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\begin{corollary*}
  Whenever $\Nbb$ is finitely coloured, there
  exists a colour class that contains arbitrarily
  long \glspl{ap}.
\end{corollary*}

What about infinite monochromatic \glspl{ap}

No, for example:
\begin{enumerate}[(1)]
  \item colour $\{1, 2\}$ red, $\{3,4,5\}$ blue,
    $\{6, 7, 8, 9\}$ red, etc
  \item Or ``just do it'': the set of \glspl{ap}
    in $\Nbb$ is countable. So let them be $A_1,
    A_2, \ldots, A_k, \ldots$. Pick $x_1 < y_1$ in
    $A_1$, and colour $x_1$ red, $y_1$ blue. Next
    go to $A_2$ and select $x_1 < y_1 < x_2 < y_2$
    in $A_2$. Colour $x_2$ red, $y_2$ blue. Keep
    going... 
\end{enumerate}

\begin{fcthm}[Strengthened Van Waerden]
  \label{svw}
  % Theorem 1.8
  Assuming:
  - $m, k \in \Nbb$
  Then:
  there exists $n$ such that whenever $[n]$ is
  $k$-coloured, there exists a monochromatic
  \gls{ap} of length $m$ together with their
  common differences, i.e. the set $\{d, a, a + d,
  \ldots, a + (m - 1) d\}$ is monochromatic.
\end{fcthm}

\begin{proof}
  By induction on the number of colours. $k = 1$
  is trivial.

  Assume that the case for $k - 1$ colours is
  true. So there exists $n$ that works for $m$ and
  $k - 1$.

  We will show that $\W(k, n(m - 1)+1)$ works for
  $m$ and $k$. If $[\W(k, n(m - 1)+1)$ are
  $k$-coloured, then there exists a monochromatic
  \gls{ap} (say red) of length $n(m - 1)+1$, say
  $a, a + d, \ldots, a + dn(m - 1)$.

  If any of $d,
  2d, \ldots, nd$ is red, then we are done: e.g.
  $a, a + rd, \ldots, a + r(m - 1)d$ for some $r
  \in [n]$.

  If not, the set $\{d, 2d, \ldots, nd\}$ is $k -
  1$-coloured.. This involves a $(k - 1)$
  colouring on $[n]$, therefore there exist $b, b
  + r, \ldots, b + (m - 1)r$ and $r$ the same
  colour. This translates to $db, db + r, \ldots,
  db + d(m - 1)r, dr$ monochromatic.
\end{proof}

\begin{remark*}
  $m = 2$ is known as Schur's Theorem: we can
  always find $a, a + d, d$ monochromatic (for
  finite colouring of $\Nbb$).
  In other words, there exists a monochromatic
  solution to $x + y = z$.

  Can deduce Schur from Ramsey for pairs: $c :
  \Nbb \to [k]$ then we induce $c' : \Nbb^{(2)}
  \to [k]$ as follows: $c'(ij) = c(j - i)$. By
  \nameref{ramsey_rsets}, there exists $i < j < k$
  such that $c'(ij) = c'(ik) = c'(jk)$. Then
  \[
    c(\ub{j - i}_{x}) = c(\ub{k - i}_{z}) =
    c(\ub{k - j}_{y})
  .\]
  Get $x + y = z$ and $x, y, z$ monochromatic.
\end{remark*}

\subsection{The Hales-Jewett Theorem}

\glssymboldefn{pn}%
Let $X$ be a finite set and $X^n$ is words of
length $n$ on the alphabet $X$.

\begin{fcdefn}[Combinatorial line]
  \glsnoundefn{cl}{combinatorial
  line}{combinatorial lines}%
  A \emph{combinatorial line} in $X\pn$ is a set
  of the following form:
  \[
    \{(x_1, \ldots, x_n) \in X\pn : \exists I
    \subseteq [n], a_i \in X, I \neq \emptyset, x_i = a_i
    ~\forall i \notin I, x_i = x_j ~\forall i, j
    \in I\}
  .\]
\end{fcdefn}

\begin{example*}
  $X = [3]$ and we want combinatorial lines in
  $[3]^2$. If $I = \{1\}$, we get:
  \begin{align*}
    L_1
    &= \{(1, 1), (2, 1), (3, 1)\} \\
    L_2
    &= \{(1, 2), (2, 2), (3, 2)\} \\
    L_2
    &= \{(1, 3), (2, 3), (3, 3)\}
  \end{align*}
  If $I = \{2\}$ we get:
  \begin{align*}
    L_4
    &= \{(1, 1), (1, 2), (1, 3)\} \\
    L_5
    &= \{(2, 1), (2, 2), (2, 3)\} \\
    L_6
    &= \{(3, 1), (3, 2), (3, 3)\}
  \end{align*}
  $I = \{1, 2\}$, then
  \[
    L_7 = \{(1, 1), (2, 2), (3, 3)\}
  .\]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/74553bee50a3488a.png}
  \end{center}
\end{example*}

\begin{example*}
  $[3]^3$, $I = \{1\}$ $(1, 2, 3), (2, 2, 3), (3,
  2, 3)$, or if $I = \{1, 3\}$ $(1, 3, 1), (2, 3,
  2), (3, 3, 3)$.
\end{example*}

\begin{note*}
  The definition of a \gls{cl} is
  invariant under permutations of $X$.
\end{note*}

\begin{fcthm}[The Hales-Jewett Theorem]
  \label{HJ}
  Assuming:
  - $m, k \in \Nbb$
  Then:
  there exists $n$ such that whenever we
  $k$-colour $[m]\pn$ there exists a monochromatic
  \gls{cl}.
\end{fcthm}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item \glssymboldefn{HJ}%
      The smallest such $n$ is call $HJ(m,
      k)$.
    \item \nameref{HJ} implies \nameref{VW}:
      \begin{proof}
        Let $c$ be a finite colouring of $\Nbb$.
        Let $n$ be large enough ($=\hj(m, k)$)and now colour
        $[m]\pn$. $m$ is the length of desired
        \gls{ap}. Define
        \[
          c'((x_1, x_j, \ldots, x_n)) = c(x_1 +
          x_2 + \cdots x_n)
        .\]
        By \nameref{HJ}, there exists a
        monochromatic line
        \begin{center}
          \includegraphics[width=0.6\linewidth]{images/62acdb07271047e5.png}
        \end{center}
      \end{proof}
  \end{enumerate}
\end{remark*}

\textbf{Exercise:} Suppose you play Naughts and
Crosses with $m$ in a line and you play it in high
enough dimensions. Show it cannot be a draw
(assuming optimal play).
Moreover, it is a first player win. \textit{Hint:
Strategy stealing.}