Under the axiom of choice, the cardinality of Borel $\sigma$-algebra $B$ is $\mathfrak{c}$. In this proof the axiom of choice is used three times: To prove
- $\omega_1$-times recursion is sufficient,
- each $|B_{\alpha}|$ is less than or equal to $\mathfrak{c}$ in recursion steps and
- $\aleph_1\times \mathfrak{c}\le \mathfrak{c}$.
Is it known how strong types of the axiom of choice we need to prove $|B|=\mathfrak{c}$? According to this, $\mathsf{DC}$ is not enough.
I am also interested in opposite implication, that is, what restrictions on $|B|$ (like $|B|\le\aleph_1\times \mathfrak{c}$) can we get using $\mathsf{DC}$? I know $\mathsf{AC}_{\omega}$ is enough to show 1.
