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.

  • $\begingroup$ Also topologies are not in general closed under complementation. $\endgroup$ Aug 18, 2018 at 19:28
  • $\begingroup$ Thank you for the confirmation. And only closed under finite intersections.I know this part. And just curious about why Borel $\sigma$-algebra is that wired. @LordSharktheUnknown $\endgroup$
    – X Leo
    Aug 18, 2018 at 19:33

1 Answer 1


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”.

  • 1
    $\begingroup$ Thank you for the confirmation. I am not mathematician and learn books by my own. Need some confirmations. $\endgroup$
    – X Leo
    Aug 18, 2018 at 19:30

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