how can we prove the following statement $\overline{A}=\cap_{v\in V} (A+v)$

Question

let $(G,+,\tau)$ a topological group , let $V$ be a fundamental neighbourhoods system for the neutral element $(e)$ , let $A$ be a subset of $G$

how can we prove that $$\overline{A}=\cap_{v\in V} (A+v)$$

Answer 1

If $x\in\bar A$ and if $v\in V$, then $-v$ is also a neighborhood of $e$ and therefore $x+(-v)$ is a neighborhood of $x$. Since $x\in\bar A$, there is some $a\in A$ such that $a\in x+(-v)$. But this is the same thing as saying that $x\in a+v$. Since this is true for all $v\in V$, $x\in\bigcap_{v\in V}(A+v)$.

On the other hand, if $x\in\bigcap_{v\in V}(A+v)$ you want to prove that $x\in\bar A$. Let $W$ be a neighborhood of $x$; you want to prove that $W\cap A\neq\emptyset$. Note that $W\supset x+(-v)$, for some $v\in V$. You know that $x\in A+v$; in other words, $x\in a+v$, for some $a\in A$. But then $a\in x+(-v)\subset W$.