Given $\mathscr{A}\subseteq\mathbb{N}$ an infinite set, consider its $h$-fold sumsets


$$h\mathscr{A}:=\left\{\sum\_{i=1}^{h} k\_i : k\_1,\ldots, k\_h\in \mathscr{A} \right\},$$


and let "$\simeq$" be the equivalence relation (in $\mathcal{P}(\mathbb{N})$) given by $\mathscr{A}\simeq \mathscr{B}$ if and only if $\mathscr{A}\triangle \mathscr{B} := (\mathscr{A}\setminus \mathscr{B})\cup (\mathscr{B}\setminus \mathscr{A})$ is finite. Being $\mathrm{d}$ the usual asymptotic density,



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> **Q1.** If $h\geqslant 2$, does $\mathrm{d}(h\mathscr{A}) = 1$ imply $h\mathscr{A} \simeq \mathbb{N}$?
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This is clearly false for $h=1$, for $\mathscr{A}:= \mathbb{N}\setminus \{n^2: n\in\mathbb{N}\}$ is an easy counterexample. For general $h$ I also think this is false (for if not, the case $h=2$ could be easily used to prove Goldbach's conjecture for suff. large numbers from Vinogradov's results), but I wasn't able to come up with counterexamples. Thus my real question is



> 
> How to construct counterexamples to **Q1**? Furthermore, can we at least guarantee from $\mathrm{d}(h\mathscr{A}) = 1$ that $(h+1)\mathscr{A} \simeq \mathbb{N}$, for all sequences $\mathscr{A}\subseteq \mathbb{N}$?
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For the second question, here's a [related question](https://math.stackexchange.com/questions/1947486/if-mathrmd-mathcala-1-and-mathcalb-is-a-basis-does-mathcala) of mine in M.SE.



Sets with $\mathrm{d}(h\mathscr{A}) = 1$ but s.t. $h\mathscr{A}$ doesn't contain all large integers
Let $(n\_k)$ be a rapidly increasing sequence. Remove from the positive integers all segments $[n\_k,2n\_k-1]$ and call the remaining set $A$. Then $2n\_k$ does not lie in $A+A$, so $A+A$ is not equivalent to $\mathbb{N}$. But $A+A$ has density 1. Indeed, $A$ contains all the numbers from $2n\_k$ to $n\_{k+1}-1$, thus $A+A$ contains all the number from $4n\_k$ to $2n\_{k+1}-2$. Also $A+A$ contains $B=2n\_k+(A\cap [1,n\_k])$ and $C=3n\_k+(A\cap [1,n\_k])$ which are disjoint sets of cardinalities $n\_k+o(n\_k)$ each. This implies that if $x\in [2n\_k,2n\_{k+1}]$ and $k$ is large, then $|(A+A)\cap [1,x]|=x+o(n\_k)=x+o(x)$.


For arbitrary $h$ remove the segments $[n\_k,hn\_k-1]$.


