Set is Open iff Neighborhood of all its Points

Theorem
Let $$\left({X, \vartheta}\right)$$ be a topological space.

Let $$V \subseteq X$$ be a subset of $$X$$ such that:
 * For all $$z \in V$$: $$V$$ is a neighborhood of all the points in $$V$$.

Then $$V$$ is an open set of $$X$$.

Proof
It can be seen that $$V$$ is the union of all open sets contained in $$V$$.

Hence the result, by the axioms of a topological space.