Intersection of two Open Sets of Neighborhood Space is Open

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

Let $U$ and $V$ be open sets of $\left({S, \mathcal N}\right)$.

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

Proof
Let $U$ and $V$ be open sets of $\left({S, \mathcal N}\right)$.

Let $x \in S$ such that $x \in U \cap V$.

Then $x \in U$ and $x \in V$, both of which are neighborhoods of $x$ by definition of open set.

By neighborhood space axiom $N 4$ it follows that $U \cap V$ is a neighborhood of $x$.

As $x$ is arbitrary, it follows that the above is true for all $x \in U \cap V$.

It follows by definition that $U \cap V$ is an open set of $\left({S, \mathcal N}\right)$.