Big Breakthrough in the exciting world of sum-free sets!

 Let \([n]\) be the set \(\{1,\ldots,n\}\). (This is standard.)

Let X be a set of integers. \(X\) is sum-free if there is NO  \(x,y,z\in X\) such that \(x+y=z\). (Note that \(x=y\) is allowed.)

Lets try to find a large sum-free set of \([n]\). One obvious candidate is 

\(\{1,3^1, 3^2,\ldots,3^{\log_3 n} \}\) (assume \(n\) is a power of 3). 

So we can get a \(\log_3 n\) sized sum free set of \([n]\). Can we do better?

YES: 

Erdos showed that every set of \(n\) reals has a sum-free subset of size \(n/3\). I have a slightly weaker version of that on slides here.

That result lead to many questions:

a) Find \(f(n)\) that goes to infinity  such that every set of \(n\) reals has a sum-free subset of size \( n/3 + f(n)\).  

b) Replace reals with other sets closed under addition: naturals, integers, some groups of interest.

c) Just care about \([n]\).

Tao and Vu had a survey in 2016 of sum-free results, see here.

We mention three results from their survey

(0) Eberhard, Green, and Manners showed that the 1/3 cannot be improved (see the survey for a more formal statement of this). So, for example, nobody will ever be able to replace 1/3 with 1/2.9.

(1) Alon and Kleitman  modified the prob argument of Erdos (in the slides) and were able to replace \( n/3\) with   \( (n+1)/3\).

(2) Bourgain showed, with a sophisticated argument,that \(n/3\) could be replaced by \((n+2)/3\)

I believe that result (2) was the best until a recent paper of Ben Bedert (see here). Did he manage to 

push it to \((n+100)/3\)? No. He did much better:

For every set of \(n\) reals there is a sum-free  subset of size \(n/3 + c\log\log n\).

Reflections

1) This is a real improvement!

2) Will it stay at \(n/3 + c\log\log n\) or will there NOW be an avalanche of results?

3) Contrast: 

Erdos's proof  I can (and have) taught to my High School Ramsey Gang (note: you DO NOT want to mess with them).

 The proof by Ben Bedert is 36 pages of dense math.

4) This leads to the obvious open question: Is there an elementary proof of some bound of the form \(n/3 + f(n)\) where \(f(n)\) goes to infinity.