# Behrend’s construction

At first sight it may seem that the greedy algorithm yields the densest subset of $\{0,1,\ldots,N\}$ that is free of arithmetic progressions of length $3$. It is not hard to show that the greedy algorithm yields the set of numbers that lack digit $2$ in their ternary development (http://planetmath.org/Base3). Density (http://planetmath.org/AsymptoticDensity) of such numbers is $O(N^{\log_{3}2-1})$.

However, in 1946 Behrend constructed much denser subsets that are free of arithmetic progressions. His major idea is that if we were looking for a progression-free sets in $\mathbb{R}^{n}$, then we could use spheres. So, consider an $d$-dimensional cube $[1,n]^{d}\cap\mathbb{Z}^{d}$ and family of spheres $x_{1}^{2}+x_{2}^{2}+\cdots+x_{d}^{2}=t$ for $t=1,\ldots,dn^{2}$. Each point in the cube is contained in one of the spheres, and so at least one of the spheres contains $n^{d}/dn^{2}$ lattice points. Let us call this set $A$. Since sphere does not contain arithmetic progressions, $A$ does not contain any progressions either. Now let $f$ be a Freiman isomorphism from $A$ to a subset of $\mathbb{Z}$ defined as follows. If $x=\{x_{1},x_{2},\ldots,x_{d}\}$ is a point of $A$, then $f(x)=x_{1}+x_{2}(2n)+x_{3}(2n)^{2}+\cdots+x_{d}(2n)^{d-1}$, that is, we treat $x_{i}$ as $i$’th digit of $f(x)$ in base $2n$. It is not hard to see that $f$ is indeed a Freiman isomorphism of order $2$, and that $f(A)\subset\{1,2,\ldots,N=(2n)^{d}\}$. If we set $d=c\sqrt{\ln N}$, then we get that there is a progression-free subset of $\{1,2,\ldots,N\}$ of size at least $Ne^{-\sqrt{\ln N}(c\ln 2+2/c+o(1)}$. To maximize this value we can set $c=\sqrt{2/\ln 2}$. Thus, there exists a progression-free set of size at least

 $Ne^{-\sqrt{8\ln 2\ln N}(1+o(1))}.$

This result was later generalized to sets not containing arithmetic progressions of length $k$ by Rankin. His construction is more complicated, and depends on the estimates of the number of representations of an integer as a sum of many squares. He proves that the size of a set free of $k$-term arithmetic progression is at least

 $Ne^{-c(\log N)^{1/(k-1)}}.$

## References

• 1 Felix A. Behrend. On the sets of integers which contain no three in arithmetic progression. Proc. Nat. Acad. Sci., 23:331–332, 1946. http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0060.10302Zbl 0060.10302.
• 2 Leo Moser. On non-averaging sets of integers. Canadian J. Math., 5:245–252, 1953. http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0050.04001Zbl 0050.04001.
• 3 Robert A. Rankin. Sets of integers containing not more than a given number of terms in arithmetical progression. Proc. Roy. Soc. Edinburgh Sect. A, 65:332–344, 1962. http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0104.03705Zbl 0104.03705.
Title Behrend’s construction BehrendsConstruction 2013-03-22 13:40:53 2013-03-22 13:40:53 bbukh (348) bbukh (348) 11 bbukh (348) Result msc 05D10 msc 11B25