Definition 5: Additive Basis
A subset A \subseteq \mathbb{N} with the property that A + A + \cdots + A = \mathbb{N} for a finite sum.
Theorem 1.1.2 (L. G. Shnirelman) Every sequence with a positive density is a basis for the natural numbers.
Proof:
(1) Let A^{(k)} = \underbrace{A + A + \dots + A}_{k\,\text{times}}
(2) Using the Corollary from Theorem 1.1.1, we have:
d(A^{(k)}) \ge 1 - (1 - d(A))^k
(3) For sufficiently large k, we have:
d(A^{(k)})>\frac{1}{2}
since 0 \le\frac{A(n)}{n}\le 1 and d(A)=\inf\limits_{n}\frac{A(N)}{n}.
(4) For this k, the number of terms of A^{(k)} in the interval [1,n] will be greater than \frac{n}{2}.
(5) If a_i ranges over the terms of A^{(k)} which do not exceed n, then there exists i_1,i_2 such that: a_{i_1} = n - a_{i_2}.
Note: If i_1,i_2 don't exist, then there are more than 2*(\frac{n}{2}+1)=n+2 values in [0,n] which is impossible.
(6) So, there exists k such that for any n, there exists a_{i_1}, a_{i_2} \in A^{(k)} such that n = a_{i_1} + a_{i_2}.
QED
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