Consider the following two facts, a modified version of which appear in this paper of Arnie Miller from the early 1980's:
$\bullet$ If $\mathbb R$ can be partitioned into $\aleph_1$ closed sets, then so can every uncountable Borel set.
$\bullet$ If $\mathbb R$ can be partitioned into $\aleph_1$ $F_\sigma$ sets, then so can every uncountable Borel set.
Equivalently, we could replace "Borel" with "analytic" or "co-analytic" here, since every (co-)analytic set can be partitioned into $\aleph_1$ Borel sets.
The statement "$\mathbb R$ can be partitioned into $\aleph_1$ closed (or $F_\sigma$) sets" is independent of ZFC. It follows easily from the Continuum Hypothesis (partition it into singletons), but is also consistent with the negation of CH (a fact first proved by Stern, and discussed in the paper of Miller that I linked to above).
My question is whether the analogous statement continues to be true further up the Borel hierarchy.
Question 1: Is it true that if $\mathbb R$ can be partitioned into $\aleph_1$ $G_\delta$ sets, then so can every uncountable Borel set?
The statement "$\mathbb R$ can be partitioned into $\aleph_1$ $G_\delta$ sets" is also independent of ZFC. Fremlin and Shelah prove in this paper that it is true if and only if $\mathrm{cov}(\mathcal M) = \aleph_1$.
Moving just a little further up the Borel hierarchy, the corresponding question for $F_{\sigma \delta}$ sets takes on a slightly different form. It is no longer just consistent that $\mathbb R$ can be partitioned into $\aleph_1$ $F_{\sigma \delta}$ sets: it is a theorem of ZFC, proved by Hausdorff proved in 1936. Thus at this level of the Borel hierarchy, our question simply becomes:
Question 2: Can every uncountable Borel set be partitioned into $\aleph_1$ $F_{\sigma \delta}$ sets?
I feel like this question must have been considered before (and that probably the answer is negative), but I can't seem to find anything. Finally, let me ask an even weaker version of the question, although I suspect that even this version may have a negative answer:
Question 3: Can every (co)analytic set be partitioned into $\aleph_1$ Borel sets of bounded complexity? (By "bounded complexity" I means that all $\aleph_1$ of the Borel sets are $\mathbf{\Sigma}^0_\alpha$ for some $\alpha < \omega_1$.)
Or, to put it another way:
Question III: Is it consistent that some (co)analytic set that cannot be partitioned into $\aleph_1$ Borel sets of bounded complexity?
