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\subsection{Infinite Ramsey Theory}

Suppose we have a \rcolring{2} of $\NN^{(2)}$
(edges of \gls{complete_graph} on $\NN$).
Can we always find an infinite monochromatic
subset?

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Give edge $ij$ colour red if $i + j$ is
      even, and colour blue if $i + j$ is odd.
      Then we could take $M = \{\text{evens}\}$.
    \item Give edge $ij$ colour red if
      $\max(\text{$n = 2^k$ dividing $i + j$})$ is
      even, and colour blue otherwise. Could take
      $M = \{\text{powers of $4$}\}$.
    \item \refsteplabel[Example 3]{lec17_example_3}
      Give edge $ij$ colour red if the number
      of primes dividing $i + j$ is even, and
      colour blue otherwise. $M = ~?$.
  \end{enumerate}
\end{example*}

\begin{flashcard}[infinite-ramsey-thm]
\begin{theorem}[Infinite Ramsey]
  \label{infinite_ramsey}
  \cloze{Whenever $\NN^{(2)}$ is \rcoled{2}, there
  exists an infinite monochromatic set.}
\end{theorem}
\fcscrap{
\begin{note*}
  Much more than asking for arbitrarilty large
  finite monohromatic subsets. For example:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/fa2166674687433b.png}
  \end{center}
\end{note*}
}
\begin{proof}
  \cloze{
  Pick $x_1 \in \NN$. We have infinitely many
  \glspl{edge} from $x_1$. So infinitely many are
  the same colour -- say all edges $x_1 y$, for
  each $y \in A_1$ have colour $c_1$. Now choose
  $x_2 \in A_1$. Again, there exists infinitely
  many $A_2 \subset A_1$ such that all
  \glspl{edge} from $x_2$ to $A_2$ have same
  colour, say coloud $c_2$. Continue inductively.
  We obtain distinct points $x_1, x_2, \ldots$ and
  colours $c_1, c_2, \ldots$ such that $x_i x_j$
  ($i < j$) has colour $c_i$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/896a83c9c8d44a4b.png}
  \end{center}
  But must have $i_1, i_2, \ldots$ with $c_{i_1} =
  c_{i_2} = \cdots$, and now see that $\{x_{i_1}, x_{i_2},
  \ldots\}$ is monochromatic.
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Called a `$2$-pass' proof.
    \item In \nameref{lec17_example_3} earlier in
      this subsection, no explicit $M$ is known.
    \item Same proof works for $k$ colours (or
      `turquoise spectacles').
  \end{enumerate}
\end{remark*}

\begin{example*}
  Any sequence $x_1, x_2, \ldots$ of reals has a
  monotone subsequence. Indeed, \rcol{2}
  $\NN^{(2)}$ by giving $ij$ ($i < j$) colour
  \emph{up}
  if $x_i < x_j$, and colour \emph{down} if $x_i \ge
  x_k$. Now apply \nameref{infinite_ramsey}.
\end{example*}
\vspace{-1em}

How about \rcolring{2} $\xr \NN^r$? For example,
$r = 3$. Give $ijk$ ($i < j < k$) colour red if
$i$ divides $j + k$, and blue if not. Could take
$M = \{\text{powers of $2$}\}$.

\begin{flashcard}[infinite-ramsey-r-sets-thm]
\begin{theorem}[Infinite Ramsey for $r$-sets]
  \label{infinite_ramsey_for_r_sets}
  \cloze{Given $r = 1, 2, \ldots$: whenever $\xr\NN^r$
  \rcoled{2}, there exists a monochromatic set.}
\end{theorem}

\begin{proof}
  \cloze{
  Induction on $r$. $r = 1$ is just pigeonhole (or
  could start with $r = 2$ using
  \cref{infinite_ramsey}). Given $r > 1$, and a
  \rcolring{2} $c$ of $\xr \NN^r$: fix $x_1 \in
  \NN$. Then $c$ induces a \rcolring{2} $c'$ of
  $\xr {(\NN \setminus\{x_1\})}^{r - 1}$. By $c'(A) =
  c(A \cup \{x_1\})$ for each $A$. So (by
  induction) there exists an infinite $A_1 \subset
  \NN \setminus \{x_1\}$ and colour $c_1$ such
  that for each $(r - 1)$-set $A \subset A_1$ we
  have $A \cup \{x_1\}$ gets colour $c_1$. Pick
  $x_2 \in A_2$ and continue as before. We obtain
  distinct points $x_1, x_2, \ldots$ and colours
  $c_1, c_2, \ldots$ such that the colour of
  $x_{i_1} \cdots x_{i_r}$ ($i _1 < i_2 < \cdots <
  i_r$) is $c_{i_1}$ for all $i_1 < \cdots < i_r$.
  But must have $i_1 < i_2 < \cdots$ with $c_{i_1}
  = c_{i_2} = \cdots$, and now $\{x_{i_1},
  x_{i_2}, \ldots\}$ is monochromatic.
}
\end{proof}
\end{flashcard}

\begin{example*}
  We say what, given points $(1, x_1), (2, x_2),
  (3, x_3)$ in $\RR^2$, can find an infinite
  subset whose induced graph (piecewise-linear,
  through those points) is increasing or
  decreasing. In fact, can also insist that the
  induced graph is convex or concave. Indeed,
  colour $\xr \NN^3$ by giving $ijk$ colour
  \emph{convex} if $(j, x_j)$ lies below the line
  through $(i, x_i)$ and $(k, x_k)$, and give
  $ijk$ the colour \emph{concave} otherwise. Then
  apply \cref{infinite_ramsey_for_r_sets} ($r =
  3$).
\end{example*}

\subsubsection*{Exact Ramsey Numbers}

Very few of the (non-trivial) $\ramseyd(s, t)$ are
known exactly. We know:
\begin{align*}
  \ramseyd(3, 3)
  &= 6&
  \ramseyd(3, 4)
  &= 9&
  \ramsey(3, 5)
  &= 14&
  \ramseyd(3, 6)
  &= 18 \\
  \ramseyd(3, 7)
  &= 23&
  \ramseyd(3, 8)
  &= 28&
  \ramsey(3, 9)
  &= 36&
  \ramseyd(4, 4)
  &= 18 \\
  \ramseyd(4, 5)
  &= 25
\end{align*}
$\ramsey(5) = \ramseyd(5, 5)$ is unknown! We do
know that $43 \le \ramsey(5) \le 48$.

For more colours, only known (non-trivial) case is
$\ramseyk_3(3, 3, 3) = 17$.

For $r$-sets, only known case is $\ramseyr^3(4, 4)
= 13$.

This is because we are asking ``exactly how much
disorder do we need to guarantee a certain amount
of order?''

``Put it on a computer?''

To show $\ramsey(5) \le 4$, to show $\ramsey(5)
\le 48$, we need to look at $2^{{47 \choose 2}} >
2^{1000} > 10^{300}$ colourings -- no chance. Even
if we use a clever symmetry argument to divide the
work by $48$, and then use another clever argument
to square root the amount of work required, we
would still need to look at over $10^{100}$
colourings, so still no chance.
