We need to recall the following:
Let $X$ be a topological space , $A\subseteq X$ and $x\in X.$ Then $x\in \overline{A}$ iff for any nbd $U$ of $x$, $U\cap A \ne \varnothing$. Also, if $X$ is a TVS with topology $\tau$, then it is known that $a+G$ and $tG$ belong to $\tau$ whenever $a\in X$, $G\in \tau$ and $t\ne 0$. We call this, the invariant property. We now prove that $ \overline{A} = \bigcap_{V\in \mathcal{V}}(A+V). $ where $\mathcal{V}$ denotes the collection of all nbs of 0.
First, let $x\in \overline{A}$ and let $V\in \mathcal{V}$. Then $-V$ is nbd of 0. Hence, $x+(-V)=x-V$ is nbd of $x$. Thus, $(x-V)\cap A \ne \varnothing$. Choose $y\in x-V$ and $y\in A$. Then it follows that $x\in A+V$. Hence, $x\in A+V$ for every $V\in \mathcal{V}$. This implies that $x\in \displaystyle{\bigcap_{V\in \mathcal{V}}(A+V)}$.
Next, let $x\in \displaystyle{\bigcap_{V\in \mathcal{V}}(A+V)}$. We want to show that $x\in \overline{A}$. Let $U$ be any nbd of $x$. Then $-x+U$ is nbd of 0 and so is $-(x+U)=x-U$. Hence, $x-U\in \mathcal{V}$. Thus, $x\in A+(x-U)$. Hence, there exist $a\in A$ and $u\in U$ such that $x=a+(x-u)$. This implies that $a=u$. Hence, $U\cap A \ne \varnothing$. Thus, $x\in \overline{A}$. This completes the proof.
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