# Can all aleph_2-dense subsets of R be isomorphic?

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.

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The consistency of All $\aleph_2$-dense sets are order-isomorphic is problem 15.14 in Arnold Miller's list of problems (which he keeps up-to-date), and there is no indication there that any progress has been made on it. That's at least strong evidence that it's still open.