Union of Open Sets of Neighborhood Space is Open

Theorem
Let $S$ be a neighborhood space.

Let $I$ be an indexing set.

Let $\left \langle{U_\alpha}\right \rangle_{\alpha \mathop \in I}$ be a family of open sets of $\left({S, \mathcal N}\right)$ indexed by $I$.

Then their union $\displaystyle \bigcup_{\alpha \mathop \in I} U_i$ is an open set of $\left({S, \mathcal N}\right)$.

Proof
Let $\displaystyle x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.

Then $x \in U_\beta$ for some $\beta \in I$.

By definition of open set, $U_\beta$ is a neighborhood of $x$.

But from Set is Subset of Union:
 * $\displaystyle U_\beta \subseteq x \in \bigcup_{\alpha \mathop \in I} U_\alpha$

By neighborhood space axiom $N 3$ it follows that $\displaystyle \bigcup_{\alpha \mathop \in I} U_\alpha$ is a neighborhood of $x$.

As $x$ is arbitrary, it follows that the above is true for all $\displaystyle x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.

It follows by definition that $\displaystyle \bigcup_{\alpha \mathop \in I} U_\alpha$ is an open set of $\left({S, \mathcal N}\right)$.