# Cardinality of all $\mathbf{\Sigma}^0_\alpha$-sets over Baire space without full choice

It is well-known that the set of all open (or closed) sets on Baire space has cardinality of the continuum. In context of choice, we can prove that the set of all $\mathbf{\Sigma}^0_\alpha$-sets over Baire space has cardinality of the continuum.

However if we does not assume the choice, how to determine the set of all $\mathbf{\Sigma}^0_\alpha$-sets over Baire space? I tried to find the proof of the set of all $\mathbf{\Sigma}^0_\alpha$-sets has cardinality of continuum. But as I see, almost proof that determines the cardinality of set of all $\mathbf{\Sigma}^0_\alpha$-sets uses full choice.

Without full choice, we can we prove that the cardinality of set of all $\mathbf{\Sigma}^0_\alpha$-sets is cardinality of the continuum? If not, there is a model of ZF + Countable choice that the cardinality of set of all $\mathbf{\Sigma}^0_\alpha$-sets is not equal to the cardinality of the continuum for some $\alpha>0$? Thanks for any help.

• It is consistent without choice that $\mathbb R$ is the countable union of countable sets. In such a situation, every set of reals is such a union. Singletons are closed, so countable sets are $F_\sigma$, and we have every set being $F_{\sigma\sigma}$ -- or, if you prefer to stick to classical $\Sigma^0_\alpha$ classes, $F_{\sigma\delta\sigma}$. – Andrés E. Caicedo May 1 '14 at 14:20
• @AndresCaicedo If we assume the countable choice then we can prove the cardinality of the set of $\Sigma^0_\alpha$-sets has the cardianlity of the continuum? – Hanul Jeon May 1 '14 at 14:21
• That's probably enough. I usually assume dependent choice, but I expect that to be an overkill. I do not have the chance to check right now but, once I have time, if there are no (other) answers, I'll let you know. – Andrés E. Caicedo May 1 '14 at 14:28
• In the presence of countable choice every Borel set has a Borel code. So it's probably enough. I'm going to wait and see what @Andres has to write on the topic. In other news, Fremlin's measure theory book (which is available freely on his site) has in the fifth volume a part about choiceless analysis. But I don't know if he discusses the cardinality of the Borel sets with or without countable choice. Another place to look through is Arnold Miller's paper about long Borel hierarchies which may or may not contain some relevant information. – Asaf Karagila May 1 '14 at 14:50

There has been some very nice work pertaining to this question. First as you probably know, under $AC$ the question of the cardinality of a pointclass becomes trivial for all non-selfdual pointclasses and they all have cardinality $2^{\aleph_0}$. So the question is interesting under $AD$. Hjorth has proved that assuming $AD+DC(\mathbb{R})$, if $\alpha < \beta$ then we have that $\lvert \boldsymbol\Sigma^0_{\alpha} \rvert < \lvert \boldsymbol\Sigma^0_{\beta} \rvert$ and in the projective hierarchy we have that $\lvert \boldsymbol\Delta^1_n \rvert < \lvert \boldsymbol\Sigma^1_n \rvert < \lvert \boldsymbol\Delta^1_{n+1} \rvert$.
• You're welcome. I used Scott Morrison's \cite plugin for that. It's quite awesome. – Asaf Karagila May 12 '14 at 20:39