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As far as I understand, it is consistent with ZF that every set of real numbers is a Borel set. Furthermore, I know that relatively weak forms of the axiom of choice suffice to prove that $|\mathcal B| = \mathfrak c$. The standard proof that constructs the Borel hierarchy requires the regularty of $\omega_1$ (so, for example, countable choice) to show that the hierarchy collapses at $\omega_1$, and then requires -- as far as I can tell -- something like the well-orderability of $\mathbb R$ to show that $\aleph_1 \times \mathfrak c = \mathfrak c$.

Can we relax this last 'requirement'? In particular, can we show that there are continuum-many Borel sets using only countable choice?

Mees de Vries
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1 Answers1

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Countable choice isn't enough. It doesn't even prove that there are only a continuum of countable sets of reals. In Solovay's model, we have ZF + DC + ``all sets of reals are Lebesgue measurable." In this model, there is no injection from the countable sets of reals $[\mathbb{R}]^{\omega}$ into $\mathbb{R}$ by this Sierpinski argument: https://mathoverflow.net/a/191293/109573.

Another example is Shelah's model of ZF + DC + ``all sets of reals have the property of Baire." This has no injection from $[\mathbb{R}]^{\omega}$ into $\mathbb{R}$ since the set in Sierpinski's argument also lacks the property of Baire. The advantage of Shelah's model over Solovay's model is that Shelah's model doesn't require an inaccessible to be built. Also, in Shelah's model, $\omega_1$ injects into $\mathbb{R},$ so $\aleph_1 \times |\mathbb{R}| = |\mathbb{R}|.$ Therefore, this cardinal equality is also not enough.

Countable choice does prove there is a surjection from $\mathbb{R}$ onto the Borel sets and in particular some non-Borel set of reals exists, see https://mathoverflow.net/a/32722/109573.

Elliot Glazer
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  • Both Moore p325 and the Consequences book state the following: "form 43 (DC) implies form 363 (there are exactly $2^{\aleph_0}$ Borel sets in $\mathbb{R}$)". Trying to track down a reference led me down to your answer. Am I missing something or does this answer show that DC does not in fact imply there are exactly $2^{\aleph_0}$ many Borel sets? – Jason Zesheng Chen May 01 '24 at 17:09
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    I don't have Moore's book but the claim is certainly not true. – Elliot Glazer May 02 '24 at 17:36
  • Yeah that was my first reaction too! But now I wonder what they had in mind... does DC imply the set of Borel codes has size exactly the continuum? – Jason Zesheng Chen May 02 '24 at 19:27
  • It’s a ZF theorem that there is exactly continuum many Borel codes. I cannot make any sense of the quoted claim except maybe what I wrote in the last paragraph of my answer. – Elliot Glazer May 03 '24 at 01:00