\lecture{27}{2024-02-02}{}


\begin{refproof}{thm:unifrprox:helper}
\gist{%
    \begin{subproof}[(2), $\impliedby$, sketch]
        Assume that $x $ is not uniformly recurrent.
        Then there is a neighbourhood $G \ni x$
        such that for all $M \in \N$
        \[
        Y_M = \{ n \in \N : \forall k < M.~T^{n+k}(x) \not\in G\} \neq \emptyset.
        \]
        Note that $Y_1 \supseteq Y_2 \supseteq Y_3 \supseteq \ldots$
        Take $\cV \in \beta\N$ containing all $Y_n$.

        We aim to show that there is no $\cU\in \beta\N$ such that $T_\cU(T_\cV(x)) = x$.
        Towards a contradiction suppose that such $\cU$ exists.

        For every $k + 1$ we have $Y_{k+1} \in \cV$.
        In particular
        \[
        \{n \in \N : T^{n+k}(x) \not\in  G\} \supseteq Y_{k+1},
        \]
        so
        \[
            (\cV n) T^{n+k}(x) \not\in G,
        \]
        i.e.~
        \[
            (\cV n) T^n(x) \not\in  \underbrace{T^{-k}(G)}_{\text{open}}.
        \]
        Thus
        \[
            \underbrace{\ulim{\cV}_n T^n(x)}_{T^\cV(x)} \not\in T^{-k}(G).
        \]
        We get that
        \[
        \forall k.~T^k(T^\cV(x)) \not\in G.
        \]
        It follows that
        $\forall \cU \in \beta\N.~T^{\cU}(T^\cV(x)) \not\in G$.
        % TODO Why? Think about this.

    \end{subproof}
}{%
    \begin{itemize}
        \item 2, $\implies$:
            \begin{itemize}
                \item TODO
            \end{itemize}
        \item 2, $\impliedby$:
            \begin{itemize}
                \item TODO
            \end{itemize}
    \end{itemize}
}
\end{refproof}
Take $X = \beta\N$,
$S \colon \beta\N \to \beta\N$,
$S(\cU ) = \hat{1}+ \cU$.
Then
\[
S^\cV(\cU) = \ulim{\cV}_n S^n(\cU) = \ulim{\cV}_n(\hat{n} + \cU) =
\ulim{\cV}_n \hat{n} + \cU = \cV + \cU.
\]

\begin{corollary}
    $\cU$ is recurrent
    iff
    \[
    \exists \cV \in \beta\N \setminus \N .~S^\cV(\cU) = \cU.
    \]

    $\cU$ is uniformly recurrent iff
    \[
    \forall  \cV.~\exists  \cW.~\cW + \cV + \cU = \cU.
    \]
    $\cU_1$ and $\cU_2$ are proximal
    iff $\exists  \cV.~\cV + \cU_1 = \cV + \cU_2$.
\end{corollary}

\begin{definition}
    We say that $I \subseteq  \beta\N$
    is a \vocab{left ideal} ,
    if
    \[
    \forall  \cU \in I.~\forall  \cV \in \beta\N.~\cV + \cU \in I.
    \]
\end{definition}

\begin{theorem}
    \label{thm:unifrprox:helper2}
    \begin{enumerate}[(1)]
        \item $\cU$ is uniformly recurrent in $\beta\N$
            iff $\cU$ belongs to a minimal\footnote{wrt.~$\subseteq $} (closed)
            left ideal in $\beta\N$.
        \item $\cU$ is an idempotent in $\beta\N$
            iff $\cU$ belong to a minimal closed
            subsemigroup of $\beta\N$.
    \end{enumerate}
\end{theorem}
\begin{proof}
    \begin{enumerate}[(1)]
        \item
            \gist{
            Note that any $\cU \in \beta\N$ yields
            %gives rise to
            a left ideal $\beta\N + \cU$.
            It is closed, since it is the image
            of $\beta\N$ under the continuous maps
            $\cV \mapsto \cV + \cU$
            and  $\beta\N$ is compact.
            }{%
                Note that $\beta\N + \cU$ is closed,
                since $\beta\N$ is compact and $\cdot  + \cU$ continuous.
            }
            $\cU$ belongs to a minimal left ideal
            iff $\beta\N + \cU$ is minimal%
            \gist{,
                since every ideal containing $\cU$
                contains  $\beta\N + \cU$.
            }{.}
            \gist{%
                Note that $\beta\N + \cV + \cU \subseteq \beta\N + \cU$
                and if $I \subsetneq \beta\N + \cU$,
                we have $\cV_0 = \cV + \cU \in I$
                and $\beta\N + \cV + \cU \subseteq \beta\N + \cU$.
                So $\cU$ belongs to a minimal left ideal iff
            }{Equivalently}
            \[
            \forall  \cV \in  \beta\N .~\beta\N + \cV + \cU = \beta\N + \cU.
            \]

            This is the case iff
            \[
            \underbrace{\forall \cV .~\exists \cW.~ \cW + \cV + \cU = \cU.}_%
            {\cV \text{ uniformly recurrent}}
            \]
            \gist{(For one direction take $\cW$ such that $\cW + \cV + \cU= \hat{0} +\cU$.
                For the other direction note that
                for every $\cV_0 $, $\cV_0 + \cU$
                can be written as $\cV_0 + \cW + (\cV + \cU)$.
                Where we take $\cW$ such that $\cW + \cV + \cU = \cU$.
            }{}

        \item This is very similar to the proof of the \yaref{lem:ellisnumakura}.

