Whole Space is Open in Neighborhood Space

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

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

Proof
Let $x \in S$.

Then by neighborhood space axiom $N 1$ there exists a neighborhood $N$ of $x$.

As $N \subseteq S$ it follows from neighborhood space axiom $N 3$ that $S$ is a neighborhood of $x$.

As this holds for all $x \in S$ it follows that $S$ is an open set of $\left({S, \mathcal N}\right)$.