The number of subsets of integers with no $k$-term arithmetic progression

10 May 2016  ·  Balogh József, Liu Hong, Sharifzadeh Maryam ·

Addressing a question of Cameron and Erd\Ho s, we show that, for infinitely many values of $n$, the number of subsets of $\{1,2,\ldots, n\}$ that do not contain a $k$-term arithmetic progression is at most $2^{O(r_k(n))}$, where $r_k(n)$ is the maximum cardinality of a subset of $\{1,2,\ldots, n\}$ without a $k$-term arithmetic progression. This bound is optimal up to a constant factor in the exponent... For all values of $n$, we prove a weaker bound, which is nevertheless sufficient to transfer the current best upper bound on $r_k(n)$ to the sparse random setting. To achieve these bounds, we establish a new supersaturation result, which roughly states that sets of size $\Theta(r_k(n))$ contain superlinearly many $k$-term arithmetic progressions. For integers $r$ and $k$, Erd\Ho s asked whether there is a set of integers $S$ with no $(k+1)$-term arithmetic progression, but such that any $r$-coloring of $S$ yields a monochromatic $k$-term arithmetic progression. Ne\v{s}et\v{r}il and R\"odl, and independently Spencer, answered this question affirmatively. We show the following density version: for every $k\ge 3$ and $\delta>0$, there exists a reasonably dense subset of primes $S$ with no $(k+1)$-term arithmetic progression, yet every $U\subseteq S$ of size $|U|\ge\delta|S|$ contains a $k$-term arithmetic progression. Our proof uses the hypergraph container method, which has proven to be a very powerful tool in extremal combinatorics. The idea behind the container method is to have a small certificate set to describe a large independent set. We give two further applications in the appendix using this idea. read more

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Combinatorics Number Theory