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\begin{proof}
  $\bN$ is Hausdorff:
  Let $\mathcal{U} \neq \mathcal{V}$ be two
  \glspl{ultra}. Then there exists $A \in
  \mathcal{U}$ such that $A \notin \mathcal{V}$.
  Then $\mathcal{U} \in C_A$ (and $C_A$ is open),
  and $A^c \in \mathcal{V}$, hence $\mathcal{V}
  \in C_{A^c}$ (and $C_{A^c}$ is open). Note $C_A
  \cap C_{A^c} = \emptyset$. So indeed $\bN$ is
  Hausdorff.

  $\bN$ is compact: want to show that every open
  cover has a finite subcover. This is equivalent
  to showing that if a collection of closed sets has
  the property that no finite subset covers $\bN$,
  then the whole collection doesn't cover $\bN$.
  This is equivalent to showing that if you have a
  collection of closed sets such that they have
  the finite intersection property (for any $J
  \subset I$ finite, $\bigcap_{i \in J} F_i \neq
  \emptyset$), then their intersection is
  non-empty.

  Further, in the first sentence we can without
  loss of generality that the open sets are basis
  sets (i.e. of the form $C_A$), and carrying this
  forward tells us that we may assume that the
  closed sets in the last sentence are of the form
  $C_A^c$, or equivalently, of the form $C_A$.

  We are given some closed, non-empty sets in
  $\bN$. Without loss of generality,
  they are all $C_A$ for some $A \neq
  \emptyset$. Suppose $(C_{A_i})_{i \in I}$ with
  the finite intersection property. First note
  $C_{A_{i_1}} \cap C_{A_{i_2}} \cap \cdots \cap
  C_{A_{i_k}} = C_{A_{i_1} \cap \cdots \cap
  A_{i_k}} \neq \emptyset$, hence $A_{i_1} \cap
  \cdots \cap A_{i_k}$.

  So let $\mathcal{F} = \{A : A \supset A_{i_1}
  \cap A_{i_2} \cap \cdots \cap A_{i_k} \text{ for
  some $(A_{i_j})_{j = 1}^n$}\}$. $\mathcal{F}$ is
  a \gls{filt} because:
  \begin{enumerate}[(1)]
    \item $\emptyset \notin \mathcal{F}$
    \item $B \supset A \implies B \in \mathcal{F}$
    \item If $A \supset \bigcap A_{i_j}$, $B
      \supset \bigcap A_{k_j}$, then $A \cap B
      \supset \bigcap A_{i_j} \cap \bigcap
      A_{k_j}$ hence $A \cap B \in \mathcal{F}$.
  \end{enumerate}
  Let $\mathcal{U}$ be an \gls{ultra} extending
  $\mathcal{F}$. Note: $\forall A_i \in
  \mathcal{U}$ if and only if $\mathcal{U} \in
  C_{A_i} \forall i$. Hence $\mathcal{U} \in
  \bigcap A_{A_i} \implies \bigcap C_{A_i} \neq
  \emptyset$. Thus $\bN$ is compact.
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\bN$ can be viewed as a subset
      $\mathcal{P}(\Nbb) \to \{0, 1\}$ or a subset
      of $\{0, 1\}^{\mathcal{P}(\Nbb)}$. The
      topology on $\bN$ comes from restricting the
      product topology on $\{0,
      1\}^{\mathcal{P}(\Nbb)}$ and also $\bN$ is a
      closed subset of $\{0,
      1\}^{\mathcal{P}(\Nbb)}$, hence it is
      compact by Tychonoff's theorem.
    \item $\bN$ is the biggest compact Hausdorff
      space in which $\Nbb$ is dense. In other
      words, if $X$ is compact and Hausdorff and
      $f : \Nbb \to X$, there exists a unique
      $\tilde{f}$ continuous that extends $f$,
      i.e. $\tilde{f} : \bN \to X$ makes the
      diagram
      \begin{picmath}
        \begin{tikzcd}
          \Nbb \ar[r, "f"] \ar[d]
          & X \\
          \bN \ar[ur, dashed, "\tilde{f}"]
        \end{tikzcd}
      \end{picmath}
      commute.
    \item $\bN$ is called the Stone-C\v{e}ch
      Compactification of $\Nbb$.
  \end{enumerate}
\end{remark*}

