\lecture{24}{2024-01-23}{Combinatorics!}

% ANKI 2


\subsection{Applications to Combinatorics} % Ramsey Theory}

\begin{definition}
    An \vocab{ultrafilter} on $\N$ (or any other set)
    is a family $\cU \subseteq \cP(\N)$
    such that
    \begin{enumerate}[(1)]
        \item $X \in \cU \land X \subseteq Y \subseteq \N \implies Y \in \cU$.
        \item $X,Y \in \cU \implies X \cap Y \in \cU$.
        \item $\emptyset \not\in \cU$, $\N \in \cU$.
        \item For all $X \subseteq \N$,
            we have $X \in \cU \lor \N \setminus X \in \cU$.
    \end{enumerate}
\end{definition}
\gist{
\begin{remark}
    \begin{itemize}
        \item If $X \cup Y \in \cU$ then $X \in \cU$ or $Y \in \cU$:
            Consider $((\N \setminus X) \cap (\N \setminus Y) = \N  \setminus (X \cup Y)$.
        \item Every filter can be extended to an ultrafilter.
            (Zorn's lemma)
    \end{itemize}
\end{remark}
}{}
\begin{definition}
    An ultrafilter is called \vocab[Ultrafilter!principal]{principal} or \vocab[Ultrafilter!trivial]{trivial}
    iff it is of the form
    \[
    \hat{n} = \{X \subseteq \N : n \in X\}.
    \]
\end{definition}
\begin{notation}
    Let $\phi(\cdot )$ be a formula, where the argument is a natural number.
    Let $\cU$ be an ultrafilter.
    We write
    \[
        (\cU n) ~ \phi(n)
    \]
    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.

    \begin{enumerate}[(1)]
        \item $(\cU n) ~ (\phi(n) \land \psi(m)) \iff (\cU n) \phi(n) \land (\cU n) \psi(n)$.
        \item $(\cU n) ~ (\phi(n) \lor \psi(m)) \iff (\cU n) ~ \phi(n) \lor (\cU n) ~ \psi(n)$.
        \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.
    Let $\cU$ be an ultrafilter.
    Then for every sequence $(x_n)$ in $X$,
    there is a unique $x \in X$,
    such that
    \[
    (\cU n)~(x_n \in G)
    \]
    for every neighbourhood%
    \footnote{$G \subseteq X$ is a neighbourhood iff $x \in \inter G$.}
    $G$ of $x$.
\end{lemma}
\begin{notation}
    In this case we write $x = \ulim{\cU}_n x_n$.
\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 $(\cU n) x_n \in Z$.

    So we can repeatedly chop the space in two pieces,
    one of them contains $\cU$-almost all $x_n$,
    Then we restrict to this piece and continue.

    For this to work, we need
    a finite collection $\cP_n$ of closed sets for every $n$,
    such that $\bigcup \cP_n = X$,
    $C \in \cP_{n+1} \implies \exists  C \subseteq D \in \cP_{n}$
    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.

}{}
    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 $\N$,
i.e.~the set of all ultrafilters on $\N$
with the topology given by open sets $V_{A} = \{ p \in \beta\N : A \in P\} $
for $A \subseteq  \N$.

This is a compact Hausdorff space.%
\footnote{cf.~\yaref{fact:bNhd}, \yaref{fact:bNcompact}}%
\todo{move facts}
We can turn it into a compact semigroup:
Consider $+ \colon \N \times  \N \to  \N$.
This gives an operation on principal ultrafilters
(we identify $n \in \N$ with the corresponding principal filter).
We want to extend this to all of $\beta\N$.
Fix the first argument to get a function $\N \to  \N, n \mapsto k+n$.
For $\cU \in \beta\N$ consider $\ulim{\cU}_n (k+n)$.
So for a fixed $k \in \N$ we get $k+ \cdot \colon\beta\N \to  \beta\N$,
i.e.~$+ \colon \N \times  \beta\N \to  \beta\N$.
Fixing the second coordinate to be $\cV \in \beta\N$,
we get a function $+\cV \colon \N \to \beta\N$.
For $ \cU \in \beta\N$
consider $\ulim{\cU}_n n + \cV$.
This gives $+ \colon \beta\N \times  \beta\N \to  \beta\N$.
% TODO ?

\[
\cU + \cV = \{X \subseteq \N : \{m \colon  \{n \colon  m+n \in X\} \in \cV \} \in \cU \}.
\]


This is not commutative,
but associative and $a \mapsto a + b$ is continuous
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 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
$T^\cU(x) \coloneqq  \ulim{\cU}_n T^n(x)$ for $x \in X$.

For fixed $x$, the map $\cU \mapsto T^{\cU}(x)$ is continuous.

(More generally, for every $f\colon  \N \to  X$
the extension $\tilde{f}\colon \beta\N \to X$ is continuous).

Note that for fixed $\cU$, the map $x \mapsto T^\cU(x)$
is not necessarily continuous.


\begin{definition}
    Let $X$ be a compact Hausdorff space
    and $T\colon X\to X$ continuous.
    A point $x \in X$ is \vocab{recurrent},
    iff for every neighbourhood $G$ of $x$,
    infinitely many $n$ satisfy $T^n(x) \in G$.

    A point $x \in X$ is \vocab{uniformly recurrent},
    if for every neighbourhood $G$ of $x$,
    there exists $M \in \N$,
    such that
    \[
    \forall n.~\exists k < M.~ T^{n+k}(x) \in G.
    \]
\end{definition}
\begin{fact}
    Let $\cU, \cV \in \beta\N$
    and $T\colon  X \to  X$ continuous
    for a compact Hausdorff space $X$.
    Then $T^{\cU}(T^{\cV}(x)) = T^{\cU + \cV}(x)$.
\end{fact}
\begin{proof}
    \begin{IEEEeqnarray*}{rCl}
        T^{\cU + \cV}(x) &=& \ulim{(\cU + \cV)}_k T^k(x)\\
                         &=& \ulim{\cU}_m \ulim{\cV}_n T^{m+n}(x)\\
                         &\overset{T^m \text{ continuous}}{=}& \ulim{\cU}_m T^m (\ulim{\cV}_n T^n(x))\\
                         &=& T^\cU(T^\cV(x)).
    \end{IEEEeqnarray*}
\end{proof}
\todo{Homework: Check the details that were omitted during the lecture.}




