
\nr 1 \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 \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)] \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 $f,g,h \in X^{\N}$. Let $f,g,h \in X^{\N}$.
@ -70,10 +94,15 @@
\nr 3 \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)] \begin{enumerate}[(a)]
\item $S_{\infty}$ is a Polish space: \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 $G_{\delta}$ Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$
with respect to $\N^\N$. with respect to $\N^\N$.
@ -111,69 +140,9 @@
Clearly there cannot exist a finite subcover Clearly there cannot exist a finite subcover
as $B$ is the disjoint union of the $B_j$. as $B$ is the disjoint union of the $B_j$.
% TODO Think about this
\end{enumerate} \end{enumerate}
\nr 4 \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} \begin{fact}
Let $X $ be a compact Hausdorff space. Let $X $ be a compact Hausdorff space.
Then the following are equivalent: Then the following are equivalent:
@ -205,7 +174,7 @@
Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish} Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish}
and $Y $ Polish. and $Y $ Polish.
Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$. 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. Clearly $d_u$ is a metric.
\begin{claim} \begin{claim}
@ -243,7 +212,7 @@ Clearly $d_u$ is a metric.
for each $y \in X_m$. for each $y \in X_m$.
Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$: Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$:
Indeed if $f \in \cC(X,Y)$ and $\eta > 0$, 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. since $f$ is uniformly continuous.
Let $g \in D_{m,n}$ be such that $\forall y \in X_m.~d(f(y), g(y)) < \frac{1}{n+1}$. 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 $d_u(f,g) \le \eta$, We have $d_u(f,g) \le \eta$,

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

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@ -8,7 +8,7 @@ Let $A \neq \emptyset$ be discrete.
For $D \subseteq A^{\omega}$, For $D \subseteq A^{\omega}$,
let 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)] \begin{enumerate}[(a)]
\item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree: \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} \begin{proof}
(i) $\implies$ (ii): (i) $\implies$ (ii):
Let $(Y,T)$ be a subflow of $(X,T)$. 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 $X$. But $Ty \subseteq Y$, so $Y$ is dense in $X$.
Since $Y$ is closed, we get $Y = X$. 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 $\cF$. is contained in $\cF$.
\begin{fact} \begin{fact}
\label{fact:hdifffilterlimit}
$X$ is Hausdorff iff every filter has at most one limit point. $X$ is Hausdorff iff every filter has at most one limit point.
\end{fact} \end{fact}
\begin{proof} \begin{proof}