Josia Pietsch
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243 lines
7.8 KiB
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243 lines
7.8 KiB
TeX
\lecture{25}{2024-01-26}{}
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Let $\beta\N$ denote the set of ultrafilters on $\N$.
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\begin{fact}
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\begin{itemize}
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\item This is a topological space,
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where a basis consist of sets
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$V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$.
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\gist{%
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(For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$
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and $\beta\N = V_\N$.)
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}{}
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\item Note also that for $A, B \subseteq \N$,
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$V_{A \cup B} = V_A \cup V_B$,
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$V_{A^c} = \beta\N \setminus V_A$.
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\end{itemize}
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\end{fact}
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\gist{%
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\begin{observe}
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\label{ob:bNclopenbasis}
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Note that the basis is clopen. In particular
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any closed set can be written as an intersection of sets
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of the form $V_A$:
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If $F$ is closed, then $U = \beta\N \setminus F = \bigcup_{i\in I} V_{A_i}$,
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so $F = \bigcap_{i \in I} V_{\N \setminus A_i}$.
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\end{observe}
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}{}
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\begin{fact}
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\label{fact:bNhd}
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$\beta\N$ is Hausdorff.
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\end{fact}
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\begin{proof}
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Let $\cU \neq \cV \in \beta\N$.
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Then there is some $A \in \cU \setminus \cV$,
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so $A^c \in \cV$,
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so $\cU \in V_A$ and $\cV \in V_A^c$.
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\end{proof}
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%\begin{remark}+
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% This even shows that $\beta\N$ is totally separated.
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% In fact, $\beta\N$ is a profinite space,
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% as the next fact shows.
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%\end{remark}
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\begin{fact}
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\label{fact:bNcompact}
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$\beta\N$ is compact.
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\end{fact}
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\begin{proof}
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Let $\{F_i\}_{i \in I}$ be non-empty
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and closed
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such that
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for any $i_1,\ldots., i_k \in I$,
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$k \in \N$,
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$\bigcap_{j=1}^k F_{i_j} \neq \emptyset$.
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We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$.
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\gist{%
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Replacing each $F_i$ by $V_{A_j^i}$ such
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that $F_i = \bigcap_{j \in J_i} V_{A_j^i}$
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(cf.~\yaref{ob:bNclopenbasis})
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we may assume that $F_i$ is of the form $V_{A_i}$.
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We get $\{F_i = V_{A_i} : i \in I\}$
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with the finite intersection property.
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}{Wlog.~$F_i = V_{A_i}$.}
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Hence
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$\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$
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has the finite intersection property.
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Then $\cF = \{A \subseteq \N : A \supseteq A_{i_1} \cap \ldots \cap A_{i_k}, k \in \N, i_1, \ldots, i_k \in I\}$
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is a filter.
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Let $\cU$ be an ultrafilter extending $\cF$.
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Then $\cU \in \bigcap_{i \in I} V_{A_i} = \bigcap_{i \in I} F_i$.
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\end{proof}
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\begin{fact}
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Consider $\N$ as a subspace of $\beta\N$
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via $\N \hookrightarrow \beta\N, n \mapsto \hat{n} \coloneqq \{A \subseteq \N : n \in A\}$.
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Then
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\begin{itemize}
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\item $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
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\item $\N \subseteq \beta\N$ is dense.
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\end{itemize}
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\end{fact}
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\begin{theorem}
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\label{thm:uflimit}
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For every compact Hausdorff space $X$,
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a sequence $(x_n)$ in $X$,
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and $\cU \in \beta\N$,
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we have that $\ulim{\cU}_n x_n = x$
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exists and is unique,
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i.e.~for all $x \in G \overset{\text{open}}{\subseteq} X$
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we have $\{n \in \N : x_n \in G\} \in \cU$.
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\end{theorem}
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\begin{proof}
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Towards a contradiction
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assume that there is no such $x$.
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For every $x$ take $x \in G_x \overset{\text{open}}{\subseteq} X$
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such that
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$\{ n \in \N : x_n \in G_x\} \not\in \cU$.
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So $\{G_x\}_{x \in X}$ is an open cover of $X$.
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Since $X$ is compact,
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there exists a finite subcover
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$G_{x_1}, \ldots, G_{x_m}$.
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But then
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\begin{IEEEeqnarray*}{rCl}
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\N &=& \{ n \in \N : x_n \in \bigcup_{i=1}^m G_{x_i}\}\\
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&=& \underbrace{\bigcup_{i=1}^m \overbrace{\{n \in \N : x_n \in G_{x_i}\}}^{\not\in \cU}}_{\not\in \cU},
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\end{IEEEeqnarray*}
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since $B_1 \cup \ldots \cup B_m \in \cU \iff \exists i < m.~B_i \in \cU$.
