%! TEX root = Ramsey.tex
% vim: tw=50
% 10/10/2024 11AM

% 1. Monochromatic sets
% 2. Partition regular systems
% 12 lectures for sections 1 and 2
% 3. Euclidean ramsey --> 4 lectures

% 3 example sheets

\newpage
\section{Ramsey's Theorem}

\begin{notation*}
  $\Nbb = \{1, 2, \ldots\}$, $[n] = \{1, 2,
  \ldots, n\}$ for a set $X$, $r \ge 1$, $X^{(r)}
  = \{A \subseteq X, |A| = r\}$.
\end{notation*}

Given a 2-colouring of $\Nbb^{(2)}$, are we
guaranteed to have an infinite monochromatic set
(i.e. $M \subseteq \Nbb$, $M$ infinite such that
the colouring is constant on $M^{(2)}$)?

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\{i, j\}$ red if $i + j$ even, odd
      otherwise. Then $M = \{n : \text{$n$
      even}\}$ works.
    \item $\{i, j\}$ red if $\max\{n : 2^n \mid i +
      j\}$ is even, blue otherwise. Then $M =
      \{4^0, 4^1, \ldots\}$ works.
    \item $\{i, j\}$ red if $i + j$ has an even
      number of distinct prime divisors, and blue
      otherwise. No explicit $M$ is known!
  \end{enumerate}
\end{example*}

\begin{fcthm}[Ramsey's Theorem for pairs]
  \label{ramsey_pairs}%
  % Theorem 1.1
  Assuming:
  - $\Nbb^{(2)}$ are 2-coloured (i.e. $c :
    \Nbb^{(2)} \to \{1, 2\}$).
  Then:
  there exists $M$ infinite monochromatic.
\end{fcthm}

\begin{proof}
  Pick $a_1 \in \Nbb$. Then there exists an
  infinite set $A_1$ such that $c(a_1 i) = c_1$
  for all $i \in A_1$. Pick $a_2 \in A_1$ and find
  $A_2$ (infinite) such that $c(a_2 i) = c_2$ for
  all $i \in A_2$. Keep on doing this. We end up
  with $a_1 < a_2 < a_3 < \cdots < a_k < \cdots$
  and $A_1 \supset A_2 \supset \cdots$ such that
  $c(a_i j) = c_i$ for all $j \in A_i$.

  One colour appears infinitely many times
  $c_{i_1} = c_{i_2} = \cdots = c_{i_k} = \cdots =
  e$. Now note $M = \{a_{i_1}, a_{i_2}, a_{i_3},
  \ldots\}$ is a monochromatic set.
\end{proof}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/e8e76d4829924506.png}
\end{center}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The same proof works for $k$ colours.
      This is referred to as a ``$2$-pass''
      proof. Alternatively: if we have colours $1,
      2, \ldots, k$, then we can consider $1$ to
      be red, and everything else to be blue. Then
      using the above result and induction, we get
      an alternative way to prove the theorem for
      greater than $2$ colours.
    \item Infinite monochromatic is \textbf{very}
      different than arbitrarily large
      monochromatic.

      For example: suppose we write $A_1 = \{1,
      2\}$, $A_2 = \{3, 4, 5\}$, $A_3 = \{6, 7, 8,
      9\}$ and so on. Say $\{i, j\}$ is red if
      there exists $k$ such that $i, j \in A_k$,
      and blue otherwise. Then there exist
      arbitrarily large monochromatic red sets,
      but no infinite monochromatic red set.
  \end{enumerate}
\end{remark*}

What about $\Nbb^{(r)}$ with $r = 3$?

\begin{example*}
  $r = 3$, $\{i, j, k\}$, $i < j < k$ red if and only if $i \mid
  j + k$. Then $M = \{2^0, 2^1, 2^2, \ldots\}$ is
  monochromatic.
\end{example*}

\begin{fcthm}[Ramsey's Theorem for $r$-sets]
  \label{ramsey_rsets}%
  % Theorem 1.2
  Assuming:
  - $\Nbb^{(r)}$ is finitely coloured.
  Then:
  there exists a monochromatic infinite set.
\end{fcthm}

\begin{proof}
  $r = 1$ pigeonhole, $r = 2$ is
  \cref{ramsey_pairs}. Prove this by induction.

