Is every sigma-algebra the Borel algebra of a topology? (original) (raw)

This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some time, I am bringing it up here on mathoverflow in the hopes that it may find an answer here.

For any topological space, one may consider the Borel sets of the space, the sigma\sigmasigma-algebra generated by the open sets of that topology. The question is whether every sigma\sigmasigma-algebra arises in this way.

Question. Is every sigma\sigmasigma-algebra the Borel algebra of a topology?

In other words, does every sigma\sigmasigma-algebra Sigma\SigmaSigma on a set XXX contain a topology tau\tautau on XXX such that Sigma\SigmaSigma is the sigma\sigmasigma algebra generated by the sets in tau\tautau?

Some candidate counterexamples were proposed on the math.SE question, but ultimately shown not to be counterexamples. For example, my answer there shows that the collection of Lebesgue measurable sets, which seemed at first as though it might be a counterexample, is nevertheless the Borel algebra of the topology consisting of sets of the form O−NO-NO−N, where OOO is open in the usual topology and NNN is measure zero. A proposed counterexample of Gerald Edgar's there, however, remains unresolved. And I'm not clear on the status of a related proposed counterexample of George Lowther's.

Meanwhile, allow me to propose here a few additional candidate counterexamples:

• Consider the collection Sigma_0\Sigma_0Sigma0 of eventually periodic subsets of omega1\omega_1omega1. A set Ssubsetomega1S\subset\omega_1Ssubsetomega1 is eventually periodic if above some countable ordinal gamma\gammagamma, there is a countable length pattern which is simply repeated up to omega1\omega_1omega_1 to form SSS. This is a sigma\sigmasigma-algebra, since it is closed under complements and countable intersections (one may find a common period among countably many eventually periodic sets by intersecting the club sets consisting of starting points of the repeated pattern).
• Consider the collection Sigma1\Sigma_1Sigma1 of eventually-agreeing subsets of the disjoint union omega1sqcupomega1\omega_1\sqcup\omega_1omega1sqcupomega1 of two copies of omega1\omega_1omega1. That is, sets Ssubsetomega1sqcupomega1S\subset \omega_1\sqcup\omega_1Ssubsetomega_1sqcupomega_1, such that except for countably many exceptions, SSS looks the same on the first copy as it does on the second. Another way to say it is that the symmetric difference of SSS on the first copy with SSS on the second copy is bounded. This is a sigma\sigmasigma-algebra, since it is closed under complement and also under countable intersection, as the countably many exceptional sets will union up to a countable set.

Please enlighten me by showing either that these are not actually counterexamples or that they are, or by giving another counterexample or a proof that there is no counterexample.

If the answer to the question should prove to be affirmative, but only via strange or unattractive topologies, then consider it to go without saying that we also want to know how good a topology can be found (Hausdorff, compact and so on) to generate the given sigma\sigmasigma-algebra.