A sequence of subsets of an infinite group Is there an infinite group $G$ such that there is not any sequence $(A_n)$  of its subsets such that always
$$A_n=A_n^{-1}, \quad A_{n+1}A_{n+1}\subsetneqq A_n$$
?
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 A: Here an extended comment. Call Property (P) the assumption of non-existence of such a sequence, so that the question is whether there exists an infinite group with (P)
My comment is the observation that if such a group exists, then it can be chosen to be finitely generated with all its proper subgroups finite ("quasi-finite").
Indeed, a group with Property (P) satisfies the following 2 properties:


*

*it admits no nondiscrete Hausdorff topology "non-topologizable"

*it's artinian (or min): it admits no properly descending chain of subgroups


Moreover Property (P) passes to subgroups.
Now I claim that if $G$ satisfies (P), then it contains an infinite quasi-finite subgroup: indeed if $G=G_0$ is not quasi-finite, then it admits a proper infinite subgroup $G_1$, and by induction we define a properly descending chain of subgroups, which has to stop, i.e. we eventually get a quasi-finite subgroup (hence with (P)).
Now a quasi-finite group is either finitely generated, or is an abelian Prüfer group $\mathbf{Z}[1/p]/\mathbf{Z}$ (Hall-Kulatilaka), but the latter is topologizable as we see by embedding it densely in the Lie group $\mathbf{R}/\mathbf{Z}$. Hence the resulting quasi-finite groups with (P) have to be finitely generated. 
To conclude, the question boils down to whether there exists an infinite, quasi-finite, finitely generated group with (P).
A: This is a partial answer. There exists an infinite group $G$ having no sequence of subsets $\{A_i\}$ such that 
$$
A_{n+1}A_{n+1}\subsetneq A_n,
\quad\hbox{and}\quad
\bigcap A_i=\{1\}.
$$
(We omit the inverse condition but add the the intersection-triviality condition.)
Indeed, such a sequence defines a Hausdorff topology on the group where $A_i$ are neigbourhoods of $1$ and all other points are isolated. The multiplication is continuous at $(1,1)\in G\times G$ with respect to this topology. It remains to note that infinite locally non-topologizable groups do exist. 
