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Source. Published p. 231, Fact 3.10 and its proof; the bounded-gap version also completes the dimension step in Lemma 3.4. (canonical PDF).

As printed (p. 231): let $c$, $\alpha$ and $\epsilon$ be fixed and let $\{iD\}_{i=1}^\infty$ be an infinite arithmetic progression. If $V$ is a finite set such that for every $i\ge1$ there is a set $\mathcal H(iD)\subseteq\mathbb R^{iD}$ satisfying (i)–(iii) of Definition 3.1, then $V$ is $\alpha$-hyper Ramsey.

Bounded-gap form proved here. Fix a finite nonempty configuration $X$ and constants $R>0$, $c>1$ and $0<\epsilon<1$. Suppose an unbounded increasing sequence of positive integer dimensions $n_i$ has bounded gaps and, for every sufficiently large $i$, has a nonempty witness $H_{n_i}\subseteq S(R,n_i)$ with $|H_{n_i}|<c^{n_i}$ such that every $X$-free subset has density less than $(1-\epsilon)^{n_i}$.

Then the same fixed target and radius admit such witnesses in every sufficiently large dimension, with $\epsilon$ replaced by a smaller positive constant. In particular, if $R^2=\rho(X)^2+\alpha$ and $n_i=iD$ for a fixed positive integer $D$, then $X$ is $\alpha$-hyper-Ramsey.

Proof.

Let $D$ bound the gaps, and for each large integer $m$ choose the largest available $n_i\le m$. Then $n_i>m-D$, so $n_i\ge m/2$ for all sufficiently large $m$. Pad every vector of $H_{n_i}$ with zero coordinates. This is an isometry into $S(R,m)$, and $|H_{n_i}|<c^{n_i}\le c^m$.

Set $1-\epsilon'=\sqrt{1-\epsilon}$, so $0<\epsilon'<\epsilon$. Every $X$-free subset has density strictly less than

$$(1-\epsilon)^{n_i}\le(1-\epsilon)^{m/2}=(1-\epsilon')^m.$$

Thus a subset whose density is at least the last quantity must contain $X$, with the weak forcing endpoint preserved. The arithmetic progression in Fact 3.10 is the case $n_i=iD$. The proof also permits a finite initial segment of dimensions to be absent.

Source precision.

The target and radius are fixed before the dimension varies. This argument cannot transfer witnesses for a varying sequence of noncongruent targets.

Bears on. #174.