Condition for Point being in Closure/Topological Vector Space

Theorem
Let $K$ be a topological field.

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

Let $A \subseteq X$.

Let $A^-$ denote the closure of $A$ in $X$.

Let $x \in X$.

Then $x \in A^-$ :


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

Proof
From the definition of the closure of $A$, $A^-$ is the set of adherent points of $A$.

So $x \in A^-$ for each open neighborhood $U$ of $x$ we have:


 * $U \cap A \ne \O$

From Classification of Open Neighborhoods in Topological Vector Space, every such $U$ has the form $U = x + V$ for some open neighborhood $V$ of ${\mathbf 0}_X$, and conversely every set of this form is an open neighborhood of $x$.

So we have $x \in A^-$ :


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