  Assume it is true for $r - 1$. Given $c :
  \Nbb^{(r)} \to [k]$, we must find $M$ (infinite
  and monochromatic). Pick $a_1 \in \Nbb$. Look at
  the $r - 1$ sets of $\Nbb \setminus \{a_1\}$.
  Define $c' : (\Nbb \setminus \{a_1\})^{(r - 1)}
  \to [k]$ via $c' (F) = c(F \cup \{a_1\})$.

  By induction there exists $A_1 \subseteq \Nbb
  \setminus \{a_1\}$ such that $c'$ is constant on
  it, say constantly equal to $c_1$.

  Now pick $a_2 \in A_1$ and induce $c' : (A_1
  \setminus \{a_2\})^{(r - 1)} \to [k]$ defined by
  $c'(F) = c(F \cup \{a_2\})$. By induction there
  exists $A_2 \subset A_1 \setminus \{a_2\}$ such
  that $c'$ is constant on it, say equal to $c_2$.

  Continuing this, we end up with $a_1, a_2,
  \ldots$ and sets $A_1, A_2, \ldots$ such that
  $A_{i + 1} \subset A_i \setminus \{a_{i + 1}\}$
  with $c(F \cup \{a_i\}) = c_i$ for all $F
  \subset A_{i + 1}$, $|F| = r - 1$.

  Some colour must appear infinitely many times:
  say $c_{i_1} = c_{i_2} = c_{i_3} = \cdots = c$.
  Check: $M = \{a_{i_1}, a_{i_2}, \ldots\}$ is
  monochromatic.
\end{proof}

\begin{example*}
  Applications:
  \begin{enumerate}[(1)]
    \item In a totally ordered set, any sequence
      has a monotone subsequence.

      \begin{proof}
        Let the sequence be $x_1, x_2, \ldots$.
        Say $\{i < j\}$ is red if $x_i \le x_j$,
        and blue otherwise. By
        \cref{ramsey_pairs}, we may find $M =
        \{i_1 < i_2 < \cdots\}$ monochromatic. If
        $M$ is red, then the sequence $x_{i_1},
        x_{i_2}, x_{i_3}, \ldots$ is increasing,
        and if $M$ is blue then the sequence is
        strictly decreasing.
      \end{proof}
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/7925b4f77e6445f8.png}
      \end{center}
    \item Using a slightly adjusted argument, we
      can insist that the function given by $(i_j,
      x_{i_j})$ is either concave or convex. We do
      this by:
      for a triple $(i_{j_1}, i_{j_2}, i_{j_k})$
      we colour it convex or concave. Then apply
      \cref{ramsey_rsets}.
  \end{enumerate}
\end{example*}

From \cref{ramsey_rsets} we can deduce:

\begin{fcthm}[Finite Ramsey]
  Assuming:
  - $r \ge 1$, $k \ge 1$, $m \ge 1$
  Then:
  there exists $n \in \Nbb$ such that whenever
  $[n]^{(r)}$ is $k$-coloured, we can find a
  monochromatic set of size $m$.
\end{fcthm}

\begin{proof}
  Suppose not. Then for each $n$ we can find $c_n
  : [n]^{(r)} \to [k]$ with no monochromatic
  $m$-sets. Note that there are only finitely many
  ways to $k$-colour $[r]^{(r)}$. So infinitely
  many $c_n$ will agree on $[r]^{(r)}$. Pick $A_1$
  such that for all $n \in A_1$,
  \[ c_n|_{[r]^{(r)}} = dr : [r]^{(r)} \to [k] .\]
  We can do the same on $[r + 1]^{(r)}$ and
  produce some $A_2 \subset A_1$ such that
  $c_n|_{[r + 1]^{(r)}}$ is constant on $A_2$.

  Continuing this, we get $\cdots \subset A_n
  \subset A_{n - 1} \subset \cdots \subset A_1$.
  They satisfy:
  \begin{enumerate}[(1)]
    \item There is no monochromatic $m$-set for
      any $d_n : [n]^{(r)} \to [k]$ (because $d_n =
      c_i|_{[n]^{(r)}}$).
    \item These $d_n$'s are nested:
      $d_j|_{[i]^{(r)}}
      = d_i$ for $j > i$.
  \end{enumerate}
  Finally: colour $\Nbb^{(r)}$ via $c(F) = d_n(F)$,
  where $n$ is any integer $n \ge \max F$. One can
  see that this is well defined, and gives a
  contradiction to \cref{ramsey_rsets}.
\end{proof}