Expression for Closure of Set in Topological Vector Space

Theorem
Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$.

Let $A \subseteq X$.

Let $\mathcal V$ be the set of open neighborhoods of ${\mathbf 0}_X$.

Then:


 * $\ds A^- = \bigcap_{V \in \mathcal V} \paren {A + V}$

where $A^-$ is the closure of $A$.

Proof
From Condition for Point being in Closure: Topological Vector Space, we have $x \in A^-$ :


 * for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$.

We have that $\paren {x + V} \cap A \ne \O$ there exists some $u \in X$ with $u \in A$ and $u \in x + V$.

This is equivalent to, $u \in A$ and there exists $v \in V$ such that $u = x + v$.

This is equivalent to, $x = u - v$ for $u \in A$ and $v \in V$.

That is, equivalent to $x \in A - V$.

So, we have that $x \in A^-$ $x \in A - V$ for all open neighborhoods $V$ of ${\mathbf 0}_X$.

So, we have:


 * $\ds A^- = \bigcap_{V \in \mathcal V} \paren {A - V}$

From Dilation of Open Set in Topological Vector Space is Open, we have that $-V$ is open $V$ is open.

Writing $A - V = A + \paren {-V}$, we therefore see that:


 * $\ds A^- = \bigcap_{V \in \mathcal V} \paren {A + V}$