w23-logic-3/inputs/lecture_25.tex

\lecture{25}{2024-01-26}{}

Let $\beta\N$ denote the set of ultrafilters on $\N$.
\begin{fact}
    \begin{itemize}
        \item This is a topological space,
            where a basis consist of sets
            $V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$.

            (For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$
            and $\beta\N = V_\N$.)

        \item Note also that for $A, B \subseteq \N$,
            $V_{A \cup B} = V_A \cup V_B$,
            $V_{A^c} = \beta\N \setminus V_A$.
    \end{itemize}
\end{fact}

\begin{observe}
    \label{ob:bNclopenbasis}
    Note that the basis is clopen. In particular
    any closed set can be written as an intersection of sets
    of the form  $V_A$:

    If $F$ is closed, then $U = \beta\N \setminus F = \bigcup_{i\in I} V_{A_i}$,
    so $F = \bigcap_{i \in I} V_{\N \setminus A_i}$.
\end{observe}

\begin{fact}
    \label{fact:bNhd}
    $\beta\N$ is Hausdorff.
\end{fact}
\begin{proof}
    Let $\cU \neq  \cV \in \beta\N$.
    Then there is some $A \in \cU \setminus \cV$,
    so $A^c \in \cV$,
    so $\cU \in V_A$ and $\cV \in V_A^c$.
\end{proof}
%\begin{remark}+
%    This even shows that  $\beta\N$ is totally separated.
%    In fact, $\beta\N$ is a profinite space,
%    as the next fact shows.
%\end{remark}
\begin{fact}
    \label{fact:bNcompact}
    $\beta\N$ is compact.
\end{fact}
\begin{proof}
    Let $\{F_i\}_{i \in I}$ be non-empty
    and closed
    such that
    for any $i_1,\ldots., i_k \in I$,
    $k \in \N$,
    $\bigcap_{j=1}^k F_{i_j} \neq \emptyset$.

    We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$.
    Replacing each $F_i$ by $V_{A_j^i}$ such
    that $F_i = \bigcap_{j \in J_i} V_{A_j^i}$
    (cf.~\yaref{ob:bNclopenbasis})
    we may assume that $F_i$ is of the form $V_{A_i}$.
    We get $\{F_i = V_{A_i} : i \in I\}$
    with the finite intersection property.
    Hence
    $\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$
    has the finite intersection property.

    Then $\cF = \{A \subseteq  \N : A \supseteq A_{i_1} \cap \ldots \cap A_{i_k}, k \in \N, i_1, \ldots, i_k \in I\}$
    is a filter.

    Let $\cU$ be an ultrafilter extending $\cF$.
    Then  $\cU \in \bigcap_{i \in I} V_{A_i} = \bigcap_{i \in I} F_i$.
\end{proof}
\begin{fact}
    Consider $\N$ as a subspace of $\beta\N$
    via $\N \hookrightarrow \beta\N, n \mapsto \hat{n} \coloneqq \{A \subseteq \N : n \in A\}$.
    Then
    \begin{itemize}
        \item $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
        \item $\N \subseteq  \beta\N$ is dense.
    \end{itemize}
    \todo{Easy exercise}
    % TODO write down (exercise)
\end{fact}

\begin{theorem}
    For every compact Hausdorff space $X$,
    a sequence $(x_n)$ in $X$,
    and $\cU \in \beta\N$,
    we have that $\ulim{\cU}_n x_n = x$
    exists and is unique,
    i.e.~for all  $x \in G \overset{\text{open}}{\subseteq} X$
    we have $\{n \in \N : x_n \in G\}  \in \cU$.
\end{theorem}
\begin{proof}
    Towards a contradiction
    assume that there is no such $x$.

    For every $x$ take $x \in G_x \overset{\text{open}}{\subseteq} X$
    such that
    $\{ n \in \N : x_n \in G_x\} \not\in  \cU$.
    So $\{G_x\}_{x \in X}$ is an open cover of $X$.
    Since $X$ is compact,
    there exists a finite subcover
    $G_{x_1}, \ldots, G_{x_m}$.

    But then
    \begin{IEEEeqnarray*}{rCl}
        \N &=& \{ n \in \N : x_n \in \bigcup_{i=1}^m G_{x_i}\}\\
           &=& \underbrace{\bigcup_{i=1}^m \overbrace{\{n \in \N : x_n \in G_{x_i}\}}^{\not\in \cU}}_{\not\in \cU},
    \end{IEEEeqnarray*}
    since $B_1 \cup \ldots \cup B_m \in \cU \iff \exists  i < m.~B_i \in \cU$.

