%! TEX root = Ramsey.tex
% vim: tw=50
% 21/11/2024 11AM

\begin{proof}
  Note $\mathcal{U} + \mathcal{V} \in C_A$ (for
  some $A$) if and only if $A \in \mathcal{U} +
  \mathcal{V}$, which happens if and only if
  $\forall_{\mathcal{U}} x, \forall_{\mathcal{V}}
  y, x + y \in A$. This is equivalent to saying
  \[
    \{x : \forall_{\mathcal{V}}, x + y \in A\} \in
    \mathcal{U}
  ,\]
  which is equivalent to $\mathcal{U} \in C_{\{x :
  \forall_{\mathcal{V}} y, x + y \in A\}}$, so the
  pre-image is open.
\end{proof}

Recall: $\bN$ is compact Hausdorff, with $\Nbb$ a
dense subset, $+$ is left continuous, associative,
and $\bN$ is non-empty.

Goal: want $\mathcal{U}$ such that $\mathcal{U} +
\mathcal{U} = \mathcal{U}$, i.e. idempotent.

\begin{proposition}[Idempotent lemma]
  \label{prop:2.16}
  % Proposition 2.16
  There exists $\mathcal{U} \in \bN$ such that
  $\mathcal{U} = \mathcal{U} + \mathcal{U}$.
\end{proposition}

\begin{proof}
  (Warning: we will use Zorn's lemma :O)

  Start with $M \subseteq \bN$ such that $M + M
  \defeq \{x + y : x, y \in M\} \subseteq M$. Seek
  $M \subseteq \bN$, non-empty, compact and minimal with the
  property that $M + M \subseteq M$ (hope to show
  $M$ is a singleton).

  Proof of existence: there exists such a set,
  namely $\bN$ itself. Look at all such $M$ -- this
  set is not empty. By Zorn's lemma, it is enough
  to show that if $(M_i)_{i \in I}$ is a
  collection of such sets that is also a chain,
  then $M = \bigcap_{i \in I} M_i$ has this
  property also ($M + M \subseteq M$, $M$ is
  compact).

  Compact: We are in a compact Hausdorff space, so
  a subspace is compact if and only if it is
  closed. Since the $M_i$ are closed we have that
  $M$ is closed, hence $M$ is compact.

  Why $M + M \subseteq M$? Let $x, y \in M$, $x, y
  \in M_i$ for all $i$. Then $x + y \in M_i + M_i
  \subseteq M_i$ for all $i$, hence $x + y \in
  M_i$ for all $i$, hence $x + y \in M$. So $M + M
  \subseteq M$.

  Also $M$ is non-empty: $(M_i)_{i \in M}$ have
  the finite intersection property (as they are a
  chain). Since they are closed, we get that the
  intersection is non-empty.

  Therefore, by Zorn's lemma, there exists a
  minimal $M$, which is non-empty, compact such
  that $M + M \subseteq M$.

  Pick $x \in M$. Look at $M + x$ and we want to
  show that this is $M$.

  Claim: $M + x = M$.

  Proof: $M + x \subseteq M + M \subseteq M$.
  Check:
  \begin{itemize}
    \item non-empty
    \item compact, as continuous image ($\bullet+
      x$) of a compact set
    \item $(M + x) + (M + x) = (M + x + M) + x
      \subseteq M + x$
  \end{itemize}
  So by minimality $M + x = M$.

  In particular, there exists $y \in M$ such that
  $y + x = x$. Consider $T = \{y \in M : y + x =
  x\}$.

  Claim: $T = M$. Since $T \subseteq M$, it's
  enough to show (by minimality) that $T$ is
  compact, non-empty and $T + T \subseteq T$.
  Indeed:
  \begin{itemize}
    \item non-empty: $y \in T$
    \item $T$ compact as $T$ is the pre-image of
      a singleton (which is compact hence closed),
      thus closed, thus compact.
    \item for $T + T \subseteq T$: $y, z \in T$,
      $y + x = x$, $z + x = x$ so $y + z + x = y +
      x = x$ hence $y + z \in T$. So $T + T
      \subseteq T$.
  \end{itemize}
  By minimality, $T = M$.

  So $\{y : y + x = x\} = M$, hence $x + x = x$.
\end{proof}

\begin{remark*}
  $M = \{x\}$.
\end{remark*}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The finite subgroup question: can we
      find a non-trivial subgroup of $\bN$? For
      example, $\mathcal{U}$, $\mathcal{U} +
      \mathcal{U} \neq \mathcal{U}$, $\mathcal{U}
      + \mathcal{U} + \mathcal{U} = \mathcal{U}$?
      Solved by Zeleyum (1996) -- No!
    \item Can an \gls{ultra} ``absorb'' another
      \gls{ultra}? That is, $\mathcal{U} \neq
      \mathcal{V}$ such that all $\mathcal{U} +
      \mathcal{U}$, $\mathcal{U} + \mathcal{V}$,
      $\mathcal{V} + \mathcal{U}$, $\mathcal{V} +
      \mathcal{V}$ are equal to
      $\mathcal{V}$? Totally open (until it was
      show that the answer is yes)!
  \end{enumerate}
\end{remark*}

\begin{proof}[Proof of \nameref{thm:2.11}]
  (If $\Nbb$ is finitely coloured, there exists
  $(x_n)_{n = 1}^\infty$ such that $\FS(x_1, x_2,
  \ldots)$ is monochromatic).

