Canonical versions of van der Waerden and Szemerédi type recurrence (conjectures and implications)

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Problem: Reconstructed conjecture, with $\mathbb N=\{1,2,\dots\}$: for every compact Hausdorff space $X$, continuous $T:X\to X$, open $U\subseteq X$, and $\ell\ge1$, there is $n\ge1$ such that either

or the sets $T^{-in}U$, $0\le i\le \ell$, are pairwise disjoint. Here $T^{-k}U=\{x:T^k x\in U\}$.

If $0\in\mathbb N$, the statement is trivial for nonempty $U$ by taking $n=0$, so the recurrence interpretation requires positive $n$.

Result: The conjecture is false, already for $\ell=2$, with $X$ a compact metric subshift and $T$ a homeomorphism.

First construct a set $A\subseteq\mathbb N$ such that:

1. $A$ contains no nontrivial three-term arithmetic progression;
2. for every $d\ge1$, there is $a\in A$ with $a+d\in A$.

Construction: build finite AP-free sets $S_m$. Given finite AP-free $S$ and $d\ge1$, only finitely many $x\in\mathbb Z$ make $S\cup\{x,x+d\}$ contain a three-term AP: any new AP must contain $x$ or $x+d$; if it contains one new point, that point is determined by two old points; if it contains both, the old point must be one of $x-d,x+d/2,x+2d$. Hence choose $x_d$ large outside this finite bad set and set

Then $S_d$ remains AP-free and contains a pair of difference $d$. Let

Any three-term AP in $A$ would already occur in some finite $S_d$, contradiction. Also every $d\ge1$ occurs as a difference in $A$.

Now let $y\in\{0,1\}^{\mathbb Z}$ be the indicator of $A$, let $\sigma$ be the shift $(\sigma z)_k=z_{k+1}$, and set

Then $X$ is compact Hausdorff, $\sigma|_X$ is continuous, and

is clopen.

For every $n\ge1$, since $n\in A-A$, choose $a,a+n\in A$. Then $\sigma^a y\in U\cap \sigma^{-n}U$. Hence $U,\sigma^{-n}U,\sigma^{-2n}U$ are not pairwise disjoint, so alternative (4) fails for every $n$.

On the other hand, if

then some $z\in X$ has $z_0=z_n=z_{2n}=1$. Since this is a clopen cylinder condition and $X$ is the orbit closure of $y$, some shift $\sigma^r y$ also has ones at coordinates $0,n,2n$. Thus

a forbidden three-term AP. Hence alternative (3) fails for every $n$.

Therefore no positive $n$ satisfies either alternative for $\ell=2$. This disproves the conjecture.

Citation: No external citation is used; the counterexample is self-contained.