The Borel sigma algebra on $\mathbb{R}$ is obtained by starting with open sets in $\mathbb{R}$ and repeatedly applying the operations of complement, countable union, and countable intersections. My question is, what would happen if you don't just allow countable unions and intersections, but also allow unions and intersections of $\aleph_1$ many sets?

Now obviously if we assume the Continuum Hypothesis, the answer is "we obtain all subsets of $\mathbb{R}$". But we just assume $ZFC$, then what sets do we know we can obtain? Can we at least obtain all Lebesgue measurable sets? Can we still obtain all subsets of $\mathbb{R}$?


This is an excellent question! Note first that, just from cardinal arithmetic considerations, whether we obtain all sets is independent (and therefore so is whether we obtain all Lebesgue measurable sets).

The right context to study this question is descriptive set theory, and indeed the problem was considered early on. For example, Sierpiński proved in Sur une classe d’ensembles, Fundamenta Mathematicae 7 (1925), 237–243, that all $\mathbf\Sigma^1_2$ sets are unions of $\aleph_1$ many Borel sets. Recall that a $\mathbf\Sigma^1_2$ set is one which is the continuous image of the complement of the continuous image of a Borel set.

[Note that it is consistent that there are $\mathbf\Sigma^1_2$ sets that are not Lebesgue measurable.]

Around 1970, Martin and Solovay proved that the conjunction of Martin's axiom, $\lnot\mathrm{CH}$, and the statement that $\omega_1$ is accessible to reals implies the converse: Every union of $\aleph_1$ many Borel sets is $\mathbf\Sigma^1_2$. The statement that $\omega_1$ is accessible to reals means that there is a real $r$ such that $\omega_1^{L[r]}=\omega_1$. On the other hand, Solovay showed that not much more can be said: It is consistent with the continuum being arbitrarily large that there is a $\mathbf\Pi^1_2$ set (that is, a set which is the complement of a $\mathbf\Sigma^1_2$ set) that is not the union of fewer than $2^{\aleph_0}$ many Borel sets.

One can nevertheless soldier on and wonder what can be said if we allow longer unions. The right tool to study the problem is now the notion of $\kappa$-Suslin set. It is simpler to work on Baire space rather than the Euclidean line, but this makes no difference since Baire space $\omega^\omega$ is homeomorphic to the irrationals. A subset $A$ of $\omega^\omega$ is $\kappa$-Suslin if and only if there is a tree $T$ on $\omega\times\kappa$ such that $A=p[T]$. Recall that a (descriptive set-theoretic) tree on a set $B$ is a set of finite sequences of elements of $B$ closed under initial segments. For $B=\omega\times\kappa$, we can identify any such sequence with two sequences of the same length, one consisting of finite numbers, and one of members of $\kappa$. A branch through $T$ is an infinite sequence all of whose finite initial segments are in $T$ or, thinking of $T$ as consisting of pairs of sequences, a branch would be a pair $(x,t)$ where $x\in\omega^\omega$ (a "real") and $t\in\kappa^\omega$; the set of all branches through $T$ is denoted $[T]$. The projection $p[T]$ is defined as the set of reals $x$ for which we can find a $t\in\kappa^\omega$ with $(x,t)\in[T]$.

Martin proved that for any (finite) $n>0$, the $\omega_n$-Suslin sets are precisely the ones that are the union of $\aleph_n$ many Borel sets. In the context of determinacy, the extent of the $\kappa$-Suslin sets has been identified precisely for various $\kappa$ (significantly beyond the $\omega_n$). This implies results about the extent of these sets under appropriate large cardinal assumptions, and has also been studied from forcing axioms.


The system of sets you've described is the $\omega_1$-Borel hierarchy. I don't know who first introduced it, but Arnie Miller has a bunch of work on it and related topics - see e.g. this article and this set of notes.

This is well outside my comfort zone, so let me just say a brief bit about one of the very basic questions (cribbing horribly from page $2$ of Miller's aforementioned paper, and using his notation): can we get all the sets, and if so when? Of course, we need to assume $\neg$CH for anything interesting to happen.

  • Steprans showed that it is consistent that $\Pi_3^*=\Sigma_3^*=\mathcal{P}(2^\omega)$, but $\Pi_2^*$ and $\Sigma_2^*$ are incomparable proper subsets of $\mathcal{P}(2^\omega)$. I don't know anything about his proof, but Miller mentions that the model it produces has $2^\omega=\aleph_{\omega_1}$, and Carlson showed that in order to have every set be $\omega_1$-Borel we must have $cf(2^\omega)=\omega_1$. So in some sense, "Every set is $\omega_1$-Borel" does have a "CHy-flavor."

  • Steprans further observed that ZFC+$\neg$CH implies that we can't have $\Sigma^*_2=\mathcal{P}(2^\omega)$. For suppose $A$ were a $\Sigma^*_2$ set of size $>\omega_1$ (which must exist since CH fails). Then as a union of $\omega_1$-many closed sets, it must contain an uncountable closed set and hence a perfect subset. But now consider any Bernstein set of cardinality continuum.

  • Finally - and this is Miller's own result - Martin's axiom at $\omega_1$ alone implies that the hierarchy doesn't collapse; however, ZFC+$\neg$CH alone doesn't obviously tell us even that $\Pi^*_2\not=\Sigma^*_2$ (Miller lists this as open).

As mentioned above, this is far outside my range of literacy; please let me know if I've made any mistakes!

  • $\begingroup$ Do we we at least get all the Lebesgue measurable sets? $\endgroup$ – Keshav Srinivasan Oct 25 '18 at 4:11
  • 2
    $\begingroup$ @Keshav Not necessarily, just on cardinal arithmetic considerations. $\endgroup$ – Andrés E. Caicedo Oct 25 '18 at 4:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.