TITLE: Can all $\aleph_2$-dense subsets of $\mathbb{R}$ be isomorphic?
QUESTION [17 upvotes]: Let $\kappa$ be an infinite cardinal. For a subset $A \subseteq \mathbb{R}$, we say that $A$ is $\kappa$-dense if $|A \cap (a, b)| = \kappa$ for every interval $(a, b)$. By Cantor, any two $\aleph_0$-dense sets are order-isomorphic. 
If the CH holds, then there are many non-isomorphic $\aleph_1$-dense sets. However, Baumgartner showed that it is consistent with $2^{\aleph_0}=\aleph_2$ that all $\aleph_1$-dense sets are isomorphic, giving a generalization of Cantor's result. My question is, can this be pushed further? Is it consistent with $2^{\aleph_0} \geq \aleph_3$ that all $\aleph_2$-dense sets of reals are isomorphic? And what about for larger cardinals still? Can we have the result for $\aleph_1$-dense sets and $\aleph_2$-dense sets simultaneously? 
Baumgartner himself asked if all $\aleph_2$-dense sets of reals could be isomorphic in his original paper, which I found linked here on MO (all thanks to François Dorais). But I have been unable to find out if his question has subsequently been answered. If anyone knows the answer, or can provide a reference, I would be grateful.

REPLY [8 votes]: In section 5 (Concluding remarks and recent developments) of the paper Baumgartner's isomorphism problem for $\aleph_2$-dense suborders of $\mathbb{R}$, by Moore and Todorcevic, the following is stated:

He (Itay Neeman) moreover tentatively announced that, in spite of the gap in the
  present paper, he was able to show that if there is a weakly compact
  cardinal, then there is a forcing extension in which $BA_{\aleph_2}$ and $MA_{\aleph_2}$
  both hold.

If the claim is true, it means that the problem is solved now!!!
Also in the notes Reflection of clubs, and forcing
principles at $\aleph_2$ by Itay Neeman, the last theorem stated the consistency of $BA_{\aleph_2}.$