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Source. Published pp. 350 and 362, compressed infinite-set extensions (published scan).

Statement. Fix integers $d,m\ge1$ and a set $K\subseteq\mathbb R^d$. If $K$ is not contained in a union of at most $m$ concentric spheres, some finite subset already has this property. Radii zero are allowed; unused spheres may be repeated.

Complete proof. Let $V$ be the finite-dimensional vector space of real polynomials in $d$ variables of total degree at most $2m$. Each $x\in K$ defines an evaluation functional $e_x\in V^*$. Select finitely many points $S\subseteq K$ whose evaluation functionals form a basis for the span of all $e_x$, $x\in K$. Then every polynomial in $V$ vanishing on $S$ also vanishes on $K$.

If $S$ lay on concentric spheres with center $a$ and radii $r_1,\ldots,r_m\ge0$, the polynomial

$$P(X)=\prod_{j=1}^m\bigl(\|X-a\|^2-r_j^2\bigr)$$

would lie in $V$ and vanish on $S$. It would therefore vanish at every $x\in K$, placing $K$ on the same union of spheres. The contrapositive proves the assertion.

Containment in a higher-dimensional ambient space does not change this property. Project a common center orthogonally onto the affine hull of the configuration. Every squared distance decreases by the same squared projection length; for each sphere actually meeting the configuration the remaining squared radius is nonnegative. Thus at most $m$ concentric spheres in the original affine hull still contain the configuration. Congruence transports the assertion by the affine Gram isometry. $\square$

This finite-dimensional polynomial argument supplies the finite-witness step that the source calls immediate. It does not assert that an infinite Ramsey set exists or extend the finite-product theorem to infinite factors.

Bears on. #174.