Showing there is a model of PA whose ordering is $\aleph_1$-like

Question

This is problem 4.5.15 of Marker's Model Theory: An Introduction; this is purely for review purposes, and is not homework.

Marker defines an ordering $(X,<)$ to be $\aleph_1$-like if $|X|$ is $\aleph_1$, but the cardinality of every initial segment of the ordering is $\leq\aleph_0$. The problem is then to show there is a model of PA whose ordering is $\aleph_1$-like.

What does a solution to this look like?

I have no hunch for how to solve this, nor even much of an idea what the right tools might be. Chapter 4 is on types and saturation, and the mention of cardinals suggests I ought to be looking at this through the lens of saturation, homogeneity, or stability; but there are no further hints and I wasn't able to see a similarity with any examples earlier in the chapter.

Answer 1

I don't think there's a particularly simple proof by general model-theoretic results. Rather, this is a situation where a particular theorem about $\mathsf{PA}$ is the key:

(MacDowell-Specker Theorem) Every model of $\mathsf{PA}$ has an elementary end extension (of the same cardinality, via downward Lowenheim-Skolem).

(See here. Interestingly, my memory is that this result only applied to countable models of $\mathsf{PA}$, but that appears incorrect; I think I'm misremembering a more complicated result about elementary extensions of models of $\mathsf{PA}$, and will mention it for contrast once I find it. Certainly though the MacDowell-Specker Theorem for countable models would be enough here.)

By iterating $\omega_1$-many times (taking unions at limit stages - this is fine since the extensions are elementary, so we're just taking the union of an elementary chain when we do this), we can build an $\aleph_1$-like model of $\mathsf{PA}$ as desired: that the cardinality is $\aleph_1$ is trivial, and that all proper initial segments are countable follows from the fact that we built our model via end extensions only.