            If $\cU$ is idempotent, then $\{\cU\}$
            is a semigroup.
            Let $C$ be a minimal closed subsemigroup of $\beta\N$.
            Then $C + \cU$ is a closed subsemigroup.
            By minimality, we get $C = C + \cU$.

            Let  $D = \{ \cV \in C .~ \cV + \cU = \cU\}$.
            We have $D \neq \emptyset$.
            $D$ is a closed semigroup,
            so $D = C$ be minimality.
            Hence $\cU + \cU = \cU$.
    \end{enumerate}
\end{proof}
\begin{corollary}
    Idempotent and uniformly recurrent elements exist.
\end{corollary}
\begin{proof}
    Use \yaref{thm:unifrprox:helper2}
    and Zorn's lemma.
\end{proof}
\begin{theorem}
    (1) $\implies$ (2) $\implies$ (3)
    where
    \begin{enumerate}[(1)]
        \item $\cU$ is uniformly recurrent and proximal to $\hat{0}$.
        \item $\cU$ is an idempotent.
        \item $\cU$ is recurrent and proximal to $\hat{0}$.
    \end{enumerate}
\end{theorem}
\begin{proof}
    (1) $\implies$ (2):
    Let $\cU$ be uniformly recurrent and proximal to $ \hat{0}$.
    Take $\cV$ such that $\cV + \cU = \cV + \hat{0} = \cV$.
    % TODO REF beginning of lecture

    Since  $\cU$ is uniformly recurrent,
    there exists $\cW$ such that $\cW + \cV + \cU = \cU$,
    i.e.~$\cW + \cV = \cU$.
    Then $\cU + \cU = \cW + \cV + \cU = \cU$.

    (2) $\implies$ (3):
    Let $\cU$ be an idempotent.
    Then $\cV + \cU = \cV$ (proximal to  $0$)
    and  $\cV + \cU = \cU$ (recurrent)
    are satisfied for  $\cV \coloneqq  \cU$.
\end{proof}

\begin{corollary}
    $\cU$ is uniformly recurrent and proximal to $0$
    iff $\cU$ is an idempotent
    and belongs to some minimal left ideal of $\beta\N$.
\end{corollary}

Finally:
\begin{refproof}{thm:unifrprox}
    Let $T\colon  X \to  X$ and $x \in X$.
    We want to find $y \in X$
    such that $y$ is uniformly recurrent
    and proximal to $x$.

    We first prove a version for ultrafilters and then
    transfer it to $X$.

    There exists a uniformly recurrent $\cV \in \beta\N$.
    So for any $\cW$,
    $\cW + \cV$ is also uniformly recurrent\gist{:
    Take $\cV_0$.
    We need to find $\cX$ such that $\cX + \cV_0 + \cW +\cV = \cW + \cV$.
    By uniform recurrence of $\cV$ we find $\cX'$
    such that  $\cX' + (\cV_0 + \cW) + \cV = \cV$.
    Then $\cX = \cW + \cX'$ works.
    }{.}
    So all elements of $\beta\N + \cV$
    are uniformly recurrent.
    It is a closed ideal and hence a closed semigroup.
    So $\beta\N + \cV$ contains a minimal closed
    semigroup.
    In particular, there exists an idempotent $\cU \in  \beta\N + \cV$.

    $\cU$ is idempotent and uniformly recurrent
    hence it is proximal to $0$.


    Now let us consider $X$.
    Take $y = T^\cU(x)$.

    \begin{claim}
        $y$ uniformly recurrent.
    \end{claim}
    \begin{subproof}
        Recall that $T^{\cV_1 + \cV_2} = T^{\cV_1} \circ T^{\cV_2}$.

        Since $\cU$ is uniformly recurrent,
        $\forall  \cV .~\exists  \cW.~\cW+ \cV+\cU= \cU$,
        i.e.~$T^{\cW + \cV + \cU} (x) = T^\cW(T^\cV(y)) = T^\cU(x) = y$.
    \end{subproof}
    \begin{claim}
        $y$ is proximal to $x$.
    \end{claim}
    \begin{subproof}
        $\cU$ is proximal to $0$.
        So $\exists  \cV.~\cV + \cU = \cV + \hat{0} = \cV$,
        i.e.~$T^{\cV}(y) = T^{\cV + \cU}(x) = T^\cV(x)$.
        Thus $x$ and $y$ are proximal.%TODO REF
    \end{subproof}
\end{refproof}
% Office hours wednesday 15:30 - 18:30 office 805
% Exam: First question: present favorite theorem (7-8 minutes, moderate length proof)