So cool. On the one hand, going from n/3 to (n/3 + log log n) seems like such a small improvement, but as the article shows, the history is formidable:
- n/3 (Erdős, 1965)
- (n+1)/3 (Alon and Kleitman, 1990)
- (n+2)/3 (Bourgain, 1997)
- n/3 + Ω(log log n) (this paper, Benjamin Bedert, https://arxiv.org/abs/2502.08624)
And the upper bound:
- n/3 + o(n) (Eberhard, Green, Manners, 2014).
Ben Green's list of 100 open problems is which this is (was?) Problem 1, is here: https://people.maths.ox.ac.uk/greenbj/papers/open-problems.p...
It's weird to spend paragraphs talking about the incremental improvements to (N+1)/3 and (N+2)/3 and then miss that the new bound is N/3 + c*log(log(n)) for some c>0, not N/3+log(log(n))
Why is the lower bound N/3 and not N/2? Doesn’t the set of all odd numbers make the lower bound N/2?
N/3 + log log N holds for any arbitrary set, not just for 1 … N or something.
An adversary first gives you any set, and then you have to find a subset.
They could give you only even numbers.
Ohhhh ok thanks!
Is the main subject here addition or sets?
It concerns the size of the largest sum-free set [0]. Take a (finite) set of integers, A. What is the largest subset of A such that no two entries sum to a third.
The previous results was not much better than |A|/3. The current, just proved, result shows that the largest subset is |A|/3 + c log(log(|A|)).
For example, the set {1,2,3} is not sum-free (1+2 = 3) but the subset {2,3} is sum-free (2+3 \notin {2,3}).