    It is clear that  $\ulim{\cU}_n x_n$
    is unique, since  $X$ is Hausdorff.
\end{proof}

\begin{theorem}
    Let $X$ be a compact Hausdorff space.
    For any $f\colon  \N \to X$
    there is a unique continuous extension $\tilde{f}\colon \beta\N \to X$.
\end{theorem}
\begin{proof}
    Let
    \begin{IEEEeqnarray*}{rCl}
        \tilde{f}\colon \beta\N &\longrightarrow & X \\
        \cU &\longmapsto & \ulim{\cU}_n f(n).
    \end{IEEEeqnarray*}
    \todo{Exercise: Check that $\tilde{f}$ is continuous.}

    $\tilde{f}$ is uniquely determined,
    since $\N \subseteq \beta\N$ is dense.
    % TODO general fact: continuous functions agreeing on a dense set
    % agree everywhere (fact section)
\end{proof}
\begin{trivial}+
    More generally,
    $\beta$ is a functor from the category of topological
    spaces to the category of compact Hausdorff spaces.
    It is left adjoint to the inclusion functor.
\end{trivial}

% RECAP
\gist{%
$\beta\N$ is equipped with $+$ which extends $+\colon  \N \times \N \to \N$,
\[
 \cU + \cV = \{A \subseteq  \N : (\cU m)\left( (\cU n) \{m+n \in A\}  \right)\}.
\]
This is associative, but not commutative.
}{}
% END RECAP
\begin{fact}
    $+\colon \beta\N\times \beta\N \to \beta\N$
    is left continuous,
    i.e.~for $\cV$ fixed,
    $\cU \mapsto \cU + \cV$ is continuous.
\end{fact}
\begin{proof}
    Fix $A$ and consider $V_A$.
    We need to show that the inverse image of  $V_A$ is open.

    We have
    \begin{IEEEeqnarray*}{rCl}
        \cU + \cV \in V_A &\iff& A \in  \cU + \cV\\
                          &\iff& (\cU_m)(\cV_n) \{m+n \in A\}\\
                          &\iff& \{m \in \N : (\cV n) m+n \in A\} \in \cU\\
                          &\iff& \cU \in V_{\{m \in \N: (\cV n) m+n \in A\}}.
    \end{IEEEeqnarray*}
\end{proof}
\begin{corollary}
    $(\beta\N,+)$
    is a %(left)
    \vocab{compact semigroup},
    i.e.~it is compact, Hausdorff, associative and left-continuous.%
    %\footnote{There is no convention on left and right.}
\end{corollary}
So we can apply the \yaref{lem:ellisnumakura}
to obtain
\begin{corollary}
    There is $\cU \in \beta\N$
    such that $\cU + \cU = \cU$.
\end{corollary}
\begin{observe}
    Principal ultrafilters $\neq \hat{0}$ are not idempotent.
    We can restrict to $\beta\N \setminus \N$
    to get an idempotent element that is not principal.
    % TODO THINK ABOUT THIS
\end{observe}

\begin{theorem}[Hindman]
    \label{thm:hindman}
    If $\N$ is partitioned into finitely many
    sets,
    then there is is an infinite subset $H \subseteq \N$
    such that all finite sums of distinct
    elements of $H$
    belong to the same set of the partition.
\end{theorem}
\begin{proof}[Galvin,Glazer]
    Let $\cU \in \beta\N \setminus \N$
    be such that $\cU + \cU = \cU$.
    Let $P$ be the piece of the partition
    that is in $\cU$.
    So $(\cU n ) n \in P$.
    Let us define a sequence $x_1,x_2,\ldots$
    \begin{itemize}
        \item $\cU$ is idempotent,
            so $(\cU n)(\cU k) n+k \in P$.
            We get
            \[(\cU n) \left( n \in P \land (\cU_k) n+k \in P \right)\].
            Pick $x_1$ that satisfies this,
            i.e.~$x_1 \in P$ and $(\cU_k) x_1+k \in P$.
        \item $\cU$ is idempotent,
            so
            \[
                (\cU n)[
                    n \in P
                    \land (\cU_k) n + k \in P
                    \land x_1 + n \in P
                    \land (\cU_k) x_1 + n + k \in P
                ]
            \]
            Take $x_2 > x_1$ that satisfies this.
        \item Suppose we have chosen $\langle x_i : i < n \rangle$.
            Since $\cU$ is idempotent, we have
            \[
                (\cU n)[
                    n \in P
                    \land (\cU_k) n + k \in  P
                    \land \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)
                    \land (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
            \]
            Chose $x_n > x_{n-1}$ that satisfies this.
    \end{itemize}
    Set $H \coloneqq \{x_i : i < \omega\}$.
\end{proof}


Next time we'll see another proof of this theorem.