w23-logic-3/inputs/lecture_26.tex

\lecture{26}{2024-01-30}{}

Let $T\colon  X \to X$ be a continuous map.
This gives $\N \acts X$.

\begin{definition}
    \label{def:unifrec}
    A point $x \in X$
    is called \vocab{uniformly recurrent}
    iff
    for each neighbourhood $G$ of $x$,
    there is $M \in \N_+$,
    such that
    \[
    \forall  n \in \N.~\exists  k < m.~T^{n+k}(x) \in G.
    \]
\end{definition}
\begin{definition}
    A pair $x,y \in X$ is \vocab{proximal}%
    \footnote{see also \yaref{def:flow}, where we defined proximal for metric spaces}
    iff for all neighbourhoods $G$ of
    the diagonal%
    \gist{\footnote{recall that the diagonal is defined to be $\Delta \coloneqq  \{(x,x) : x \in X\}$}}{}
    infinitely many $n$ satisfy $(T^n(x), T^n(y)) \in G$.
\end{definition}

\begin{theorem}
    \label{thm:unifrprox}
    Let $X$ be a compact Hausdorff space and $T\colon  X \to X$
    continuous.
    Consider $(X,T)$.%TODO different notations
    Then for every $x \in X$
    there is a uniformly recurrent $y \in X$
    such that $y $ is proximal to $x$.
\end{theorem}


We do a second proof of \yaref{thm:hindman}:
\begin{refproof}{thm:hindmanfurstenberg}[Furstenberg]
\gist{%
    A partition of $\N$ into $k$-many pieces can be viewed
    as a function $f\colon \N \to k$.

    Let $X = k^\N$ be the set of all such functions.
    Equip $X$ with the product topology.
    Then $X$ is compact and Hausdorff.

    Let $T\colon  X \to X$ be the shift
    given by \gist{%
    \begin{IEEEeqnarray*}{rCl}
        T\colon k^{\N} &\longrightarrow & k^{\N} \\
        (y\colon \N \to  k)&\longmapsto &
        \begin{pmatrix*}[l]
            \N &\longrightarrow & k \\
            n &\longmapsto & y(n+1),
        \end{pmatrix*}
    \end{IEEEeqnarray*}
    i.e.~}{}%
    $T(y)(n) = y(n+1)$.

    Let $x $ be the given partition.
    We want to find an infinite set $H$ for $x$ as in the theorem.
    Let $y$ be uniformly recurrent and proximal to $x$.

    \begin{itemize}
        \item % Gist: proximal
            Since  $x$ and $y$ are proximal,
            we get that for every $N \in \N$,
            there are infinitely many $n$ such that
            $T^n(x)\defon{N} = T^n(y)\defon{N}$.%
            \footnote{%
                Consider $G_N = \{(a,b) \in X^2 : a\defon N = b\defon N\}$
                This is a neighbourhood of the diagonal.%
            }
        \item % Gist: unif. recurrent
            Consider the neighbourhood
            \[
            G_n \coloneqq \{z \in X: z\defon{n} = y\defon{n}\}
            \]
            of $y$.
            By the uniform recurrence of $y$,
            we get that%
            \footnote{Note that here we might need to choose
                a bigger $N$ than the $M$ in \yaref{def:unifrec},
                but $2M$ suffices.}%
            \[
            \forall n.~\exists N% \gg n
            .~\forall r.~(y(r), y(r+1), \ldots, y(r+N - 1)
            \text{ contains }
            (y(0), y(1), \ldots, y(n))
            \text{ as a subsequence.}
            \]
    \end{itemize}

    Consider $y(0)$.
    We will prove that this color works and construct a corresponding $H$.

    \begin{itemize}
        \item % Step 1
            Let $G_0 \coloneqq  [y(0)]$
            and let $N_0$ be such that
            \[
            \forall  r.~(y(r), \ldots, y(r + N_0 - 1)) \text{ contains $y(0)$.}
            \]
            By proximality, there exist infinitely many $r$
            such that  $(y(r), \ldots, (y(r+N_0-1)) = (x(r), \ldots, x(r+N_0-1))$.
            Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.

        \item%
            % Step 2
            Let $G_{n_0} = [(y(0), \ldots, y(h_0)]$.
            Choose $N_1$.

            For all $r$,
            $(y(r), \ldots, y(r+N-1))$ contains
            $(\underbrace{y(0)}_{= C}, \ldots, \underbrace{y(h_0)}_{= C})$.

            Pick $r > h_0$ such that
            $(x(r), \ldots, x(r+N-1))$ contains
            $(y(0), \ldots y(h_0))$.
            Let $(x(r+s), \ldots, x(r+s+h_0)) = (y(0), \ldots, y(h_0))$.
            Then set $h_1 = r + s$.

            Then $x(h_0) = c$, $x(h_1) = y(0) = c$
            and $x(h_0+h_1) = y(h_0) = c$.


