# general topology – Union of open sets is open

The proof is generally pretty straightforward, but my only difficulty is with a small subtlety. The statement is:

Let $$(X,d)$$ be a matrix space and $${U_i}_{i in I}$$ a collection of open sets. Prove that $$bigcuplimits_{i in I} U_i$$ is open.

My attempt is:

If $$I = emptyset$$, then $$bigcuplimits_{i in I} U_i = emptyset$$, which is open, so we proceed under the assumption that $$I neq emptyset$$. Furthermore, if $$U_i = emptyset$$ for all $$i in I$$, then $$bigcuplimits_{i in I} U_i = emptyset$$, which is open, so assume $$U_i neq emptyset$$ for at least one $$i in I$$. So, let $$x in bigcuplimits_{i in I} U_i$$. Then $$x in U_{i_0}$$ for some $$i_0 in I$$. As $$U_{i_0}$$ is open, there exists $$epsilon > 0$$ such that $$B_{epsilon} (x) subset U_{i_0} subset bigcuplimits_{i in I} U_i$$, so $$x$$ an interior point of $$bigcuplimits_{i in I} U_i$$. Therefore, $$bigcuplimits_{i in I} U_i$$ is open.

Every proof of this fact I have every seen completely omits the first two cases I wrote down, where $$I = emptyset$$ or $$U_i = emptyset$$ for all $$i in I$$. Are these even necessary to consider? Are they just too trivial to comment on?