Schnirelmann density

Remarks

Always $0 \le \sigma(A) \le 1$, and $A(n) \ge \sigma(A)\, n$ for every $n \ge 1$. The element $0$, if present, does not count. One has $\sigma(A) = 1$ if and only if $A$ contains every positive integer.

Example: for the set $A$ of odd positive integers, $A(n)/n = 1/2$ for even $n$ and $A(n)/n = (n+1)/(2n) > 1/2$ for odd $n$; so $\sigma(A) = 1/2$. Contrast: for the even positive integers, $A(1)/1 = 0$, so $\sigma(A) = 0$, although their natural density is $1/2$.

Schnirelmann's inequality: if $0 \in A$ and $0 \in B$, then $\sigma(A + B) \ge \sigma(A) + \sigma(B) - \sigma(A)\sigma(B)$, where $A + B = \set{a + b : a \in A, b \in B}$.