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\setcounter{subsection}{1}%
\subsection{Ultrafilters}

\textbf{Aim:}

\glssymboldefn{FS}%
\begin{fcthm}[Hindman's Theorem]
  \label{thm:2.11}
  % Theorem 2.11
  Assuming:
  - $\Nbb$ is finitely coloured
  Then:
  there exists infinitely many $(x_n)_{n \ge 1}$
  such that
  \[
    \FSinternal(X_1)
    \defeq \left\{
      \sum_{i \in I} x_i : \emptyset \neq I
      \subseteq \Nbb, I \text{ finite}
    \right\}
  \]
  is monochromatic.
\end{fcthm}

This is the first infinite \gls{pr} system in the
course.

\begin{definition*}[Filter]
  \glsnoundefn{filt}{filter}{filters}%
  A filter is a non-empty collection $\mathcal{F}$ of
  subsets of $\Nbb$ satisfying:
  \begin{enumerate}[(a)]
    \item $\emptyset \notin \mathcal{F}$.
    \item If $A \in \mathcal{F}$, $A \subset B$,
      then $B \in \mathcal{F}$ (`upset').
    \item If $A \in \mathcal{F}$, $B \in
      \mathcal{F}$ then $A \cap B \in \mathcal{F}$
      (closed under finite intersections).
  \end{enumerate}
\end{definition*}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\mathcal{F} = \{A \subseteq \Nbb : 1
      \in A\}$ is a \gls{filt}.
    \item $\mathcal{F} = \{A \subseteq \Nbb : 1, 2
      \in A\}$ is a \gls{filt}.
    \item
      \glsnoundefn{cofilt}{cofinite filter}{cofinite
      filter}%
      $\mathcal{F} = \{A \subseteq \Nbb :
      \text{$A^c$ is finite}\}$ is a \gls{filt},
      called the \emph{cofinite filter}.
    \item $\mathcal{F} = \{A \subseteq \Nbb :
      \text{$A$ is infinite}\}$. This is not a
      \gls{filt}, since the intersection
      $\{\text{odd numbers}\} \cap \{\text{even
      numbers}\}$ is $\emptyset$, which is not in
      $\mathcal{F}$.
    \item $\mathcal{F} = \{A \subseteq \Nbb :
      \{\text{even numbers}\} \setminus A \text{
      is finite}\}$ is a \gls{filt}.
  \end{enumerate}
\end{example*}

\begin{definition*}[Ultrafilter]
  \glsnoundefn{ultra}{ultrafilter}{ultrafilters}%
  An \emph{ultrafilter} is a maximal \gls{filt}.
\end{definition*}

\begin{example*}
  \phantom{}
  \begin{enumerate}
    \item \glsnoundefn{princ}{principal}{principal}%
      $\mathcal{U} = \{A \subseteq \Nbb : x \in
      A\}$, $B \notin \mathcal{U}$, then $B^c$
      will contain $x$, so $B^c \in \mathcal{U}$,
      but $B^c \cap B = \emptyset$ so we cannot
      extend $\mathcal{U}$ by adding $B$ to it. So
      $\mathcal{U}$ is maximal. This is called the
      \emph{principal filter} at $x$.
    \item In the examples above: (1) is an
      \gls{ultra}, (2) is not as (1) extends it,
      (3) is not as (5) extends it, and (5) is not
      as $\mathcal{F}' = \{A \subseteq \Nbb :
      \{\text{multiples of $4$}\} \setminus A
      \text{ is finite}\}$ extends it.
  \end{enumerate}
\end{example*}

\begin{proposition}[]
  \label{prop:2.12}
  % Proposition 2.12
  \begin{iffc}
    \lhs
    $\mathcal{U}$ is an \gls{ultra}
    \rhs
    for all $A \subseteq \Nbb$, either $A \in
    \mathcal{U}$ or $A^c$ is in $\mathcal{U}$.
  \end{iffc}
\end{proposition}

\begin{iffproof}
  \leftimpl
  If I try to extend $\mathcal{U}$ by adding in
  some $A \notin \mathcal{U}$, then since $A^c \in
  \mathcal{U}$, we would also have to have $A \cap
  A^c \in \mathcal{U}$, which violated one of the
  properties of being a \gls{filt}.
  \rightimpl
  Suppose $\mathcal{U}$ is an \gls{ultra} and there
  exists $A$ such that $A, A^c$ are not in
  $\mathcal{U}$.

  By maximality, if $A$ is not in $\mathcal{U}$
  then there exists $B \in \mathcal{U}$ such that
  $A \cap B = \emptyset$. Indeed, suppose not.
  Then
  \[
    \mathcal{F}' = \{S : S \supseteq A \cap B
    \text{ for some $B \in \mathcal{U}$}\}
  \]
  extends it (the only way this can fail to be a
  \gls{filt} is if $\emptyset \in \mathcal{F}'$,
  which would require a $B$ such that $A \cap B =
  \emptyset$).

