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\begin{proof}
  $X = \{1, 2, \ldots, m\}$, where the numbers are
  the names of points.
  \begin{center}
    \includegraphics[width=0.4\linewidth]{images/0119b1d7955d4d9c.png}
  \end{center}
  We find a copy of $\sqrt{m} X$ of the form
  \[
    (1, 2, \ldots, m), (2, 3, \ldots, m, 1),
    (3, 4, \ldots, m, 1, 2), \ldots
    (m, 1, 2, \ldots, m - 1)
  .\]
  We will show by induction on $r$ and all $k$ at
  once that we can find a copy of $X$ with $\{1,
  2, \ldots, r\}$ monochromatic.

  $r = 1$ is trivial as it is just a point.

  $r = 2$ is 2 points at a specified distance
  (which we showed is \gls{eram}).

  Assume true for $r$ and all $k$. By
  \nameref{prop:3.7}, there exists $S$ and $N$
  such that we have a $X^N$ (copy) on which the
  colouring is $\{1, 2, \ldots, r\}$-invariant on
  $X^N$ (for any $N$). We will choose $N$ to be as
  big as we want.

  \textbf{The clever bit:}
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/9cdf25da7b7e48af.png}
  \end{center}
  We will colour $(m - 1)$ sets in $[N]$ say $a_1
  < a_2 < \cdots < a_{m - 1}$ as follows.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/b1637d680d2f4b8e.png}
  \end{center}
  $c'(\{a_1, \ldots, a_{m + 1}\}) = (c(w_1),
  c(w_2), \ldots, c(w_r))$ is a $k^r$ colouring of
  $[N]^{(m - 1)}$.

  As $N$ can be taken as big as needed, by Ramsey
  there exists a $m$-monochromatic set. By
  relabeling, we may assume that this set is $\{1,
  2, \ldots, m\}$ coordinates.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/e8408da002054225.png}
  \end{center}
  Now we look at the following:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/786a1bb0410a45f1.png}
  \end{center}
  with this we note that the colour of $y_i$ is
  the same as the colour of $x_{i + 1}$.

  Now look at: $(1, 2, \ldots, m)$, $2, 3, \ldots,
  m, 1)$, \ldots, $(r + 1, \ldots, r, r - 1)$.
  monochromatic copy of $\{1, 2, \ldots, r + 1\}$.
  They all have the same colour (ignoring this).
\end{proof}

\begin{remark*}
  Same proof works for any cyclic set: i.e. a set
  $\{x_1, \ldots, x_n\}$ such that the map $x_i
  \mapsto x_{i + 1}$ (modulo $n$) is a symmetry of
  the set.

  Or equivalently, there exists a cyclic
  transitive symmetry
  group on $X$.
\end{remark*}

\begin{example*}
  Given by rotation $120^\circ$ and reflection.
  Generates order $6$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/6f8d1f0d632443ef.png}
  \end{center}
  This is \gls{eram}.
\end{example*}


\nonexaminableon

A soluble group is ``built'' up from cyclic
groups.

\begin{theorem*}[Kriz]
  If $X$ hs a soluble, transitive symmetry group,
  then $X$ is \gls{eram}.
\end{theorem*}

Rival conjecture to the \gls{spher} conjecture
(2010, Leader, Russell, Walters):

$X$ is \gls{eram} if and only if $X$ is
subtransitive (subtransitive means that it can be
embedded in a transitive set, i.e. it can be
embedded in a set that has a transitive isometry
group).

Why are they rival?

\gls{spher} does not imply sub-transitive.

sub-transitive does imply \gls{spher}: let $X$ be
sub-transitive. Embed into $Y$ transitive. There
exists a unique minimal sphere containing $Y$,
which is preserved by the isometry group.

For \gls{spher} doesn't imply sub-transitive: the
\emph{kite}.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/c014d8add226433b.png}
\end{center}
This is not sub-transitive.

What happens if the kite is \gls{eram}? It would
disprove the 2010 conjecture. If not \gls{eram},
it would disprove the original conjecture.

It is believed that the transitive conjecture is
true.

It could also be the case that neither is true :?.

\nonexaminableoff