Union of Open Sets of Neighborhood Space is Open

Theorem

Let $I$ be an indexing set.

Let $\family {U_\alpha}_{\alpha \mathop \in I}$ be a family of open sets of $\struct {S, \NN}$ indexed by $I$.

Then their union $\ds \bigcup_{\alpha \mathop \in I} U_i$ is an open set of $\struct {S, \NN}$.

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

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

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

$\ds U_\beta \subseteq x \in \bigcup_{\alpha \mathop \in I} U_\alpha$

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

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

It follows by definition that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is an open set of $\struct {S, \NN}$.

$\blacksquare$

1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$