Does $A_{n+1} + A_{n+1} \subset A_n$ for $n = 0,1,2,\dots$ imply $A_1 + A_2 + \cdots \subset A_0$?

Question

Let $\{A_n\}$ be a local base of a topological vector space satisfying $A_{n+1} + A_{n+1} \subset A_n$ for $n = 0,1,2,\dots$ where the sum is defined as usual: $A + B = \{a + b\mid a\in A,\,b\in B\}$. Then, we can easily check that $A_1 + A_2 + \cdots + A_n \subset A_0$ for every finite $n$. I am wondering if it holds when $n$ goes to $\infty$ but cannot figure out how to prove or disprove it. Would you give me any hint?

Answer 1

This is more of a long comment on infinite sums of sets.

So according to the discussion in the comments, "$A_1+A_2+\cdots$" is just being used as a short-hand for "$\bigcup_n \sum_{i=1}^n A_i$", the set of finite sums of elements of the $A_i$, in which case the result is clear. I'll denote this set as $^{fin}\sum_{i=1}^\infty A_i$.

But what if we take "$A_1+A_2+\cdots$" to mean the set of converging (limits of ) series $\sum_i a_i$ with $a_i\in A_i$? I'll denote this set as $\sum_{i=1}^\infty A_i$. Note that $$^{fin}\sum_{i=1}^\infty A_i\subseteq\sum_{i=1}^\infty A_i\subseteq\overline{\left(^{fin}\sum_{i=1}^\infty A_i\right)}$$ In general these sets are different. But is the largest one still contained in $A_0$? Is the one in the middle contained in $A_0$?

The answers are NO (or at least "not necessarily") and YES, respectively. Indeed, let us first prove a small lemma.

Lemma: If $A$ and $B$ are open nhoods of $0$ in TVS $V$ and $A+A\subseteq B$, then $A+\overline{A}\subseteq B$.

$\triangleright$ Indeed, suppose $x\in A$ and $\overline{y}\in\overline{A}$. Since $A$ is open, there is another nhood $U$ of $0$ such that $x+U\subseteq A$. Since $\overline{y}\in\overline{A}$, there exists $y\in A$ with $\overline{y}-y\in U$. Then $$x+\overline{y}=\underbrace{x+\underbrace{(\overline{y}-y)}_{\in U}}_{\in A}+\underbrace{y}_{\in A}$$ belongs to $A+A\subseteq B$. $\triangleleft$

Cool. So we know $$^{fin}\sum_{i=2}^\infty A_i\subseteq A_1.$$ Taking closures, $$\sum_{i=2}^\infty A_i\subseteq\overline{\left(^{fin}\sum_{i=2}^\infty A_i\right)}\subseteq\overline{A_1}$$ and adding with $A_1$ and using the lemma $$\sum_{i=1}^\infty A_1=A_1+\sum_{i=2}^\infty A_i\subseteq A_1+\overline{A_1}\subseteq A_0$$ (the first equality needs to be verified, but it should be easy).

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However in general we might have $\overline{\left(^{fin}\sum_{i=1}^\infty A_i\right)}\not\subseteq A_0$. For example, take $A_n=(-2^{-n},2^{-n})$ in $\mathbb{R}$. Then $\overline{\left(^{fin}\sum_{i=1}^\infty A_i\right)}=[-1,-1]$ but $A_0=(-1,1)$.