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\begin{fcthm}[]
  % Theorem 1.5
  Assuming:
  - $\Nbb^{(r)}$ abitrarily coloured
  Then:
  we can find an infinite set $M$ and $I \subseteq
  [r]$ such that for any $x_1 < x_2 < \cdots <
  x_r$ in $M$, and $y_1 < y_2 < \cdots < y_r$ in
  $M$ we have $c(x_1 x_2 \cdots x_r) = c(y_1 y_2
  \cdots y_r)$ if and only if $x_i = y_i$ for all
  $i \in I$.
\end{fcthm}

\begin{example*}
  In the previous theorem:
  \begin{enumerate}[(i)]
    \item $I = \emptyset$
    \item $I = \{1, 2\}$
    \item $I = \{1\}$
    \item $I = \{2\}$
  \end{enumerate}
\end{example*}

These $2^r$ colourings are call the ``canonical
colourings'' of $\Nbb^{(r)}$.

\begin{proof}
  Exercise. Note that this proof \emph{is}
  examinable (because the ideas are exactly the
  same as those in the previous theorem).
\end{proof}

\subsection{Van der Waerden's Theorem}

We will colour $\Nbb$.

\textbf{Aim 1:} Whenever we $2$-colour $\Nbb$, we
find a monochromatic arithmetic progression of
length $m$ for any $m$.

\glsnoundefn{ap}{arithmetic
progression}{arithmetic progressions}%
The abbreviation A.P. can be used to mean
``arithmetic progression'', i.e. a sequence of the
form $\{a, a + d, \ldots, a + (m - 1)d\}$.

\textbf{Aim 2:} For any $m$, there exists $n$ such
that whenever $[n]$ are $2$-coloured, there exists
a monochromatic \gls{ap} of length $m$.

This is equivalent to Aim 1, by using a \gls{pbc}
argument like before:

If Aim 2 is not true, then we can find $c_n : [n] \to \{1, 2\}$
such that infinitely many agree on $\{1\}$. Of
those infinitely many agree on $\{2\}$, etc. Keep
going (as before), and then get a colouring of
$\Nbb$ without a monochromatic \gls{ap} of length
$m$.

The other direction is easier.

We will show something a bit stronger (because it
turns out to be easier): we will prove Aim 2 but
with $k$ colours.

This is in contrast with the earlier theorems,
where the proofs were slightly easier to think about
with just 2 colours.

\begin{fcdefn}[Focused]
  \glsadjdefn{foc}{focused}{\gls{ap}}%
  \glsadjdefn{cfoc}{colour-focused}{\gls{ap}}%
  Let $A_1, A_2, \ldots, A_k$ be \glspl{ap} of
  length $m$. Say
  \[
    A_i = \{a_i, a_i + d_i, \ldots, a_i + (m - 1)
    d_i\}
  .\]
  We say that these \glspl{ap} are \emph{focused}
  if $a_i + m d_i = f$ for all $i \in [k]$.

  If we have a colouring of $\Nbb$ and $A_1,
  \ldots, A_k$ are \gls{foc} \glspl{ap} if each
  $A_i$ is monochromatic but they all have
  different colours, then we call them
  \emph{colour-focused}.
\end{fcdefn}

\begin{example*}
  $\{4, 8\}$ and $\{10, 11\}$ are \gls{foc} at
  $12$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/e5261c48f8dd451b.png}
  \end{center}
\end{example*}

\begin{fcthm}[]
  \label{thm:1.5}
  % Theorem 1.5
  Assuming:
  - $\Nbb$ is $k$-coloured
  Then:
  we can find a
  monochromatic \gls{ap} of length $3$
  (equivalently, for any $k$ we find an $n$ that
  works).
\end{fcthm}

\begin{proof}
  \textbf{Claim:} For any $r \le k$ there exists
  an $n$ such that if $[n]$ is $k$-coloured then
  either:
  \begin{itemize}
    \item there exists a monochromatic \gls{ap} of
      length $3$
    \item there exists $r$ \gls{cfoc} \glspl{ap}
      of length $2$
  \end{itemize}
  The proof is by induction on $r$.

  Base case $r = 1$ we can take $n = k + 1$, since
  2 numbers will have the same colour.

  Suppose the result is true for $r - 1$ and let
  $n$ be an $n$ that satisfies the property in the
  claim.

  We will show that $N = (k^{2n} + 1)2n$ works for
  $r$. Let $c : [(k^{2n} + 1)2n] \to [k]$ be a
  colouring. We will split the ground set into
  $k^{2n} + 1$ blocks of length $2n$. Call the
  blocks $B_i$.

  If there exists a monochromatic \gls{ap} of
  length $3$ in this
  colouring, then we are done. So assume not.

  By the induction hypothesis, the first half of
  each $B_i$ has $r -
  1$ \gls{cfoc} \glspl{ap} of length $2$. Because
  $|B_i| = 2n$, each block also contains their
  focus.

  For a set $|M| = 2n$, there are exactly $k^{2n}$
  ways to $k$-colour it. So there exists two
  blocks $B_p$ and $B_{p + t}$ that are identically
  coloured.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/e07964cb02814928.png}
  \end{center}
  Let $\{a_i, a_i + d_i\}$ be the $r - 1$ \gls{cfoc}
  \glspl{ap} in $B_p$. Then $\{a_i + 2n t, a_i +
  d_i + 2nt\}$ are the corresponding ones in $B_{p
  + t}$. Let $f$ be the focus in $B_p$, so
  therefore $f + nt$ is the focus in $B_{p + t}$.

  Now take $\{a_i, a_i + d_i + 2nt\}$ for $i \in
  [r - 1]$, and $\{f, f + 2nt\}$. Since $a_i + 2d_i
  = f$, we have $a_i + 2d_i + 4nt = f + 4nt$. So
  all $r$ of these sequences are \gls{foc} at $f +
  4nt$.

  We know that $\{a_i, a_i + d_i + nt\}$ and $\{f,
  f + 2nt\}$ are monochromatic by the choice of
  $B_p$, $B_{p + t}$. Why colour focused? $\{a_i,
  a_i + 2nt\}$ have different colours by induction
  hypothesis. Also, because $c$ was assumed to
  have no monochromatic \gls{ap} of length $3$,
  the colours of $\{f, f + 2nt\}$ must be
  different to all the colours of the above $r -
  1$ \glspl{ap} of length $2$. Thus we have $r$
  colour-focused \glspl{ap} of length $2$ in
  $[(k^{2n} + 1)2n]$.
\end{proof}

\begin{remark*}
  The idea of looking at all the possible
  colouring of a set is referred to as the
  ``product argument''.
\end{remark*}

\glssymboldefn{W}%
The Van der Waerden number $W(k, m)$ is the
smallest $n$ such that whenever $[n]$ is
$k$-coloured, there exists a monochromatic
\gls{ap} of length $m$.

Proof above claims that $W(k, 3) \le
k^{k^{\iddots^{k^{4k}}}}$ for a tower of size $k -
1$. ``tower-type bound''.