Union of Open Sets of Metric Space is Open

Theorem
Let $$M = \left({A, d}\right)$$ be a metric space.

The union of any collection of open subsets of $$M$$ is open in $M$.

Proof
Let $$I$$ be any indexing set.

Let $$U_i$$ be open in $M$ for all $$i \in I$$.

Let $$x \in \bigcup_{I} U_i$$.

Then $$x \in U_k$$ for some $$k \in I$$.

Since $$U_k$$ is open in $M$, $$\exists \epsilon > 0: N_{\epsilon} \left({x}\right) \subseteq U_k \subseteq \bigcup_{I} U_i$$.

The result follows.