If $A_{\alpha}$ are open, what about finite unions? Countable unions? All unions?
My answer:
Any union of open sets is open. If $x\in\cup_{\alpha}A_{\alpha}$, then by definition of union, $x\in A_{\alpha}$ for some particular $\alpha$. Since $A_{\alpha}$ is open, then there exists an open set $V$ such that $a\in V\subset A_{\alpha}$. Then (again by definition of union) $V\subset\cup_{\alpha}A_{\alpha}$. The same argument holds for infinitely many open sets.
I am not sure if I am right but any suggestions would be greatly appreciated.
Best Answer
Yes, the union of any collection of open sets is open. This is one of the axioms of topology.
Your proof is essentially correct, but you can improve it by noting that since $A_\alpha$ is open you can just take $V=A_\alpha$ as your open set.