Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:40:43Z http://mathoverflow.net/feeds/question/72259 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72259/partitioning-mathbbr-into-aleph-1-borel-sets Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets François G. Dorais 2011-08-06T19:22:51Z 2011-11-14T01:59:18Z <p>I just ran into this deceptively simple looking question.</p> <blockquote> <p>Is it always possible to partition $\mathbb{R}$ (or any other standard Borel space) into precisely $\aleph_1$ Borel sets?</p> </blockquote> <p>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.</p> <p>On the other hand, PFA or MM might give a negative answer but I don't see a good handle from that end either.</p> http://mathoverflow.net/questions/72259/partitioning-mathbbr-into-aleph-1-borel-sets/72263#72263 Answer by Andreas Blass for Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets Andreas Blass 2011-08-06T20:09:52Z 2011-08-06T20:09:52Z <p>It suffices to express $\mathbb R$ as the union of <code>$\aleph_1$</code> (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 <code>$\Pi^1_1$</code> set $A$ (say the set of codes of well-orderings of $\omega$) and its complement. A classical theorem says that any <code>$\Pi^1_1$</code> set is a union of <code>$\aleph_1$</code> Borel sets, and so is every <code>$\Sigma^1_1$</code> set. Apply that to $A$ and to $\mathbb R-A$ to get $\mathbb R$ as a union of <code>$\aleph_1$</code> Borel sets. No countably many of them cover $\mathbb R$ because $A$ is not Borel and thus not a countable union of Borel sets.</p> http://mathoverflow.net/questions/72259/partitioning-mathbbr-into-aleph-1-borel-sets/72274#72274 Answer by Péter Komjáth for Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets Péter Komjáth 2011-08-07T07:17:00Z 2011-08-07T13:42:17Z <p>There is a Hausdorff gap, a sequence <code>$\{f_\alpha,g_\alpha:\alpha\lt\omega_1\}$</code> of $\omega\to\omega$ functions such that $f_\alpha\lt^*f_\beta&lt;^*g_\beta&lt;^*g_\alpha$ 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 <code>$H_\alpha=\{f:f_\alpha\lt^* f \lt^* g_\alpha\}$</code> then <code>$\{H_\alpha:\alpha\lt\omega_1\}$</code> is a decreasing sequence of nonempty $F_\sigma$ with empty intersection. </p> http://mathoverflow.net/questions/72259/partitioning-mathbbr-into-aleph-1-borel-sets/80858#80858 Answer by Liang Yu for Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets Liang Yu 2011-11-14T01:59:18Z 2011-11-14T01:59:18Z <p>Here is another example from recursion theory:</p> <p>Take a chain $\{x_{\alpha}\}_{\alpha&lt;\omega_1}$ from Turing degrees.</p> <p>For each $\alpha&lt;\omega_1$, let $A_{\alpha}$ be the collection of the reals neither in $\bigcup_{\beta&lt;\alpha}A_{\beta}$ nor Turing-computing $x_{\alpha}$.</p> <p>Then $\{A_{\alpha}\}_{\alpha&lt;\omega_1}$ is a Borel partition of $\mathbb{R}$.</p>