Question

This was of course motivated by this question.

Suppose $\kappa<\theta$ are uncountable regular cardinals. Given a structure ${\mathcal A}=(H_\theta,\in,<,\dots)$ where < is a well-ordering, let $C_{\mathcal A}=${$\sup(M\cap\kappa)\mid M\prec{\mathcal A}\mbox{ and }|M|<\kappa$}. Then $C_{\mathcal A}$ contains a club. Under what (reasonably general) circumstances can we guarantee that $C_{\mathcal A}$ actually is (respectively, is not) a club?

Answer 1

The idea in my previous answer can, I think, be upgraded to solve the whole problem, as follows. Again, fix Skolem functions for $\mathcal A$ as given by the well-ordering $<$, and again let $D$ be a set of fewer than $\kappa$ ordinals $\delta$, each of which is $\sup(\kappa\cap M_\delta)$ for some $M_\delta\prec\mathcal A$ with $|M_\delta|<\kappa$. I need to show that $\sup(D)$ is also in $C_{\mathcal A}$. For each $\delta\in D$, define $N_\delta$ to be the Skolem hull of $\kappa\cap\bigcup_{\xi\in D, \xi\leq\delta}M_\xi$. With appropriate gratitude for the hypothesis that $\kappa$ is regular, note that $N_\delta$ is an elementary submodel of $\mathcal A$ of size $<\kappa$ and that the sequence $\langle N_\delta\rangle$ is an elementary chain. Let $N$ be its union, and note that it, too, is an elementary submodel of $\mathcal A$ of size $<\kappa$. So it suffices to show that $\sup(D)=\sup(\kappa\cap N)$. The $\leq$ direction here is obvious, as $N$ includes $\kappa\cap M_\delta$ for each $\delta\in D$. To complete the proof, suppose the $\geq$ direction failed. Then we would have $\sup(D)<\sup(\kappa\cap N)$, so there would be an ordinal $\alpha\in\kappa\cap N$ with $\sup(D)\leq\alpha$. By construction, we would have some $\delta\in D$ with $\alpha\in N_\delta$, and so $\alpha$ would be of the form $f(\vec\beta)$ for some Skolem function $f$ and some ordinals $\beta_i$ in $\kappa\cap M_{\xi_i}$ for certain $\xi_i\leq\delta$. For each $i$, we have $\beta_i<\xi_i\leq\delta$, and, since there are only finitely many $i$ (as Skolem functions are finitary), we can find $\gamma<\delta$ with all $\beta_i<\gamma$. Increasing $\gamma$ if necessary, we can arrange that $\gamma\in M_\delta$. In $\mathcal A$, we can define the function $g$ sending each ordinal $\nu<\kappa$ to the supremum of all $f(\vec\eta)<\kappa$ for $\eta$ bounded by $\nu$; the values of this function are $<\kappa$ by regularity. As an elementary submodel of $\mathcal A$, $M_\delta$ is closed under $g$ and, in particular, contains $g(\gamma)$. But (again by elementarity) $g(\gamma)$ majorizes $f(\vec\beta)=\alpha>\sup(D)\geq\delta$. That contradicts the fact that $\delta$ is the supremum of $\kappa\cap M_\delta$, and this contradiction completes the proof.

Answer 2

This isn't really an answer but rather just clearing out some of the underbrush. If $\kappa=\aleph_1$ (or, more generally, if $\kappa$ is a successor cardinal and all ordinals below its predecessor are named in $\mathcal A$), then $C_{\mathcal A}$ is a club, thanks to the presence of a well-ordering $<$ in $\mathcal A$. The well-ordering gives us definable Skolem functions for $\mathcal A$, and, in the following, "Skolem hull" refers to those particular Skolem functions. When $\kappa=\aleph_1$ (or under the weaker assumption indicated above), then, for any $M\prec\mathcal A$, the ordinal $\sup(M\cap\kappa)$ is also the first ordinal not in $M$. Suppose we have a set $D$ of such ordinals, bounded below $\kappa$. For each $\delta\in D$, we have some $M_\delta\prec\mathcal A$ such that $\delta$ is the first ordinal not in $M_\delta$, but there's no obvious connection between these $M_\delta$'s for different $\delta$'s. We can improve that situation by considering, instead of $M_\delta$, the Skolem hull $N_\delta$ of $\delta$ (in $\mathcal A$ or equivalently in its elementary submodel $M_\delta$). So $\delta\subseteq N_\delta\subseteq M_\delta$ and therefore $\delta$ is the first ordinal not in $N_\delta$. Unlike the $M_\delta$'s, these $N_\delta$'s form a chain, and so their union $N$ is an elementary submodel of $\mathcal A$. The first ordinal not in it is $\sup D$, which is therefore in $C_{\mathcal A}$. Thus $C_{\mathcal A}$ is a closed subset of $\kappa$, and it's obviously unbounded.

There remains the "real" question of what happens when my additional hypotheses are not satisfied, so that $M\cap\kappa$ can have gaps below its supremum.