\begin{notation*}
  Let $p$ be a statement and $\mathcal{U}$ an
  \gls{ultra}. We write
  \[
    \forall_{\mathcal{U}} x, p(x)
  \]
  to mean $\{x : p(x)\} \in \mathcal{U}$ (``$p(x)$
  holds for ($\mathcal{U}$-)most $x$'').
\end{notation*}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If $\mathcal{U}$ is \gls{princ}, then
      $\mathcal{U} = \tilde{n}$ so
      $\forall_{\mathcal{U}} x, p(x) \iff p(n)$.
    \item If $\mathcal{U}$ is not \gls{princ} then
      let's consider $\forall_{\mathcal{U}} x, (4 <
      x)$. This says $\{x : x > 4\} \in
      \mathcal{U}$. If not true, then $\{x : x >
      4\}^c \in \mathcal{U}$, i.e. $\{1, 2, 3, 4\}
      \in \mathcal{U}$. Then $\{1\} \cup \{2\}
      \cup \{3\} \cup \{4\} \in \mathcal{U}$ hence
      for some $1 \le i \le 4$,
      $\{i\} \in \mathcal{U}$, so $\mathcal{U}$ is
      \gls{princ}, contradiction.
  \end{enumerate}
\end{example*}

\begin{fcprop}[]
  \label{prop:2.15}
  % Proposition 2.15
  Assuming:
  - $\mathcal{U}$ an \gls{ultra}
  - $p, q$ statements
  Then:
  \begin{enumerate}[(i)]
    \item $\forall_{\mathcal{U}} x, (p(x) \text{
      and }
      q(x))$ if and only if $\forall_{\mathcal{U}}
      x, p(x)$ and $\forall_{\mathcal{U}} x, q(x)$
    \item $\forall_{\mathcal{U}} x, (p(x) \text{ or
      } q(x))$ if and only if
      $\forall_{\mathcal{U}} x, p(x)$ or
      $\forall_{\mathcal{U}} x, q(x)$
    \item $\forall_{\mathcal{U}} x, p(x)$ is false
      if and only if $\forall_{\mathcal{U}} x, \neg
      p(x)$
  \end{enumerate}
\end{fcprop}

\begin{proof}
  Let $A = \{x : p(x)\}$, $B = \{x : q(x)\}$.
  \begin{enumerate}[(i)]
    \item $A \cap B \in \mathcal{U}$ if and only
      if $A \in \mathcal{U}$ and $B \in \mathcal{U}$
    \item $A \cup B \in \mathcal{U}$ if and only
      if $A \in \mathcal{U}$ or $B \in \mathcal{U}$
    \item $A \notin \mathcal{U}$ if and only if
      $A^c \in \mathcal{U}$ \qedhere
  \end{enumerate}
\end{proof}

\begin{warning*}
  $\forall_{\mathcal{U}} x, \forall_{\mathcal{V}}
  y, p(x, y)$ is not necessarily the same as
  $\forall_{\mathcal{V}} y, \forall_{\mathcal{U}}
  x,
  p(x, y)$ (there is even a counterexample in the
  case $\mathcal{U} = \mathcal{V}$!).
\end{warning*}

\begin{example*}
  Let $\mathcal{U}$ be a non-principal
  \gls{ultra}. Let $p(x, y) = x < y$. Then:
  \begin{itemize}
    \item $\forall_{\mathcal{U}} x
      \forall_{\mathcal{U}} y, x < y$ is true
      (since $\forall_{\mathcal{U}} y, x < y$ is
      always true)
    \item $\forall_{\mathcal{U}} y
      \forall_{\mathcal{U}} x, x < y$ is false
      (since $\forall_{\mathcal{U}} x, x < y$ is
      always false)
  \end{itemize}
\end{example*}

\begin{moral*}
  Don't swap quantifiers!!
\end{moral*}

\textbf{Cool fact:} we can ``add'' \glspl{ultra}.

