Expression for Closure of Set in Topological Vector Space/Corollary

Theorem
Let $K$ be a topological field.

Let $\struct {X, \tau}$ be a topological vector space over $K$.

Let $A \subseteq X$. Let $\BB$ be a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$.

Then:
 * $\ds A^- = \bigcap_{V \in \BB} \paren {A + V}$

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

Proof
Let $\VV$ be the set of open neighborhoods of ${\mathbf 0}_X$ in $\struct {X, \tau}$.

From Intersection is Decreasing, we have:
 * $\ds \bigcap_{V \in \mathcal V} \paren {A + V} \subseteq \bigcap_{V \in \BB} \paren {A + V}$

Conversely, suppose that:
 * $\ds x \in \bigcap_{V \in \BB} \paren {A + V}$

Let $O \in \VV$.

Since $\BB$ is a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$, there exists $U \in \BB$ such that $U \subseteq O$.

Since $x \in A + V$ for each $V \in \BB$, we in particular have $x \in A + U$.

Since $U \subseteq O$, we have that $x \in A + O$.

Since $O \in \VV$ was arbitrary, we have:
 * $\ds x \in \bigcap_{V \in \BB} \paren {A + V}$