# If $A+A+A$ contains the extremes, does it contain the middle?

Let $$b \ge 1$$ and $$A\subseteq [0,b]$$ be a set of integers (all intervals will be of integers).

Write $$hA := \underbrace{A + \ldots + A}_{h\text{ summands}} = \{ \sum_{i=1}^h a_i ~|~a_i \in A,\, \forall 1\leq i\leq h \}$$.

Suppose that $$[0,b]\cup [2b,3b]\subseteq 3A$$. This implies, in particular, that $$0,1,b-1,b$$ belong to $$A$$, to ensure that $$0 = 0+0+0, 1 = 0+0+1, 3b-1 = b+b + (b-1)$$, and $$3b = b+b+b$$ belong to $$3A$$. My question is:

If $$[0,b]\cup [2b,3b]\subseteq 3A$$, does that imply that $$3A = [0,3b]$$?

At first I thought it could be false (e.g., a thin set with $$|A| \asymp b^{1/3+o(1)}$$ elements concentrated near $$0$$ and $$b$$, with middle mostly empty), but I wasn't able to formalize a counterexample. The positive answer is motivated by checking a few small cases by computer (with $$b$$ up to around $$15$$) and the following little heuristic, which generalizes nicely to the question:

Does having $$[0,b]\cup [(h-1)b,hb]\subseteq hA$$ imply that $$hA = [0,hb]$$?

for $$h\geq 3$$.

Heuristic: If we consider, for example, the set $$A_{\alpha}:=[0,\,\alpha b]\cup[(1-\alpha)b,\,b] \subseteq [0,b]$$ for some $$0<\alpha\leq \frac{1}{2}$$, we check by induction that $$hA = \bigcup_{k=0}^{h} [(k-k\alpha)b,\, (k + (h-k)\alpha) b].$$ Thus, to have $$[0,b]\cup[(h-1)b,hb]\subseteq hA$$ we must take $$\alpha \geq 1/(h+1)$$. However, the distance between consecutive intervals in $$hA$$ is $$((k+1)-(k+1)\alpha)b - (k + (h-k)\alpha) b = (1 - (h+1)\alpha)b,$$ so if $$\alpha \geq 1/(h+1)$$ then $$hA = [0,hb]$$.

• From context it seems that by $3A$ you mean $A+A+A=\{a+a'+a''\mid a, a', a'' \in A\}$, and not $\{3a\mid a\in A\}$? Aug 6 at 2:24
• @Aaron yes, I will clarify that in an edit. Aug 6 at 2:25
• @mathworker21 oops, overlooked redundancy! Thanks. Aug 6 at 3:12
• Your question is very similar to this one. The answer is negative: letting $A=\{0,1,2,3,8,11,26,38,56,69,85,89,171,175,191,204,222,234,249,252,257,258,259,260\}$, the set $3A$ has a unique gap which resides in the interval $[260,520]$. This example was found by @Jukka Kohonen, see comments to his own answer.
– Seva
Aug 6 at 8:10
• @Seva Wow, now I know why I couldn't prove that the answer is affirmative. Aug 6 at 8:35