Consider the following variant of [Freiling's Axiom of Symmetry, $\mathsf{AS}$](https://en.wikipedia.org/wiki/Freiling%27s_axiom_of_symmetry), 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](http://www.cs.nyu.edu/pipermail/fom/2011-August/015674.html) 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.
> 
> 
> 



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
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). 


