Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood/Corollary

Corollary
Let $F$ be a topological field.

Let $X$ be a topological vector space over $F$. Let $V$ be an open neighborhood of ${\mathbf 0}_X$.

Then there exists an open neighborhood $W$ of ${\mathbf 0}_X$ such that:


 * $W^- \subseteq V$

where $W^-$ denotes the closure of $W$.

Proof
From Finite Topological Space is Compact, we have:


 * $\set { {\mathbf 0}_X}$ is compact.

Since $V$ is open, $X \setminus V$ is closed with:


 * $\set { {\mathbf 0}_X} \cap \paren {X \setminus V} = \O$

So from Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood, there exists an open neighborhood $W$ of ${\mathbf 0}_X$ such that:


 * $\paren {\set { {\mathbf 0}_X} + W} \cap \paren {\paren {X \setminus V} + W} = \O$

That is:


 * $W \cap \paren {\paren {X \setminus V} + W} = \O$

Since $W$ is open:


 * $\paren {X \setminus V} + W$ is open

from Sum of Set and Open Set in Topological Vector Space is Open.

From Open Set Disjoint from Set is Disjoint from Closure:


 * $W^- \cap \paren {\paren {X \setminus V} + W} = \O$

Since ${\mathbf 0}_X \in W$:


 * $X \setminus V \subseteq \paren {X \setminus V} + W$

From Set Intersection Preserves Subsets:


 * $W^- \cap \paren {X \setminus V} = \O$

From Empty Intersection iff Subset of Complement:


 * $W^- \subseteq V$