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I've been told once or twice that the following holds:

There is a model of $ZFC+MA+\neg CH$ in which there is a $\mathfrak{c}$-universal linear order embedded in $(\omega^\omega, \le^\ast)$

Moreover, several people have attributed the construction of such a model to Hugh Woodin. The context that seems the most natural for this construction to appear is in his work on automatic-continuity, however for the life of me, I can't seem to find it.

It might also be the case that the embedding is into $\mathcal{P}(\omega)/fin$.

Does anybody know of a reference for this? Or if Woodin did indeed construct it (I only ask because my searches have yielded little fruit)?

Edit:

In this context a $\mathfrak{c}$-universal linear order $(L, \prec)$ has the following property: $\vert L \vert \le \mathfrak{c}$ and for every linear order $(\ell, <) $, with $\vert \ell \vert \le \mathfrak{c}$ there is an embedding $\varphi_\ell:\ell \rightarrow L$ respecting the linear order of $\ell$.

Thought I should share what I can find/know:

So far I've been able to find models for the following

$ZFC+ \neg CH$ and there is such a $\mathfrak{c}$-universal linear order in $(\omega^\omega,\le^\ast)$.

This is due to Laver, http://www.sciencedirect.com/science/article/pii/S0049237X08716306

In addition, there is also

$ZFC+MA+\neg CH$ and no such order is embedded in $(\omega^\omega,\le^\ast)$

There seem to be several models for this one, Laver cites two, one from Solovay and one from Kunen. I can only assume the Solovay model is the same constructed in Theorem 5.7 (p. 201, "Hausdroff Gaps and Limits", by Frankiewicz and Zbierski)

In addition there is

$CH\implies $ $(\omega^\omega,\le^\ast)$ contains a $\mathfrak{c}$-universal linear order

Which should be contained in (or atleast hinted at in) "Model Theory", by Chang and Keisler. If not there, the lack of $(\omega,\omega)$-gaps in $(\omega^\omega,\le^\ast)$ should produce it rather quickly.

So yeah, it would seem that the only link missing in this puzzle is the one I can't find the reference for.

edited Mar 14 2012 at 14:00

asked Mar 12 2012 at 22:53

Michael Blackmon
608●6●16

	Of course $(\omega^\omega, \leq ^ \ast)$ is never a linear order. I guess you meant to say that under $CH$ it contains a $\mathfrak{c}$-universal linear order. – Ramiro de la Vega Mar 14 2012 at 13:52
	Yes, of course! Thank you. – Michael Blackmon Mar 14 2012 at 13:59

Juris Steprans · Answer 1 · 2012-05-02 11:08:02Z UTC

2

Francisco Kibedi has written down a version of this for the $c$ universal linear order consistent with MA for $\sigma$ linked partial orders. His thesis is concerned with maximal such linear orders in the sense of Hausdorff's pantachies.

answered May 2 2012 at 11:08

Juris Steprans
2,338●4●13

KP Hart · Answer 2 · 2012-05-02 09:26:11Z UTC

You may want to have a look at this paper (PDF) by Baumgartner, Frankiewicz and Zbierski; it establishes the consistency of $\mathrm{MA}_{\sigma\text{-linked}}+\neg\mathrm{CH}$ plus ``every Boolean algebra of size at most $\mathfrak{c}$ is embeddable into $\mathcal{P}(\omega)/\mathit{fin}$''. The method is like that of Laver's, building a universal algebra inside $\mathcal{P}(\omega)/\mathit{fin}$ at the even stages of the iteration and getting Martin's Axiom for $\sigma$-linked posets at the even stages.

$\mathfrak{c}$-universal linear order

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