A variant of Freiling's Axiom of Symmetry and a weak form of the Continuum Hypothesis in models where all sets of reals are Lebesgue measurable

Question

Consider the following variant of Freiling's Axiom of Symmetry, $\mathsf{AS}$, which will be denoted $A_{< 2^{\aleph_0}}$:

• given any function $f$ from $\mathbb{R}$ into the families of of subsets of $\mathbb{R}$ of size $< 2^{\aleph_0}$ there are $x_1,x_2 \in \mathbb{R}$ such that $x_1 \notin f(x_2)$ and $x_2 \notin f(x_1)$.

Does $\mathsf{ZF} + A_{<2^{\aleph_0}} + \mathsf{LM}$ imply $\neg \mathsf{WCH}$? If not, what is the current state of research in discovering models of $\mathsf{ZF}$ in which $\mathsf{LM} +\lnot \mathsf{WCH}$ holds?

Here

• $\mathsf{LM}$ is the statement that every set of reals is Lebesgue measurable,
• $\mathsf{WCH}$ is the statement that every uncountable subset of $\mathbb R$ can be put into 1-1 correspondence with $\mathbb R$.

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For background to this question, in his 3 August 2011 FOM post to Timothy Chow, Ali Enayat stated that

$\mathsf{ZF} + \mathsf{AS} + \mathsf{LM} + \mathsf{WCH}$ holds in Solovay's model since in Solovay's model every uncountable subset of $\mathbb{R}$ has a perfect subset.

Answer 1

Assuming $AD$, a version of the continuum hypothesis holds: every set of reals is either countable or of size continuum (this is a consequence of the perfect set property). So assuming $AD$, your $A_{<2^{\aleph_0}}$ and Freiling's $AS$ are equivalent. In particular, both can hold in Solovay's model, so the answer to your question is "no."

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Your second question seems much more interesting. Off the top of my head, I don't see a reason why we can't have a model of ZF+DC+LM+$\neg$WCH (except of course for the small fact that I don't know how to build one).

Answer 2

Every model of ZF + AS + LM + WCH is a model of ZF + $A_{<2^{\aleph_0}}$ + LM + WCH, since WCH implies that any set of reals with cardinal $<2^{\aleph_0}$ is countable.