        \item%
            % Step 3
            Let $G_{h_0 + h_1} = [y(0), \ldots, y(h_0+h_1)]$.
            Let $r > h_0 + h_1$.
            Choose $N_2$ large enough
            such that
            $(y(0), \ldots, y(h_0+h_1))$
            is contained in $(x(r), \ldots, x(r+N-1))$.
            Let $(y(0), \ldots, y(h_0+h_1)) = (x(r+s), \ldots, x(r+s+N-1))$.
        \item Repeat this:
            Inductively choose $h_i$ such that $x(s+h_i) = y(s+h_i) = c$
            for all sums  $s$ of subsets of $\{h_0,\ldots,h_{i-1}\}$.
            To do this, find $N_i$ such that every $N_i$ consecutive
            terms of $y$ contain $(y(0), \ldots, y(\sum_{j < i} h_j))$.
            Then find $h_i > h_{i-1}$ such that
            $(x(h_i), \ldots, x(\sum_{j < i} h_j)) = (y(0), \ldots, y(\sum_{j < i} h_j))$.
    \end{itemize}
}{
    \begin{itemize}
        \item View partition as $f\colon  \N \to k$. Consider $X \coloneqq k^{\N}$ (product topology, compact and Hausdorff).
            Let $x \in X$ be the given partition.
        \item $T\colon  X \to X$ shift: $T(y)(n) \coloneqq  y(n+1)$.
        \item Let $y$ proximal to $x$, uniformly recurrent.
        \begin{itemize}
            \item proximal $\leadsto$ $\forall N$.~$T^n(x)\defon{N} = T^n(y)\defon{N}$
                for infinitely many $n$.
            \item uniform recurrence $\leadsto$
                \[
                    \forall n .~\exists N.~\forall r.y\defon{\{r,\ldots,r+N-1\}}
                    \text{ contains } y\defon{\{0,\ldots,n\}} \text{ as a subsequence.}
                \]
                (consider neighbourhood $G_n = \{z \in X : z\defon{n}  = y\defon{n} \}$).
        \end{itemize}
        \item Consider $c \coloneqq y(0)$. This color works:
            \begin{itemize}
                \item $G_0 \coloneqq  y\defon{\{0\}}$,
                take $N_0$ such that $y\defon{\{r, \ldots, r + N_0 - 1\}} $
                contains $y(0)$ for all $r$ (unif.~recurrence).
                $y\defon{\{r,\ldots,r+N_0 - 1\} } = x\defon{\{r,\ldots,r+N_0 -1\} }$
                for infinitely many $r$ (proximality).
                Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
                \item $G_1 \coloneqq y\defon{\{0,\ldots,h_0\} }$,
                take $N_1$ such that $y\defon{\{r,\ldots,r +N_1-1\}}$
                contains $y\defon{\{0,\ldots,h_0\} }$
                for all $r$ (unif.~recurrence).
                So among ever $N_1$ terms, there are two of distance $h_0$
                where $y$ has value  $c$.
                So $\exists  h_1 > h_0$ such that $x(h_1) = x(h_1 + h_0) = c$
                (proximality).

            \item Repeat:
                Choose $h_i$ such that
                for all sums $s$ of subsets of  $\{h_0,\ldots, h_{i-1}\}$,
                $x(s+h_i) = y(s+h_i) = c$:
                Find $N_i$ such that every $N_i$ consecutive
                terms of $y$ contain a segment that coincides
                with the initial segment of $y$
                up  to the largest $s$,
                then find a segment of length $N_i$ beyond $h_{i-1}$
                where $x$ and $y$ coincide.
            \end{itemize}
    \end{itemize}
}
\end{refproof}

In order to prove \yaref{thm:unifrprox},
we need to rephrase the problem in terms of $\beta\N$:
\begin{theorem}
    \label{thm:unifrprox:helper}
    Let $X$ be a compact Hausdorff space. % ?
    Let $T\colon  X \to X$ be continuous.
    \begin{enumerate}[(1)]
        \item $x \in X$ is recurrent
            iff $T^\cU(x) = x$ for some  $\cU \in  \beta\N \setminus \N$.
        \item $x \in X$ is uniformly recurrent
            iff for every $\cV \in  \beta\N$,
            there is $\cU \in \beta\N$
            with $T^{\cU}(T^{\cV}(x)) = x$.
        \item $x, y  \in X$ are proximal
            iff there is $\cU \in  \beta\N$
            such that $T^\cU(x) = T^\cU(y)$.
            % TODO compare with the statement for the ellis semigroup.
    \end{enumerate}
\end{theorem}


\gist{%
\begin{refproof}{thm:unifrprox:helper}[sketch]
    We only prove (2) here, as it is the most interesting point.%
    \todo{other parts will be in the official notes}

    \begin{subproof}[(2), $\implies$]
        Suppose that $ x$ is uniformly recurrent.
        Take some $\cV \in \beta\N$.
        Let $G_0$ be a neighbourhood of $x$.
        Then $x \in G \subseteq G_0$,
        where $G$ is a closed neighbourhood,
        i.e.~$X \in \inter G$.

        Let $M$ be such that
        \[
        \forall  n .~ \exists  k < M.~T^{n+k}(x) \in G.
        \]
        So there is a $k < M$ such that
        \[
            (\cV n) T^{n +k}(x) \in G.
        \]


        Hence
        \[
            (\cV n) T^n(x) \in  \underbrace{T^{-k}(\overbrace{G}^{\text{closed}})}_{\text{closed}}.
        \]

        Therefore $\ulim{\cV}_n T^n(x) \in T^{-k}(G)$.
        So $T^k(T^\cV(x)) \in G \subseteq G_0$.


        We have shown that for every open neighbourhood $G$ of $x$,
        the set $Y_G = \{k \in \N : T^k(T^\cV(x)) \in G\} \neq \emptyset$.
        The sets $\{Y_G : G \text{ open neighbourhood of } G\}$
        form a filter basis,\footnote{The sets and their supersets form a filter.}
        since $Y_{G_1} \cap Y_{G_2} = Y_{G_1 \cap G_2}$.
        Let $\cU$ be an ultrafilter containing all the  $Y_G$.

        Then
        \[
            (\cU k) R^k(T^\cV)(x)) \in G
        \]
        i.e.~$T^\cU(T^\cV(x)) \in \overline{G}$.

        Since we get this for every neighbourhood,
        it follows that $T^\cU ( T^\cV(x)) = x$.
    \end{subproof}

    \phantom\qedhere
\end{refproof}
}{}