There are many answers to show examples that a sigma algebra is not necessarily a topology. In answer, it is shown that on every uncountable set there is a $\sigma$-algebra that isn't a topology. In detail:
Let $S$ be any uncountable set, and let $\mathcal{A}$ be the
collection of all subsets of $S$ which are either countable or have
countable complement.This collection is evidently closed under complementation. If I have
a countable union of elements of $\mathcal{A}$, all of which are
countable, then the union is countable. Otherwise, at least one
element is cocountable, hence so is the union. A similar argument
works for intersections. So $\mathcal{A}$ is a $\sigma$-algebra.
I want to confirm that the basic reason behind this answer. I thought it is because:
$\sigma$-algebra is closed under finite and infinite countable unions; while a topology is closed under finite, infinite countable unions, and infinite uncountable unions. And it is the need of infinite uncountable unions makes some topologies that are not $\sigma$-algebras. Am I right?
Also, I want to confirm when we say "topologies are also closed under arbitrary unions", the arbitrary here means all possible unions: finite unions, infinite countable unions and infinite uncountable unions.
Best Answer
Yes, all topologies are closed under arbitrary unions, not just finite and countable ones, and that's the basic reason behind the existence of topologies which are not $\sigma$-algebras.
And, yes, arbitray unions means all possible unions. I don't understand why is it that you classify these unions into several types. The term “arbitrary” simply means “without restrictions”.