Let $(H_n)$ be a sequence of nonempty subsets of $\Bbb Z$ such that always $\{a-b\mid a,b\in H_{n+1}\}\subsetneqq H_n$.
Can we deduce that there is some $n$ such that $\{a-b\mid a,b\in H_{n}\} = H_n$?
Note: $\subsetneqq$ means proper subset.
Let $(H_n)$ be a sequence of nonempty subsets of $\Bbb Z$ such that always $\{a-b\mid a,b\in H_{n+1}\}\subsetneqq H_n$.
Can we deduce that there is some $n$ such that $\{a-b\mid a,b\in H_{n}\} = H_n$?
Note: $\subsetneqq$ means proper subset.
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I use the answer given by Rory Daulton to your previous question. Take for $n\geq 1$ $H_n=\{2^{n-1}\}\cup 2^n \mathbb{Z}$. If $a,b\in H_{n+1}$, then $a,b\in 2^n\mathbb{Z}$, hence $a-b\in 2^n\mathbb{Z}\subset H_n$. We also have by the above that $2^{n-1}$ is not in $\{a-b, a,b\in H_{n+1}\}$. Hence your hypothesis are satisfied. But for any $n\geq 1$, $2^{n-1}-2^n$ for example, is in $\{a-b, a,b\in H_{n}\}$, but not in $H_n$.