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Is the following consistent?

CH fails and there is a partial order $\preceq$ on $\omega_2$ such that for every $\alpha < \omega_2$, $(\alpha, \preceq)$ can be extended to a separable linear order $(\alpha, \preceq_{\alpha})$ but there is no separable linear order extending $(\omega_2, \preceq)$.

This is related to an earlier question of Noah Schweber: What kind of compactness does "expanding $\mathbb{R}$ by constants" have?

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  • $\begingroup$ Do you know what happens if you try to unify the linearizations of the $\alpha$-suborders by using a uniform ultrafilter on $\omega_2$? In other words, fix a separable linearization for each $\alpha$, and then say $\beta\lhd\gamma$ if $\beta$ is below $\gamma$ in the $\alpha$-linearization for $U$-almost all $\alpha$ above. This gives a linear order, but I don't think it has to be separable. $\endgroup$ May 31, 2017 at 23:11

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