I just ran into this deceptively simple looking question.



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> Is it always possible to partition $\mathbb{R}$ (or any other standard Borel space) into precisely $\aleph\_1$ Borel sets?
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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.



Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets
It suffices to express $\mathbb R$ as the union of $\aleph\_1$ (not necessarily disjoint) Borel sets such that no countably many of them cover $\mathbb R$, because then you can list them in an $\omega\_1$-sequence and subtract from each one the union of the previous ones. Partition $\mathbb R$ into a non-Borel $\Pi^1\_1$ set $A$ (say the set of codes of well-orderings of $\omega$) and its complement. A classical theorem says that any $\Pi^1\_1$ set is a union of $\aleph\_1$ Borel sets, and so is every $\Sigma^1\_1$ set. Apply that to $A$ and to $\mathbb R-A$ to get $\mathbb R$ as a union of $\aleph\_1$ Borel sets. No countably many of them cover $\mathbb R$ because $A$ is not Borel and thus not a countable union of Borel sets.


