%! TEX root = Ramsey.tex
% vim: tw=50
% 29/10/2024 11AM

\begin{fcdefn}[]
  If I have a \gls{cl} $L$ in $[m]^n$, then I can
  order $x \le y$ if and only if $x_i \le y_i$ for
  all $i$. Let $L^-$ denote the first point in
  this ordering, and let $L^+$ denote the last
  point in this ordering.
\end{fcdefn}

\begin{fcdefn}[]
  Let $L_1, \ldots, L_k$ be \glspl{cl}. We call them
  focussed if $L_i^+ = f$ for all $i \in [k]$. For
  a fixed colouring, they are colour focused if
  they are focused and $L_i \setminus \{L_i^+\}$
  monochromatic for each $i$, and they have
  different colours.
\end{fcdefn}

\begin{example*}
  $[4]^2$
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/87f85e6e054b4463.png}
  \end{center}
  are $3$ colour-focused \glspl{cl} $[4]^2$.
\end{example*}

\begin{proof}[Proof of \nameref{HJ}]
  The proof is by induction on the size of the
  alphabet, i.e. $m$.

  $m = 1$ is trivially true.

  Assume $m \ge 1$ and assume $\hj(m - 1, k)$
  exists for all $k$.

  \textbf{Claim:} for every $1 \le r \le k$ there
  exists $n$such that in $[m]^n$
  \begin{enumerate}[(1)]
    \item either there exists monochromatic
      \glspl{cl}
    \item there exists $r$ colour-focused
      \glspl{cl}
  \end{enumerate}
  Then we are done by $r = k$ and looking at the
  focus.

  Now we prove the claim:

  $r = 1$: look in $[m - 1]^{n'} \subset [m]^n$
  We can take $\hj(m - 1, k)$.

  Now assume $r > 1$ and that $n$ is suitable for
  $r - 1$. We will show that $n + \hj(m - 1,
  k^{m^n})$ is suitable for $r$. Let $N = \hj(m -
  1, k^{m^n})$ for convenience.

  Need: given $c$ a $k$-colouring of $[m]^{n + N}$
  with no monochromatic \glspl{cl}, we can find $r$
  colour-focused \glspl{cl}. Look at $[m]^{n + N}$ as
  $[m]^n \times [m]^N$, with $[m]^n = \{a_1,
  \ldots, a_{m^n}\}$.

  Let us colour $[m]^N$ as follows:
  \[
    c' : [m]^N \to k^{m^n} \qquad c'(b) = (c(a_1,
    b), c(a_2, b), \ldots, c(a_{m^n}, b))
  .\]
  By \nameref{HJ} there exists a \glspl{cl} $L$ with
  active coordinates $I$ such that
  \[
    c(a, b) = c(a, b') \qquad \forall a \in [m]^n,
    \forall b, b' \in L \setminus \{L^+\}
  .\]
  But now this induces $c'' : [m]^n \to k$ where
  $c''(a) = c(a, b)$ for any $b \in L \setminus
  \{L^+\}$. By the definition of $n$, there exist
  $r - 1$ colour-focused \glspl{cl} (for $c''$) $L_1,
  L_2, \ldots, L_{r - 1}$, focused at $f$, and
  with active coordinates $I_1, I_2, \ldots, I_{r
  -1}$.

  Finally: look at the \glspl{cl} that start at $L_i^-
  \times L^-$ and active coordinates $I_i \cup I$.
  These give $r - 1$ \glspl{cl}, and the \glspl{cl} that
  starts at $f \times L^-$ with active coordinates
  $I$. All focused at $f \times L^+$. Then done.
\end{proof}

\begin{fcdefn}[$d$-dimensional space]
  \glsnoundefn{dpps}{$d$-point parameter
  set}{$d$-point parameter sets}%
  A \emph{$d$-dimensional space} $S \subseteq X^n$
  or a \emph{$d$-point parameter set} is a set
  such that there exists $I_1, I_2, \ldots, I_d
  \subseteq [n]$ disjoint and $a_i \in X ~\forall
  i \in [n] \setminus (I_1 \cup \cdots \cup I_d)$,
  and $x \in S$ if and only if:
  \begin{itemize}
    \item $x_i = a_i$ for all $i \in [n] \setminus
      (I_1 \cup \cdots \cup I_d)$
    \item $x_i = x_i$ if $i, j \in I_d$ for some
      $l \in [d]$
  \end{itemize}
\end{fcdefn}

\begin{fcthm}[The Extended Hales-Jewett Theorem]
  % Theorem 1.10
  Assuming:
  - $m, k, d$ positive integers
  Then:
  there exists $n$ such that whenever $[m]^n$ is
  $k$-coloured, there exists a \gls{dpps}
  monochromatic.
\end{fcthm}

\begin{proof}
  In $X^{n'd} = (X^d)^{n'} = Y^{n'}$ what does a
  \gls{cl} in $Y^{n'}$ look like?
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/6c1bfb2518d94972.png}
  \end{center}
  So a monochromatic line in $Y^{n'}$ is a
  monochromatic \gls{dpps} in $X^{n'd}$.

  Letting $n = d\hj(m^d, d)$ works.
\end{proof}

\begin{fcdefn}[Homothetic copy]
  \glsnoundefn{hc}{homothetic copy}{homothetic
  copies}%
  Let $S$ be a finite set of points in $X^n$. A
  homothetic copy of $S$ is a set of the form $x +
  \lambda S$.
\end{fcdefn}

\begin{fcthm}[Gallai's Theorem]
  \label{gallai}
  % Theorem 1.11
  Assuming:
  - finite $S \subset \Nbb^d$
  - $k$-colouring of $\Nbb^d$
  Then:
  there exists a
  monochromatic \gls{hc} of $S$.
\end{fcthm}

\begin{proof}
  $S = \{S_1, \ldots, S_m\}$. Let $c : \Nbb^d \to
  k$ be a colouring.

  We colour $[m]^n$
  (for $n$ large enough) as follows:
  \[
    c'((x_1, \ldots, x_n)) = c(S_{x_1} + S_{x_2} +
    \cdots + S_{x_m})
  .\]
  By \nameref{HJ}, there exists a monochromatic
  \gls{cl} in $[m]^n$ with active coordinates $I$.
  Then
  \[
    c \left( \sum_{\text{inactive}} S_i + |I|S_j \right)
  \]
  has the same colour for all $j \in [m]$.

  Done as
  this is a copy of $S$ translate by
  $\sum_{\text{inactive}} S_i$, and dilation
  factor $|I|$.
\end{proof}
