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Hindman (Furstenberg)
If $\N$ is partitioned into finitely many
then there is is an infinite subset $H \subseteq \N$
such that all finite sums of distinct
elements of $H$
belong to the same set of the partition.
Let $X$ be a compact Hausdorff space and $T\colon X \to X$
Consider $(X,T)$.%TODO different notations
Then for every $x \in X$
there is a uniformly recurrent $y \in X$
such that $y $ is proximal to $x$.
\item View partition as $f\colon \N \to k$. Consider $X \coloneqq k^{\N}$ (product topology, compact and Hausdorff).
Let $x \in X$ be the given partition.
\item $T\colon X \to X$ shift: $T(y)(n) \coloneqq y(n+1)$.
\item Let $y$ proximal to $x$, uniformly recurrent.
\item proximal $\leadsto$ $\forall N$.~$T^n(x)\defon_N = T^n(y)\defon_N$
for infinitely many $n$.
\item uniform recurrence $\leadsto$
\forall n .~\exists N.~\forall r.y\defon{\{r,\ldots,r+N-1\}}
\text{ contains } $y\defon{\{0,\ldots,n\}}$ \text{ as a subsequence.}
(consider neighbourhood $G_n = \{z \in X : z\defon{n} = y\defon{n} \}$).
\item Consider $c \coloneqq y(0)$. This color works:
\item $G_0 \coloneqq y\defon{\{0\}}$,
take $N_0$ such that $y\defon{\{r, \ldots, r + N_0 - 1\}} $
contains $y(0)$ for all $r$ (unif.~recurrence).
$y\defon{\{r,\ldots,r+N_0 - 1\} } = x\defon{\{r,\ldots,r+N_0 -1\} }$
for infinitely many $r$ (proximality).
Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
\item $G_1 \coloneqq y\defon{\{0,\ldots,h_0\} }$,
take $N_1$ such that $y\defon{\{r,\ldots,r +N_1-1\}}$
contains $y\defon{\{0,\ldots,h_0\} }$
for all $r$ (unif.~recurrence).
So among ever $N_1$ terms, there are two of distance $h_0$
where $y$ has value $c$.
So $\exists h_1 > h_0$ such that $x(h_1) = x(h_1 + h_0) = c$
\item Repeat:
Choose $h_i$ such that
for all sums $s$ of subsets of $\{h_0,\ldots, h_{i-1}\}$,
$x(s+h_i) = y(s+h_i) = c$:
Find $N_i$ such that every $N_i$ consecutive
terms of $y$ contain a segment that coincides
with the initial segment of $y$
up to the largest $s$,
then find a segment of length $N_i$ beyond $h_{i-1}$
where $x$ and $y$ coincide.