# Lebesgue Measurability and Weak CH

Let $LM$ denote "all subsets of $\Bbb{R}$ are Lebesgue measurable", and

$WCH$ (weak continuum hypothesis) denote "every uncountable subset of $\Bbb{R}$ can be be put into 1-1 correspondence with $\Bbb{R}$".

[Warning: in other contexts, weak CH means something totally different , i.e., it sometimes means $2^{\aleph_{0}} < 2^{\aleph_{1}}$].

We know that $LM$ and $WCH$ both hold in Solovay models. By forcing a (Ramsey) ultrafilter over a Solovay model one can also arrange $WCH+\neg LM$ (due to joint work of Di Prisco and Todorčević, who showed that the perfect set property holds in the generic extension).

This prompts my question ($DC$ below is dependent choice).

Question: Is it known, relative to appropriate large cardinal axioms, whether there is a model of $ZF+LM+DC+\neg WCH$?

My question arose from an FOM-question of Tim Chow, and my answer to it; see also Chow's response.

• What's your definition of uncountable: $\aleph_0 < |S|$ or $|S| \not\leq \aleph_0$ ?
– user5810
Aug 3, 2011 at 23:27
• I don't think there's a difference in this case - any non-finite set $S\subseteq\mathbb{R}$ is also Dedekind non-finite. This is because $S$ comes with a natural linear ordering, the restriction of the ordering on $\mathbb{R}$. Let $L_S=\lbrace x\in S:$ There are more than finitely many $y\in S$ with $y>x$ as real numbers$\rbrace$, and $R_S=\lbrace x\in S:$ There are more than finitely many $y\in S$ with $y<x$ as real numbers$\rbrace$. Then either $L_S$ has no greatest member (in which case we can get an embedding of $\omega$ into $L_S$), or $R_S$ has no least member (cont'd.) Aug 3, 2011 at 23:52
• (in which case we can get an embedding of $\omega^*$ into $L_S$). Either way, we wind up with a subset of $S$ of cardinality $\aleph_0$. So if $S\subseteq \mathbb{R}$ is such that $\vert S\vert\not\leq\aleph_0$, then $\aleph_0<\vert S\vert$. Aug 3, 2011 at 23:53
• Assuming $L_S$ has no greatest member, it does not follow that there is an embedding of $\omega$ into $L_S\hspace{.05 in}$. See consequences.emich.edu/CONSEQ.HTM, form 13.
– user5810
Aug 4, 2011 at 0:47
• @Ricky and Noah: first of all, thanks for your comments. Next: I have added $DC$, since talk of $LM$ without it is rather silly (and of course with $DC$ the two notions of uncountability that Ricky asked about collapse to one). Aug 4, 2011 at 1:12

This is an expansion of my comments above. In the paper with Zapletal that I reference, we assume a proper class of Woodin cardinals and force over $L(\mathbb{R})$ with a partial order of countable approximations to a certain kind of MAD family (which Jindra named an "improved" MAD family). Although I have yet to write out the details, I believe that the resulting model satisfies LM (it clearly satisfies DC), and that, in this model, $\mathbb{R}$ cannot be injected into the generic MAD family. The arguments I have in mind are straightforward applications of the arguments given in the paper.
The paper of Horowitz and Shelah referenced in my second comment works from the assumption of a strongly inaccessible cardinal and, as I understand it, adds the construction of a generic MAD family to Solovay's argument. As shown in their paper, DC + LM hold in the resulting model. I wrote to Haim and asked if $\mathbb{R}$ can be injected into the generic MAD family in this model, and he said no. He says he'll update their paper to include a proof of this (so they'll probably have a proof out before we do).