  Let $A_1, A_2, \ldots, A_k$ be the colour
  classes ($A_1 \sqcup A_2 \sqcup \cdots \sqcup
  A_k = \Nbb$) and $\mathcal{U}$ an idempotent
  \gls{ultra}.

  Claim: $A_i \in \mathcal{U}$ for some $i$ (this
  is because \glspl{ultra} are \emph{prime}: whenever we
  have a finite union lying in the \gls{ultra} we
  have at least one of the components lying in the
  \gls{ultra}, else have $A_i^c \in \mathcal{U}$
  for each $i$, hence $A_1^c \cap \cdots \cap
  A_k^c \in \mathcal{U}$, but this is $(A_1 \cup
  \cdots \cup A_k)^c$, contradicting the fact that
  $A_1 \cup \cdots \cup A_k \in \mathcal{U}$).

  Let $A = A_i \in \mathcal{U}$. Therefore we have
  $\forall_{\mathcal{U}} y, y \in A$. Then:
  \begin{enumerate}[(1)]
    \item $\forall_{\mathcal{U}} x,
      \forall_{\mathcal{U}}  y, y \in A$
    \item $\forall_{\mathcal{U}} x,
      \forall_{\mathcal{U}} y, x \in A$
    \item $A \in \mathcal{U} + \mathcal{U} =
      \mathcal{U}$ gives that
      $\forall_{\mathcal{U}} x,
      \forall_{\mathcal{U}} y, x + y \in A$.
  \end{enumerate}
  Then (1) and (2) and (3) give:
  \[
    \forall_{\mathcal{U}} x, \forall_{\mathcal{U}}
    y, \FS(x, y) \subseteq A
  .\]
  Now fix $x_1 \in A$ such that
  \[
    \forall_{\mathcal{U}} y, \FS(x_1 x) \subseteq
    A
  .\]
  Assume we have found $x_1, \ldots, x_n$ such
  that
  \begin{align*}
    \forall_{\mathcal{U}} y, \FS(x_1, \ldots, x_n,
    y)
    &\subseteq A \\
    \ub{\{y : \FS(x_1, \ldots, x_n, y) \subseteq
    A\}}_{\eqdef B}
    &\in \mathcal{U} = \mathcal{U} + \mathcal{U}
  \end{align*}
  Then have $\forall_{\mathcal{U}} x,
  \forall_{\mathcal{U}} y, x + y \in B$.
  \begin{enumerate}[(1)]
    \item $\forall_{\mathcal{U}} x,
      \forall_{\mathcal{U}}  y, \FS(x_1, \ldots,
      x_n, y) \subseteq A$
    \item $\forall_{\mathcal{U}} x,
      \forall_{\mathcal{U}} y, \FS(x_1, \ldots,
      x_n, x) \subseteq A$
    \item  $\forall_{\mathcal{U}} x,
      \forall_{\mathcal{U}} y, \FS(x_1, \ldots,
      x_n, y) \subseteq A$.
  \end{enumerate}
  Then (1), (2) and (3) give:
  \[
    \forall_{\mathcal{U}} x, \forall_{\mathcal{U}}
    y, \FS(x_1, \ldots, x_n, y) \subseteq A
  .\]
  Thus fix $x_{n + 1} \gg x_n$. Have
  $\forall_{\mathcal{U}} y, \FS(x_1, \ldots, x_n,
  y) \subseteq A$. Then done by induction.
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Very few other infinite \gls{pr}
      equations are known. In particular, there
      does not exist a ``\hyperref[rado]{Rado}-type'' theorem of
      iff.
    \item The consistency theorem no longer holds.
      \[ \FS_{12}(x_1, x_2, \ldots) = .cb\sum_{i
      \in I} x_i + 2\sum_{i \in J} x_i, \max I <
      \min J \]
      is \gls{pr} (special case of Milliken-Taylor
      theorem).
      It was shown in 1995 that
      $\mathrm{FS}_{12}(x_1, x_2, \ldots)$ and
      $\FS(x_1, x_2, \ldots)$ are incompatible.
    \item Trivially from Hindman, $\{x_n\}_{n =
      1}^\infty \cup \{x_i + x_i\}$ is \gls{pr}.
      Any proof of this without Hindman? Not
      known!
  \end{enumerate}
\end{remark*}
