2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
  • $\begingroup$ Related: Can measurable spaces be generalized via Topologic spaces? $\endgroup$ Feb 1 at 3:51

2 Answers 2

5
$\begingroup$

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”.

$\endgroup$
1
  • 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
1
$\begingroup$

I know this is an old question, but I'll add a bit of color in case anyone comes across it. As others have pointed out, the difference is that a $\sigma$-algebra is closed under complements and countable unions (which implies countable intersections), while a topology is closed under arbitrary unions and finite intersections.

Note that there is no rule saying that a particular topology can't be closed under countable intersections and complements or that a particular $\sigma$-algebra can't be closed under arbitrary unions. They're just not required to be.

There is no obstruction to a $\sigma$-algebra containing a topology (i.e. $T$ is a subset of it). In fact, for a given $T$, the intersection of all such $\sigma$-algebras is the Borel algebra of $T$.

Nor is there an obstruction to a $\sigma$-algebra being contained in a topology. In that case, every set in the $\sigma$-algebra is clopen in $T$. Intuitively, we can think of each such set as a connected component in $T$ (however, there also could be other clopen sets which aren't in the $\sigma$-algebra).

In order for a topology $T$ on $\Omega$ also to be a $\sigma$-algebra on $\Omega$ we require that all open sets in $T$ are clopen. This tells us that it is closed under complements and thus under arbitrary (not just countable) intersections. Such a topology is called a partition topology and has as basis a partition of $\Omega$.

In order for a $\sigma$-algebra on $\Omega$ also to be a topology on $\Omega$, we require that it be closed under arbitrary unions, not just countable ones. It then is a partition topology as described.

$\endgroup$
1
  • 1
    $\begingroup$ Ah, you are absolutely right. I was being too glib. The correspondence is between $T$ and $T'$, not a homeomorphism between $\Omega$ and $\Omega'$. My bad. Thanks. I'll correct it. $\endgroup$
    – Kensmosis
    Feb 1 at 2:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .