Empty Set is Open in Neighborhood Space

Theorem
Let $\left({S, \mathcal N}\right)$ be a neighborhood space.

Then the empty set $\varnothing$ is an open set of $\left({S, \mathcal N}\right)$.

Proof
Suppose $\varnothing$ were not an open set of $\left({S, \mathcal N}\right)$.

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

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

Hence the result.