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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.

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What's your definition of uncountable: $\aleph_0 < |S|$ or $|S| \not\leq \aleph_0$ ? –  Ricky Demer Aug 3 '11 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.) –  Noah S Aug 3 '11 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 $\aleph_0<\vert S\vert$. –  Noah S Aug 3 '11 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. –  Ricky Demer Aug 4 '11 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). –  Ali Enayat Aug 4 '11 at 1:12
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