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Source. Published pp. 221–223, Lemma 3.4 and its proof. (canonical PDF).

If $V\subseteq\mathbb R^{d_1}$ is a finite $\alpha_V$-hyper-Ramsey set and $T\subseteq\mathbb R^{d_2}$ is a finite $\alpha_T$-hyper-Ramsey set, then the product $V*T\subseteq\mathbb R^{d_1+d_2}$ is $(\alpha_V+\alpha_T)$-hyper-Ramsey (Lemma 3.4, p. 221). Both slacks are positive, as Definition 3.1 requires.

Proof.

The intrinsic product radius satisfies $\rho(V*T)^2=\rho(V)^2+\rho(T)^2$ by circumradius_continuity. If either factor is a singleton, the product is congruent to the other factor. Radius enlargement in definitions adds the singleton's positive squared slack and proves the assertion. Hence assume $t=|T|\ge2$ and both factors have at least two points.

Take witnesses $H_V(n),H_T(m)$ with cardinality bases $c_V,c_T>1$ and avoiding density bases $a_V=1-\epsilon_V$, $a_T=1-\epsilon_T$ in $(0,1)$. Choose

$$\tau=\frac{-\log a_V}{2(t-1)\log c_T}>0, \qquad m=\lfloor\tau n\rfloor, \qquad N_n=n+m.$$

For all large $n$, both witnesses exist. Their product $H$ lies on the sphere of squared radius $\rho(V)^2+\alpha_V+\rho(T)^2+\alpha_T$ and has cardinality less than $\max(c_V,c_T)^{N_n}$.

Let $W\subseteq H$ contain no copy of $V*T$. For each $v\in H_V(n)$, let $W_v=\{t'\in H_T(m):(v,t')\in W\}$. If $W_v$ contains $b_v$ distinct copies of $T$, delete one point from each of those copies. The remaining set is $T$-free and has size at least $|W_v|-b_v$. The witness property therefore gives

$$b_v>|W_v|-a_T^m|H_T(m)|.$$

This argument counts all copies; it does not require them to be disjoint. Summing over fibers gives, for $B=\sum_vb_v$,

$$B>|W|-a_T^m|H_V(n)|\,|H_T(m)|.$$

For every fixed copy $T'\subseteq H_T(m)$, the set of $v$ with $\{v\}*T'\subseteq W$ is $V$-free. There is at least one such candidate copy in $H_T(m)$, since the full witness contains $T$. Consequently

$$\begin{aligned} B&<\binom{|H_T(m)|}{t}\,a_V^n|H_V(n)|\\ &\le |H_T(m)|^{t-1}\,a_V^n|H|\\ &<c_T^{m(t-1)}a_V^n|H| \le a_V^{n/2}|H|. \end{aligned}$$

The last inequality uses $m\le\tau n$ and the definition of $\tau$. Combining the two counts yields

$$\frac{|W|}{|H|}<a_V^{n/2}+a_T^{\lfloor\tau n\rfloor}.$$

For large $n$, $\lfloor\tau n\rfloor\ge\tau n/2$. Hence the right side is at most $2e^{-hn}$, where

$$h=\min\{-\tfrac12\log a_V,-\tfrac\tau2\log a_T\}>0.$$

Since $N_n\le(1+\tau)n$, it is at most $e^{-hN_n/[2(1+\tau)]}$ for all sufficiently large $n$. This is the required exponential avoiding-density bound on the sequence $N_n$. Its successive gaps are at most $1+\lceil\tau\rceil$. Apply the bounded-gap form of fact_3_10 to obtain witnesses in every large ambient dimension on exactly the same sphere. Its squared slack above the intrinsic product radius is exactly $\alpha_V+\alpha_T$.

Source precision.

The source's displayed construction covers the dimensions $n+\lfloor\tau n\rfloor$; it does not by itself cover every large integer. The bounded-gap step supplies that implication. The source equation for $\tau$ has an exponent $|T|-1$, so singleton factors were separated before using it. No subset-inheritance assertion at a smaller intrinsic radius enters this proof.

Bears on. #174.