Condition for Point being in Closure/Topological Vector Space
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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 $x \in A^-$ if and only if:
- 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 $x \in A^-$ if and only if 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 $x \in A^-$ if and only if:
- for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$.
$\blacksquare$