  Then $B \subseteq A^c$, so $A^c \in
  \mathcal{U}$, contradicting the initial
  assumption.
\end{iffproof}

\begin{remark*}
  If $\mathcal{U}$ is an ultrafilter and $A \in
  \mathcal{U}$, $A = B \cup C$. Then either $B$ or
  $C$ is in $U$. Indeed, suppost not. Then $B^c,
  C^c \in \mathcal{U}$, hence $B^c \cap C^c \in
  \mathcal{U}$, i.e. $B^c \cap C^c = (B \cup C)^c$
  is in $\mathcal{U}$. Hence $A \cap A^c =
  \emptyset \in \mathcal{U}$, a contradiction.
\end{remark*}

\begin{fcprop}[]
  \label{prop:2.13}
  % Proposition 2.13
  Assuming:
  - $\mathcal{F}$ a \gls{filt}
  Then:
  there exists an \gls{ultra} $\mathcal{U}$ extending
  $\mathcal{F}'$.
\end{fcprop}

\begin{proof}
  By Zorn's lemma, it is enough to show that any
  chain of \glspl{filt} extending $\mathcal{F}$
  has an upper bound.

  Let $(F_i)_{i \in I}$ be a chain of
  \glspl{filt} containing $\mathcal{F}$, i.e. for all $i \neq j$ either
  $F_i \subseteq F_j$ or $F_j \subseteq F_i$. Let
  $F = \bigcup_{i \in I} F_i$. Need to show $F$ is
  a \gls{filt}:
  \begin{enumerate}[(1)]
    \item $\emptyset \notin F$ since $\emptyset
      \notin F_i$ for each $i$.
    \item If $A \in F$ and $A \subseteq B$ then $A
      \in F_i$ for some $i$, and then we have $B
      \in F_i$ for this same $i$ (as $F_i$ is a
      \gls{filt}), so $B \in F$.
    \item If $A, B \in F$ then say $A \in F_i$, $B
      \in F_j$. Since $(F_i)$ is a chain, we can
      suppose without loss of generality that $F_i
      \subseteq F_j$. Then $A, B \in F_j$, so $A
      \cap B \in F_j \subseteq F$, so $A \cap B
      \in F$.
  \end{enumerate}
  and also clearly $F$ extends $\mathcal{F}$. So
  $F$ is an upper bound.
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Any \gls{ultra} that extends the
      \gls{cofilt} cannot be \gls{princ}. Suppose
      $\mathcal{F}$ is the \gls{cofilt} and
      $\mathcal{U}$ is a \gls{filt} extending it.
      If $\mathcal{U} = \mathcal{U}_x$ then
      $\{x\}^c \in \mathcal{F} \subseteq
      \mathcal{U}$ but we also have $\{x\} \in
      \mathcal{U}$, contradiction.
    \item If $\mathcal{U}$ is non-\gls{princ},
      then it must extend the \gls{cofilt}. If
      $\mathcal{U}$ were to contain a finite set
      $A = \{x_1\} \cup \cdots \cup \{x_i\}$, so
      there exists $i$ such that $\{x_i\} \in
      \mathcal{U}$ (contradiction). If $A$ is a
      set in the \gls{cofilt}, then $A^c$ is
      finite, so $A^c \notin \mathcal{U}$, hence
      $A \in \mathcal{U}$.
    \item The Axiom of Choice is absolutely needed
      for the existence of non-\gls{princ}
      \glspl{ultra}.
  \end{enumerate}
\end{remark*}

\begin{definition*}[$beta N$]
  \glssymboldefn{bN}%
  The set of \glspl{ultra} on $\Nbb$ is called
  $\beta\Nbb$. We define a topology on $\beta\Nbb$
  as the one induced by the following base of open
  sets
  \[
    C_A \defeq \{\mathcal{U} : A \in \mathcal{U}\}
  .\]
\end{definition*}

We can see that $\bigcup_A C_A = \beta\Nbb$ and $C_A
\cap C_B = C_{A \cap B}$ because $A \cap B \in
\mathcal{U}$ if and only if $A, B \in
\mathcal{U}$.

Open sets are $\bigcup_{i \in I} C_{A_i}$.

Closed sets are $\bigcap_{i \in I}
C_{A_i}$ (using the fact that $\left(\bigcup_{i \in I}
C_{A_i}\right)^c = \bigcap_{i \in I} C_{A_i}^c =
\bigcap_{i \in I} C_{A_i^c}$).

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\bN \setminus C_A = C_{A^c}$.
    \item We can view $\Nbb$ embedded in $\bN$ by
      identifying $n \in \Nbb$ with the
      \gls{princ} \gls{ultra} at $n$, i.e.
      $\tilde{n} = \{A : n \in A\}$, $\{\tilde{1},
      \tilde{2}, \ldots\} \leftrightarrow \Nbb$.

      Each point in $\Nbb$ is isolated because
      $C_{\{n\}} = \tilde{n}$. Under this, $\Nbb$
      is dense in $\bN$: for $C_A$ an open set and $n
      \in A$ we have $\tilde{n} \in C_A$.
  \end{enumerate}
\end{remark*}

\begin{proposition}
  \label{prop:2.14}
  % Proposition 2.14
  $\bN$ is a compact Hausdorff space.
\end{proposition}