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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 $1$ and $b-1$ belong to $A$, to ensure that $1 = 0+0+1$ and $3b-1 = b+b + (b-1)$ 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 $hA = [0,hb]$.

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 $hA = [0,hb]$.

Let $A\subseteq \mathbb{Z}_{\geq 0}$ be a finite set of non-negative integers with $0\in A$. Write $b := \max\{a\in A\}$, so that $A \subseteq [0,b]$ (here our intervals are always 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 $[0,b]\cup [2b,3b]\subseteq 3A = A+A+A$. This implies, in particular, that $1$ and $b-1$ belong to $A$, to ensure that $1 = 0+0+1$ and $3b-1 = b+b + (b-1)$ 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 $hA = [0,hb]$.

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 $1$ and $b-1$ belong to $A$, to ensure that $1 = 0+0+1$ and $3b-1 = b+b + (b-1)$ 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 $hA = [0,hb]$.

Let $A\subseteq \mathbb{Z}_{\geq 0}$ be a finite set of non-negative integers, with $0\in A$ and $\gcd(A) = 1$. Write $b := \max\{a\in A\}$, so that $A \subseteq [0,b]$ (here our intervals are always 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 $[0,b]\cup [2b,3b]\subseteq 3A = A+A+A$. This implies, in particular, that $1$ and $b-1$ belong to $A$, to ensure that $1 = 0+0+1$ and $3b-1 = b+b + (b-1)$ 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 $hA = [0,hm]$?

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 $hA = [0,hb]$.

Let $A\subseteq \mathbb{Z}_{\geq 0}$ be a finite set of non-negative integers with $0\in A$. Write $b := \max\{a\in A\}$, so that $A \subseteq [0,b]$ (here our intervals are always 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 $[0,b]\cup [2b,3b]\subseteq 3A = A+A+A$. This implies, in particular, that $1$ and $b-1$ belong to $A$, to ensure that $1 = 0+0+1$ and $3b-1 = b+b + (b-1)$ 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 $hA = [0,hb]$.