TITLE: Is there a difference between allowing only countable unions/intersections, and allowing arbitrary (possibly uncountable) unions/intersections?
QUESTION [11 upvotes]: As in the title, I am asking if there is a difference between allowing set-theoretic operations over arbitrarily many sets, and restricting to only countably many sets.
For example, the standard definition of an topology on a set $X$ requires that arbitrary unions of open sets are open.  Do I lose anything significant if I restrict this to just unions of countably many (open) sets?
I cannot come up with an example where it makes a difference.

REPLY [4 votes]: *Edit:*I originally let $X$ be just a Hausdorff space, but in that case the two aren't guaranteed to be different.
Here's an example to see a where it makes a difference.  Let $X$ be an uncountable Polish space.  If you look at the smallest collection of subsets of $X$ that a) contains all the open sets of $X$ and b) is closed under complements and countable unions (hence also countable intersections), you get the Borel $\sigma$-algebra of $X$.
But if you look at the smallest collection of subsets of $X$ that a) contains all the open sets and b) is closed under complements and arbitrary unions (hence also arbitrary intersections), this is $\cal{P}$$(X)$. This is because it includes all closed sets, hence all singletons, and then we can take an arbitrary union to get any subset of $X$.