Theorem 1.13 in Rudin's book

Theorem 1.13. Let $X$ be a topological vector space. If $A\subset X$ then $\bar{A}=\bigcap (A+V)$, where $V$ runs through all neighborhoods of $0$.

Proof. $x\in \bar{A}$ if and only if $(x+V)\cap A\neq\varnothing$ for every neighborhood $V$ of $0$, and this happens if and only if $x\in A-V$ for every such $V$. Since $-V$ is a neighborhood of $0$ if and only if $V$ is one, the proof is complete.

I don't understand why $x\in A-V$? And how to conclude $\bar{A}=\bigcap (A+V)$? I don't really understand the idea of proofs in general. Can someone help me? Thanks.

Solution 1:

$x \in \overline A$ if and only if every neighborhood of $x$ intersects $A$. The proof simply transaltes this into and equivalent condition using basic facts about topological vector spaces.

Let $V$ be neighborhood of $0$ and let $x \in \overline A$. Then $-V$ is a neighborhood of $0$ and $(x+(-V)) \cap A \neq \emptyset$ (because $x+(-V)$ is a neighborhood of $x$). If $y \in (x+(-V)) \cap A$ then there exists $w \in -V$ such that $y=x+w$. Hence, $x=y-w \in A+V$. We have proved that $x \in \cap (A+V)$ where the intersection is over all neighborhoods $V$ of $0$.

Conversely, if $x$ belongs to this intersection and $V$ is any neighborhood of $x$ then $x-V$ is a neighborhood of $0$ so $x \in A+(x-V)$. So $x=a+x-v$ for some $a \in A$ and $v \in V$ which gives $a=v$. But this shows that $a \in V \cap A$. Since $A$ intersects every neighborhood of $x$ it follows that $x \in \overline A$.