In this answer on MO, the user Gene S. Kopp gives an example of a relatively "big" set $A\subset \mathbb{N}$ with relatively "small" gaps that fails to be an asymptotic finite basis. I'm having a problem in trying to understand one of the steps.
Firstly, let $A_n$ for $n\geqslant 1$ and $A$ be as in the link. Given that $|A_n| = \frac{2^{2^n}- 2\cdot2^{2^{n-1}}}{2^n} > \frac{2^{2^n}-2^{2^{n-1}}}{2^{n-1}}$, we will have, in fact, that:
$$A(x) \geqslant \sum_{k\leqslant \log_2\log_2{\lfloor x\rfloor}} |A_k| > \sum_{k\leqslant \log_2\log_2{\lfloor x\rfloor}} \frac{2^{2^k}- 2^{2^{k-1}}}{2^{k-1}} \geqslant \frac{\lfloor x\rfloor-2}{\log_2{\lfloor x\rfloor}}\gg \frac{x}{\log{x}}.$$
I'll skip show that the gaps are in fact $\ll \sqrt{x}~$ and that it contains infinitely many non-multiples of $m$ for every $m>1$ (I haven't tried to do this, but I think it's reasonable...). Now the claim is:
Claim [G.S. Kopp]. $2^{2^n}$ cannot be written as a sum of fewer than $n−\log_2{n}$ elements of $A$.
So, to prove this, we suppose the converse: take $a_1,\ldots,a_k \in A$ with $a_1\leqslant\cdots\leqslant a_k$ such that $2^{2^n} = a_1+\ldots +a_k$ and suppose $k< n-\log_2{n}$.
Now clearly $a_k\in A_n$, otherwise we would have $a_1+\ldots +a_k < k\cdot 2^{2^{n-1}} < 2^{2^n}$. From $a_k \leqslant 2^{2^n}-2^{2^{n-1}}-2^n+1$ follows $2^{2^n}-a_k \geqslant 2^{2^{n-1}}+2^n-1>2^{2^{n-1}}$, and one may deduce by a similar argument that $a_{k-1}\in A_n\cup A_{n-1}$. However, I wasn't able to show in the same way that $a_{k-2}\in A_n\cup A_{n-1}\cup A_{n-2}$, and I have no clue how to do this is general. So, my actual question is:
My actual question: How do I prove that $\displaystyle a_j \in \bigcup_{i=n-k+j}^n A_i$ ?
The remainder of the counterexample seems to follow almost directly once this part is done.