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.
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 would be interested in taking a look at: