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Source. Theorem 28 with its proof and the sentence before it, p. 583, of P. Erdős, R. L. Graham, P. Montgomery, B. L. Rothschild, J. Spencer and E. G. Straus, Euclidean Ramsey Theorems, III, Infinite and Finite Sets (Keszthely 1973), Colloq. Math. Soc. János Bolyai 10, North-Holland (1975), 559--583, as identified on the source card.

Statement

Theorem 28 (p. 583). Let $R(1,1,x)$ hold and let $S(x)\subset E^2$ be a set with a minimal number of elements such that every two-coloring of $S(x)$ yields a monochromatic $(1,1,x)$-triple. Then $|S(x)|\to\infty$ as $x\to1$.

Similarly, let $R(1,1,\bar x)$ hold and let $\bar S(x)\subset E^2$ be a set with a minimal number of elements such that every proper two-coloring of $\bar S(x)$ yields a $(1,1,x)$-triple whose two vertices on the $x$-side are colored alike and opposite to the third vertex. Then $|\bar S(x)|\to\infty$ as $x\to2$.

The sentence before the theorem (p. 583) draws the moral: even for triples with commensurable distances, Conjectures 3 or 4 cannot be proved by coloring finite subsets of $E^2$ with a bounded number of elements. The theorem's limits are read here as taken over the $x$ for which the hypothesis holds.

Proof pointer

P. 583. If $|S(x_n)|=N$ along a sequence $x_n\to1$, minimality keeps every point of $S(x_n)$ within $2N$ of a fixed point (otherwise the set splits into two parts more than $2$ apart, each triple lying in one part). A convergent subsequence then has a limit set $S$ in which every two-coloring would have a monochromatic $(1,1,1)$-triple, contradicting the fact that $R(1,1,1)$ is false. The second part runs the same way from the falsity of $R(1,1,\bar2)$.

Read depth. Claims checked: the statement was read clause by clause on p. 583; the proof was read for its structure only.

Bears on

• Problem 173: a limit on one method, not a result on the question. Any proof that every isosceles $(1,1,x)$-triangle with $x$ near $1$ is Ramsey cannot use witness configurations of bounded size.