\begin{definition*}[Addition of ultrafilters]
  Let $\mathcal{U}, \mathcal{V}$ be \glspl{ultra}.
  Then we define
  \[
    \mathcal{U} + \mathcal{V}
    = \{A \subseteq \Nbb : \forall_{\mathcal{U}}
    x, \forall_{\mathcal{V}} y, x + y \in A\}
  .\]
\end{definition*}

\begin{example*}
  $\tilde{m} + \tilde{n} + \widetilde{m + n}$.
\end{example*}

Proof that $\mathcal{U} + \mathcal{V}$ is a
\gls{filt}:
\begin{enumerate}[(1)]
  \item $\emptyset \notin \mathcal{U} + \mathcal{V}$
  \item If $A \in \mathcal{U} + \mathcal{V}$ and
    $B \supset A$ then trivially $B \in
    \mathcal{U} + \mathcal{V}$
  \item Intersections: if $A \in \mathcal{U} +
    \mathcal{V}$ and $B \in \mathcal{U} +
    \mathcal{V}$, then:
    \begin{itemize}
      \item $\forall_{\mathcal{U}} x,
        \forall_{\mathcal{V}} y, x + y \in A$
      \item $\forall_{\mathcal{U}} x,
        \forall_{\mathcal{V}} y, x + y \in B$
    \end{itemize}
    hence by \cref{prop:2.15}(i) applied twice, we
    have $\forall_{\mathcal{U}} x,
    \forall_{\mathcal{U}} y, x + y \in A \cap B$.
\end{enumerate}
Hence $\mathcal{U} + \mathcal{V}$ is a \gls{filt}.

Now check that it is an \gls{ultra}:

Suppose $A \notin \mathcal{U} + \mathcal{V}$, i.e.
$\forall_{\mathcal{U}} x, \forall_{\mathcal{V}} y,
x + y \in A$ is false. By \cref{prop:2.15}(iii)
applied twice, this is equivalent to
$\forall_{\mathcal{U}} x, \forall_{\mathcal{V}} y,
x + y \in A^c$. Hence $A^c \in \mathcal{U} +
\mathcal{V}$.

So $\mathcal{U} + \mathcal{V}$ is indeed an
\gls{ultra}.

\begin{remark*}
  The addition of \glspl{ultra} is associative,
  i.e.
  \[
    \mathcal{U} + (\mathcal{V} + \mathcal{W}) =
    (\mathcal{U} + \mathcal{V}) + \mathcal{W}
  .\]
  To show this, we claim $A \in \text{LHS}$ if and
  only if
  \[
    \forall_{\mathcal{U}} x, \forall_{\mathcal{V}}
    y, \forall_{\mathcal{W}} z, x + y + z \in A
  .\]
  By similar reasoning, one can also show $A \in
  \text{RHS}$ if and only if the above holds,
  hence $A \in \text{LHS}$ if and only if $A \in
  \text{RHS}$, which establishes the desired
  equality.

  Let $A \in \mathcal{U} + (\mathcal{V} +
  \mathcal{W})$. So $\forall_{\mathcal{U}}x,
  (\forall_{\mathcal{V} + \mathcal{W}} y, x + y
  \in A)$. Let $B_x = \{y : x + y \in A\}$. TODO
\end{remark*}

Last piece of the puzzle: $+$ is left continuous:
we show that $\mathcal{U} \mapsto \mathcal{U} +
\mathcal{V}$ is continuous (for any fixed
$\mathcal{V}$).