TITLE: Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets
QUESTION [14 upvotes]: I just ran into this deceptively simple looking question.

Is it always possible to partition $\mathbb{R}$ (or any other standard Borel space) into precisely $\aleph_1$ Borel sets?

On the one hand, this is trivial if the Continuum Hypothesis holds. Less trivially, this also follows from $\mathrm{cov}(\mathcal{M}) = \aleph_1$, $\mathrm{cov}(\mathcal{N}) = \aleph_1$, $\mathfrak{d} = \aleph_1$, and similar hypotheses. However, I can't think of a general argument that allows one to split $\mathbb{R}$ into precisely $\aleph_1$ pairwise disjoint nonempty Borel pieces.
On the other hand, PFA or MM might give a negative answer but I don't see a good handle from that end either.

REPLY [10 votes]: Here is another example from recursion theory:
Take a chain $\{x_{\alpha}\}_{\alpha<\omega_1}$ from Turing degrees.
For each $\alpha<\omega_1$, let $A_{\alpha}$ be the collection of the reals neither in $\bigcup_{\beta<\alpha}A_{\beta}$ nor Turing-computing $x_{\alpha}$.
Then $\{A_{\alpha}\}_{\alpha<\omega_1}$ is a Borel partition of $\mathbb{R}$.