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It is clear that $\ulim{\cU}_n x_n$
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is unique, since $X$ is Hausdorff.
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\end{proof}
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\begin{theorem}
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Let $X$ be a compact Hausdorff space.
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For any $f\colon \N \to X$
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there is a unique continuous extension $\tilde{f}\colon \beta\N \to X$.
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\end{theorem}
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\begin{proof}
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Let
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\begin{IEEEeqnarray*}{rCl}
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\tilde{f}\colon \beta\N &\longrightarrow & X \\
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\cU &\longmapsto & \ulim{\cU}_n f(n).
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\end{IEEEeqnarray*}
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\todo{Exercise: Check that $\tilde{f}$ is continuous.}
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$\tilde{f}$ is uniquely determined,
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since $\N \subseteq \beta\N$ is dense.
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% TODO general fact: continuous functions agreeing on a dense set
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% agree everywhere (fact section)
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\end{proof}
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\begin{trivial}+
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$\beta$ is a functor from the category of topological
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spaces to the category of compact Hausdorff spaces.
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It is left adjoint to the inclusion functor.
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\end{trivial}
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% RECAP
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\gist{%
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$\beta\N$ is equipped with $+$ which extends $+\colon \N \times \N \to \N$,
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\[
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\cU + \cV = \{A \subseteq \N : (\cU m)\left( (\cU n) \{m+n \in A\} \right)\}.
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\]
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This is associative, but not commutative.
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}{}
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% END RECAP
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\begin{fact}
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$+\colon \beta\N\times \beta\N \to \beta\N$
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is left continuous,
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i.e.~for $\cV$ fixed,
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$\cU \mapsto \cU + \cV$ is continuous.
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\end{fact}
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\begin{proof}
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Fix $A$ and consider $V_A$.
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We need to show that the inverse image of $V_A$ is open.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\cU + \cV \in V_A &\iff& A \in \cU + \cV\\
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&\iff& (\cU_m)(\cV_n) \{m+n \in A\}\\
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&\iff& \{m \in \N : (\cV n) m+n \in A\} \in \cU\\
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&\iff& \cU \in V_{\{m \in \N: (\cV n) m+n \in A\}}.
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{corollary}
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$(\beta\N,+)$
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is a %(left)
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\vocab{compact semigroup},
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i.e.~it is compact, Hausdorff, associative and left-continuous.%
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%\footnote{There is no convention on left and right.}
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\end{corollary}
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So we can apply the \yaref{lem:ellisnumakura}
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to obtain
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\begin{corollary}
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There is $\cU \in \beta\N$
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such that $\cU + \cU = \cU$.
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\end{corollary}
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\begin{observe}
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Principal ultrafilters $\neq \hat{0}$ are not idempotent.
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We can restrict to $\beta\N \setminus \N$
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to get an idempotent element that is not principal.
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% TODO THINK ABOUT THIS
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\end{observe}
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\begin{theorem}[Hindman]
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\label{thm:hindman}
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\label{thm:hindmanfurstenberg}
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If $\N$ is partitioned into finitely many
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sets,
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then there is is an infinite subset $H \subseteq \N$
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such that all finite sums of distinct
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elements of $H$
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belong to the same set of the partition.
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\end{theorem}
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\begin{refproof}{thm:hindman}[Galvin,Glazer]
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Let $\cU \in \beta\N \setminus \N$
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be such that $\cU + \cU = \cU$.
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Let $P$ be the piece of the partition
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that is in $\cU$.
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So $(\cU n ) n \in P$.
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Let us define a sequence $x_1,x_2,\ldots$
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\begin{itemize}
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\item $\cU$ is idempotent,
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so $(\cU n)(\cU k) n+k \in P$.
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We get
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\[(\cU n) \left( n \in P \land (\cU_k) n+k \in P \right)\].
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Pick $x_1$ that satisfies this,
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i.e.~$x_1 \in P$ and $(\cU_k) x_1+k \in P$.
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\item $\cU$ is idempotent,
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so
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\[
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(\cU n)[
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n \in P
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\land (\cU_k) n + k \in P
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\land x_1 + n \in P
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\land (\cU_k) x_1 + n + k \in P
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]
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\]
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Take $x_2 > x_1$ that satisfies this.
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\item Suppose we have chosen $\langle x_i : i < n \rangle$.
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Since $\cU$ is idempotent, we have
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\begin{IEEEeqnarray*}{rCl}
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(\cU n)&& n \in P\\
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&\land& (\cU_k) n + k \in P\\
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&\land& \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)\\
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&\land& (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
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\end{IEEEeqnarray*}
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Chose $x_n > x_{n-1}$ that satisfies this.
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\end{itemize}
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Set $H \coloneqq \{x_i : i < \omega\}$.
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\end{refproof}
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Next time we'll see another proof of this theorem.
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