Is there an example of a sigma algebra that is not a topology?

Question

Is there an example of a sigma algebra that is not a topology? If this is not the case, is it possible to prove that all sigma algebras are topologies?

Answer 1

If a $\sigma$-algebra contains all one-point sets then the topology it generates is the discrete topology, so the only thing we need to ensure is that we have a $\sigma$-algebra in which not every set is measurable.

The very first example that comes to mind is already an example. As there are non-Borel sets in $[0,1]$ the Borel sets are not the discrete topology, but the smallest topology that contains the Borel sets is the discrete topology by the above paragraph.

(I'm ignoring choice issues here, as I lack the expertise).

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Added:

A similar argument works for the Lebesgue $\sigma$-algebra.

In a positive direction: Every $\sigma$-algebra on a countable set is a topology.

To sum up:

• On finite or countable sets every $\sigma$-algebra is a topology.
• In view of Pete's answer below, on every uncountable set there is a $\sigma$-algebra that isn't a topology, namely the countable-cocountable $\sigma$-algebra. This example is probably the optimal one in terms of simplicity.

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For the sake of completeness:

• Pete L. Clark's answer in this thread uses choice in the very weak form "the countable union of countably many countable sets is countable", as noted by Nate Eldredge.

• Andrés Caicedo argues beautifully in this MO-thread why some choice is necessary to ensure that there exist non-Borel sets.

• If I interpret François G. Dorais's answer here correctly, there are models of ZF with the property that the $\sigma$-algebra generated by the singleton subsets of a set is equal to the power set. (Please correct me if this is too naïve a rendering)

• Asaf Karagila adds in a comment below: "[...] an even freakier statement is given in Jech's The Axiom of Choice, Ch. 5, Exercise 14. There is an extension of every transitive model, with the same $\aleph$-cardinals, and for every $\alpha$ there is a set $X$ which is a countable union of countable sets, and $P(X)$ can be partitioned into $\aleph_{\alpha}$ nonempty sets."

• Jay adds: As John B. S. Haldane might have said, "Set theory without the axiom of choice is not only queerer than we suppose, but queerer than we can suppose."

Answer 2

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.

It is not a topology because it contains all the singleton sets but not all subsets of $S$ -- in particular, it contains no set which is uncountable with uncountable complement.