Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets - MathOverflow

Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets

2011-08-06T19:22:51Z

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.

Answer by Andreas Blass for Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets

2011-08-06T20:09:52Z

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.

Answer by Péter Komjáth for Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets

2011-08-07T07:17:00Z

There is a Hausdorff gap, a sequence $\{f_\alpha,g_\alpha:\alpha\lt\omega_1\}$ of $\omega\to\omega$ functions such that $f_\alpha\lt^*f_\beta<^*g_\beta<^*g_\alpha$ hold for $\alpha\lt\beta\lt\omega_1$ (here $f\lt^* g$ denotes eventual dominance, i.e., that $f(n)\lt g(n)$ holds for large $n\lt\omega$) and there is no function $f:\omega\to\omega$ such that $f_\alpha\lt^* f\lt^* g_\alpha$ holds for $\alpha\lt\omega_1$. If we identify the reals with ${}^\omega\omega$ and set $H_\alpha=\{f:f_\alpha\lt^* f \lt^* g_\alpha\}$ then $\{H_\alpha:\alpha\lt\omega_1\}$ is a decreasing sequence of nonempty $F_\sigma$ with empty intersection.

Answer by Liang Yu for Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets

2011-11-14T01:59:18Z

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 $\mathbb{R}$.