Empty Set is Open in Neighborhood Space

Theorem
Let $\struct {S, \NN}$ be a neighborhood space.

Then the empty set $\O$ is an open set of $\struct {S, \NN}$.

Proof
Suppose $\O$ were not an open set of $\struct {S, \NN}$.

Then $\exists x \in \O$ such that $\O$ is not a neighborhood of $\O$.

By definition of empty set, such an $x$ does not exist.

Hence the result.