Josia Pietsch
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178 lines
5.8 KiB
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\lecture{24}{2024-01-23}{Combinatorics!}
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\subsection{Applications to Combinatorics} % Ramsey Theory}
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% TODO Define Ultrafilter
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\begin{definition}
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An \vocab{ultrafilter} on $\N$ (or any other set)
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is a family $\cU \subseteq \cP(\N)$
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such that
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\begin{enumerate}[(1)]
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\item $X \in \cU \land X \subseteq Y \subseteq \N \implies Y \in \cU$.
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\item $X,Y \in \cU \implies X \cap Y \in \cU$.
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\item $\emptyset \not\in \cU$, $\N \in \cU$.
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\item For all $X \subseteq \N$,
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we have $X \in \cU \lor \N \setminus X \in \cU$.
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\end{enumerate}
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\end{definition}
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\begin{remark}
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\begin{itemize}
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\item If $X \cup Y \in \cU$ then $X \in \cU \lor Y$ or $Y \in \cU$:
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Consider $((\N \setminus X) \cap (\N \setminus Y) = \N \setminus (X \cup Y)$.
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\item Every filter can be extended to an ultrafilter.
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(Zorn's lemma)
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\end{itemize}
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\end{remark}
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\begin{definition}
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An ultrafilter is called \vocab[Ultrafilter!principal]{principal} or \vocab[Ultrafilter!trivial]{trivial}
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if it is of the form
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\[
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\hat{n} = \{X \subseteq \N : n \in X\}.
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\]
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\end{definition}
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\begin{notation}
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Let $\phi(\cdot )$ be a formula, where the argument is a natural number.
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Let $\cU$ be an ultrafilter.
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We write
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\[
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(\cU n) ~ \phi(n)
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\]
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for $\{ n \in \N : \phi(n)\} \in \cU$.
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We say that $\phi(n)$ holds for \vocab{$\cU$-almost all} $n$.
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\end{notation}
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\begin{observe}
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Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas.
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\begin{enumerate}[(1)]
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\item $(\cU n) (\phi(n) \land \psi(m)) \iff (\cU n) \phi(n) \land (\cU n) \psi(n)$.
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\item $(\cU n) (\phi(n) \lor \psi(m)) \iff (\cU n) \phi(n) \lor (\cU n) \psi(n)$.
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\item $(\cU n) \lnot \phi(n) \iff \lnot (\cU n) \phi(n)$.
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\end{enumerate}
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\end{observe}
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\begin{lemma}
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\label{lem:ultrafilterlimit}
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Let $X $ be a compact Hausdorff space.
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Let $\cU$ be an ultrafilter.
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Then for every sequence $(x_n)$ in $X$,
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there is a unique $x \in X$,
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such that
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\[
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(\cU_n) (x_n \in G)
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\]
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for every neighbourhood%
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\footnote{$G \subseteq X$ is a neighbourhood iff $x \in \inter G$.}
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$G$ of $x$.
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\end{lemma}
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\begin{notation}
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In this case we write $x = \cU-\lim_n x_n$.
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\end{notation}
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\begin{refproof}{lem:ultrafilterlimit}[sketch]
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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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So we can repeatedly chop the space in two pieces,
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one of them contains $\cU$-almost all $x_n$,
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Then we restrict to this piece and continue.
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For this to work, we need
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a finite collection $\cP_n$ of closed sets for every $n$,
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such that $\bigcup \cP_n = X$,
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$C \in \cP_{n+1} \implies \exists C \subseteq D \in \cP_{n}$
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and
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$C_1 \supseteq C_2 \supseteq \ldots$, $C_i \in \cP_i $ $\implies | \bigcap_{i} C_i| = 1$.
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It is clear that we can do this for metric spaces,
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but such partition can be found for compact Hausdorff spaces as well.
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\end{refproof}
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Let $\beta \N$ be the Čech-Stone compactification of $\N$,
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i.e.~the set of all ultrafilters on $\N$
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with the topology given by open sets $V_{A} = \{ p \in \beta\N : A \in P\} $
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for $A \subseteq \N$.
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This is a compact Hausdorff space.%
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\footnote{cf.~\yaref{fact:bNhd}, \yaref{fact:bNcompact}}%
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\todo{move facts}
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We can turn it into a compact semigroup:
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Consider $+ \colon \N \times \N \to \N$.
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This gives an operation on principal ultrafilters
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(we identify $n \in \N$ with the corresponding principal filter).
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We want to extend this to all of $\beta\N$.
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Fix the first argument to get a function $\N \to \N, n \mapsto k+n$.
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For $\cU \in \beta\N$ consider $\cU-\lim_n (k+n)$.
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So for a fixed $k \in \N$ we get $k+ \cdot \colon\beta\N \to \beta\N$,
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i.e.~$+ \colon \N \times \beta\N \to \beta\N$.
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Fixing the second coordinate to be $\cV \in \beta\N$,
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we get a function $+\cV \colon \N \to \beta\N$.
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For $ \cU \in \beta\N$
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consider $\cU-\lim_n n + \cV$.
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This gives $+ \colon \beta\N \times \beta\N \to \beta\N$.
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% TODO ?
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\[
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\cU + \cV = \{X \subseteq \N : \{m \colon \{n \colon m+n \in X\} \in \cV \} \in \cU \}.
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\]
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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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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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but not a $\Z$-action.}
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For any $\cU \in \beta\N$, we define $T^{\cU}$ by
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$T^\cU(x) \coloneqq \cU-\lim_n T^n(x)$ for $x \in X$.
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For fixed $x$, the map $\cU \mapsto T^{\cU}(x)$ is continuous.
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(More generally, for every $f\colon \N \to X$
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the extension $\tilde{f}\colon \beta\N \to X$ is continuous).
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Note that for fixed $\cU$, the map $x \mapsto T^\cU(x)$
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is not necessarily continuous.
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\begin{definition}
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Let $X$ be a compact Hausdorff space
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and $T\colon X\to X$ continuous.
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A point $x \in X$ is \vocab{recurrent},
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iff for every neighbourhood $G$ of $x$,
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infinitely many $n$ satisfy $T^n(x) \in G$.
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A point $x \in X$ is \vocab{uniformly recurrent},
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if for every neighbourhood $G$ of $x$,
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there exists $M \in \N$,
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such that
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\[
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\forall n.~\exists k < M.~ T^{n+k}(x) \in G.
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\]
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\end{definition}
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\begin{fact}
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Let $\cU, \cV \in \beta\N$
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and $T\colon X \to X$ continuous
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for a compact Hausdorff space $X$.
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Then $T^{\cU}(T^{\cV}(x)) = T^{\cU + \cV}(x)$.
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\end{fact}
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\begin{proof}
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\begin{IEEEeqnarray*}{rCl}
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T^{\cU + \cV}(x) &=& (\cU + \cV)-\lim_k T^k(x)\\
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&=& \cU-\lim_m \cV-\lim_n T^{m+n}(x)\\
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&\overset{T^m \text{ continuous}}{=}& \cU-\lim_m T^m (\cV-\lim_n T^n(x))\\
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&=& T^\cU(T^\cV(x)).
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\end{IEEEeqnarray*}
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\end{proof}
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\todo{Homework: Check the details that were omitted during the lecture.}
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