Substructure such that $\operatorname{cof}(\sup(B \cap \omega_m))= \omega_1$

Question

Let $\mathcal{L}=\{\in, \preceq\}$. Let $\mathcal{A}=(V_\theta, \in)$ (where $\theta > \omega$) be an $\mathcal{L}$-structure which interprets $\preceq^\mathcal{A}$ by some fixed well-ordering on $V_\theta$.

Does there exist an elementary substructure $\mathcal{B}$ of $\mathcal{A}$ satisfying the following: $\operatorname{cof}(\sup(B \cap \omega_m))= \omega_1$ for all $m \in \mathbb{Z}^+$?

For each $S \subseteq V_\theta$, let $\mathcal{B}_S$ be the Skolem hull of $A$ in $\mathcal{A}'$. Observe each $\mathcal{B}_S$ is an elementary substructure of $\mathcal{A}$. Perhaps we form an elementary chain from some subcollection of these $\mathcal{B}_S$ structures and take the union, but what would work? And how would I be able to know it satisfies the property?

Answer 1

Yes, this is true (modulo the detail that I think you want $\theta\ge\omega_\omega$, not $\theta>\omega$, since only the former ensures $\omega_m\in \mathcal{A}$ for all $m\in\omega$).

Fix $\theta\ge\omega_\omega$. The proof breaks into two key facts, one in the construction stage and one in the verification stage. The latter is essentially trivial but it is worth highlighting.

The first key fact we use is the following:

$(*)\quad$ Suppose $\mathcal{M}$ is a countable elementary substructure of $\mathcal{A}$ with underlying set $M$. Then there is a countable elementary substructure $\mathcal{M}'$ of $\mathcal{A}$ such that $\mathcal{M}\preccurlyeq\mathcal{M}'$ and $\sup(\mathcal{M}'\cap\omega_m)>\sup(\mathcal{M}\cap\omega_m)$ for all $m$.

Proof: just apply Lowenheim-Skolem to build a countable elementary substructure of $\mathcal{A}$ containing the countable set $M\cup\{\sup(M\cap\omega_m): m\in\omega\}$. (Note that for each $m$, $\sup(M\cap \omega_m)$ is a single ordinal, so this whole set is indeed countable.) $\quad\Box$

It may help to think of $(*)$ as addressing $\omega$-many requirements, one for each $\omega_m$ ($m<\omega$). We then apply $(*)$ $\omega_1$-many times to build the desired model $\mathcal{B}$ as the union of an elementary chain of countable models of length $\omega_1$: let $\mathcal{B}_0$ be any countable elementary submodel of $\mathcal{A}$, let $\mathcal{B}_\lambda=\bigcup_{\alpha<\lambda}\mathcal{B}_\alpha$ for $\lambda$ a countable limit, and let $\mathcal{B}_{\alpha+1}=\mathcal{B}_\alpha'$ per $(*)$ above. We then set $$\mathcal{B}=\bigcup_{\alpha<\omega_1}\mathcal{B}_\alpha.$$

Since $\mathcal{B}$ is constructed as the union of an elementary chain we have $\mathcal{B}\preccurlyeq\mathcal{A}$, so we just need to show that for each $m$ we have $cf(\mathcal{B}\cap\omega_m)=\omega_1$ as desired. This will follow from the following general observation:

$(\dagger)\quad$ Suppose $\gamma$ is an uncountable ordinal and, for each $\eta<\omega_1$, $W_\eta$ is a countable sub-ordering of $\gamma$ such that $\delta<\eta<\omega_1\implies W_\delta\subset W_\eta$ and $\sup(W_\delta)<\sup(W_{\delta+1})$. Then $\bigcup_{\eta<\omega_1}W_\eta$, as a suborder of $\gamma$, has cofinality $\omega_1$ regardless of the specific $\gamma$ we started with.

(Note that the above is false if we replace "cofinality" with "ordertype" - this, together with the proof of $(\dagger)$ in the first place, is a good exercise.) To analyze $\mathcal{B}$ we apply $(\dagger)$ "$\omega$-many times," once for for each $m<\omega$: take $\gamma=\omega_m$ and $W_\eta=\mathcal{B}_\eta\cap\omega_m$ to get $cf(\mathcal{B}\cap\omega_m)=\omega_1$ as desired.

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Incidentally, more complicated versions of this question - involving "big but locally small" models, which raise problems for greedy strategies like the above - raise the following interesting general question:

Suppose $\mathcal{X}\preccurlyeq \mathcal{T}$ with $\mathcal{T}$ a "reasonably closed" transitive set (possibly + other structure). Under what conditions does $\mathcal{X}$ have a top extension $\mathcal{Y}$ with $\mathcal{X}\prec\mathcal{Y}\preccurlyeq\mathcal{T}$?

Here $\mathcal{Y}$ is a top extension of $\mathcal{X}$ if every element of $\mathcal{Y}\setminus\mathcal{X}$ has rank greater than that of every element of $\mathcal{X}$. E.g. nontrivial forcing extensions are always end-extensions but never top-extensions of their ground models. In a sense the question above is trivial ("exactly when there is some $a\in\mathcal{T}$ of high rank and some family of Skolem functions such that [stuff]"), but getting it to gel nicely can be rather messy if memory serves.