$\aleph_1$ Borel partition Can we prove the existence or non-existence of an $\aleph_1$ Borel partition (with possibily unbounded Borel ranks) of the Baire space $\omega^\omega$ ? In which axiomatic system?
Leo Harrington proved that, assuming $\text{AD}$, we cannot have an $\omega_1$-sequence of distinct Borel sets of a fixed bounded rank, I was wondering whether we could say something similar regarding an $\aleph_1$ Borel partition (of possibly unbounded Borel ranks).
Thanks
 A: I remember this being asked here before, but I can't find the duplicate, so here goes:
Yes, this is provable in $\mathsf{ZF}$. Basically the idea is to assign to each real $r$ a countable ordinal $o(r)$ so that $(i)$ $o^{-1}(\alpha)$ is Borel for each ordinal $\alpha$ and $(ii)$ for each $\alpha$ there are reals with $o(r)>\alpha$. Then $$\omega^\omega=\bigsqcup_{\alpha<\omega_1}o^{-1}(\alpha)$$ will be a partition of the desired type.
One way to do this is by using reals to code relations on $\omega$. A binary relation $R\subseteq\omega^2$ can be represented as a real (e.g. by hitting $R$ with a bijection $\omega^2\rightarrow\omega$ we get an infinite binary sequence). Now let $o(r)$ be the ordertype of the relation coded by $r$ if that relation is a well-ordering of $\omega$, and (say) $17$ otherwise.
Another example comes from computability theory: let $o(r)$ (usually denoted "$\omega_1^r$" or "$\omega_1^{CK}(r)$") be the smallest countable ordinal with no "$r$-computable" copy.  Again, this does the job.
In each case however the Borel sets we get have unbounded Borel rank, so there is no tension with Harrington's theorem.
