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.