Arbitrary unions of open sets 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.
 A: 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.
A: Your argument looks fine for showing that an arbitrary union of open sets is open. Now that you have established that, you know the answer is "yes" to your following questions. If an arbitrary union of open sets is open, that simply means that any union of open sets is open. The arbitrariness means that it doesn't matter at all how many sets are in the union. $3$ sets, $10^{10^{10^{10^{\dots}}}}$ sets, a union indexed by the cardinality of $\Bbb{R}^\Bbb{R}$, etc. As long as each set in the union is open, the resulting set will be open. The size of the union simply doesn't matter, as you've shown.
A: This depends on your definition of open.  In a general topological space the fact that the union of open sets is open is an assumption of the definition.  So you wouldn't need to make your argument at all.  On the other hand, if you are in $\mathbb R^n$ then the definition of open probably involves a distance metric.  In that case you have to use the distance metric in your argument you can't just use an arbitrary $V$.
