Josia Pietsch de89e2dc1d
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Let $\beta\N$ denote the set of ultrafilters on $\N$.
\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 $A, B \subseteq \N$,
$V_{A \cup B} = V_A \cup V_B$,
$V_{A^c} = \beta\N \setminus V_A$.
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 $F = \bigcap_{i \in I} V_{\N \setminus A_i}$.
$\beta\N$ is Hausdorff.
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 $\cV \in V_A^c$.
% This even shows that $\beta\N$ is totally separated.
% In fact, $\beta\N$ is a profinite space,
% as the next fact shows.
$\beta\N$ is compact.
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 $F_i = \bigcap_{j \in J_i} V_{A_j^i}$
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.
}{Wlog.~$F_i = V_{A_i}$.}
$\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\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 $\cU \in \bigcap_{i \in I} V_{A_i} = \bigcap_{i \in I} F_i$.
Consider $\N$ as a subspace of $\beta\N$
via $\N \hookrightarrow \beta\N, n \mapsto \hat{n} \coloneqq \{A \subseteq \N : n \in A\}$.
\item $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
\item $\N \subseteq \beta\N$ is dense.
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 $\{n \in \N : x_n \in G\} \in \cU$.
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
\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},
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.
Let $X$ be a compact Hausdorff space.
For any $f\colon \N \to X$
there is a unique continuous extension $\tilde{f}\colon \beta\N \to X$.
\tilde{f}\colon \beta\N &\longrightarrow & X \\
\cU &\longmapsto & \ulim{\cU}_n f(n).
\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)
$\beta$ is a functor from the category of topological
spaces to the category of compact Hausdorff spaces.
It is left adjoint to the inclusion functor.
$\beta\N$ 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.
$+\colon \beta\N\times \beta\N \to \beta\N$
is left continuous,
i.e.~for $\cV$ fixed,
$\cU \mapsto \cU + \cV$ is continuous.
Fix $A$ and consider $V_A$.
We need to show that the inverse image of $V_A$ is open.
We have
\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\}}.
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.}
So we can apply the \yaref{lem:ellisnumakura}
to obtain
There is $\cU \in \beta\N$
such that $\cU + \cU = \cU$.
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.
If $\N$ is partitioned into finitely many
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.
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 $x_1,x_2,\ldots$
\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,
(\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.
Set $H \coloneqq \{x_i : i < \omega\}$.
Next time we'll see another proof of this theorem.