A Lowenheim-Skolem-like Result for Structures of Set Theory

Question

While reading Kanamori's proof (page 59-60 in $\textit{Higher Infinite}$) that $\kappa$ is $\Pi_1^1$-indescribable if and only if $\kappa$ is weakly compact; it appears that he uses the following result:

Let $\bar R$ denote a new predicate symbol. If $\kappa$ and $\alpha$ are inaccessible, there there exists a transitive $X \supsetneq V_\alpha$ such that $|V_\alpha| = |X|$, and $\langle X, \in, R \cap X \rangle \prec \langle V_\kappa, \in, R \rangle$, where of course $R$ is the interpretation of $\bar R$ in $V_\kappa$ and $\prec$ denote elementary substructure.

A Skolem Hull argument is one way of proving the downward Lowenheim-Skolem theorem for structures in general. However the problem is simultaneously making $|X| = |V_\alpha|$ and making $X$ transitive.

For instance a possible approach is to take any $X_0' \supsetneq V_\alpha$ with $X_0' \subseteq H(\alpha^+)$ and $|X_0'| = \alpha$. (Here $H(\delta)$ are the sets with transitive closure with cardinality less than $\delta$. ) Let $X_0$ be the transitive closure of $X_0'$. $|X_0| = \alpha = \beth_\alpha = |V_\alpha|$ since $\alpha$ is inaccessible. Taking the Skolem hull with respect to some complete set of Skolem functions in $V_\kappa$, one gets $H(X_0) \prec V_\kappa$. Let $X_1' = H(X_0)$. Then $|X_1'| = \alpha = |V_\alpha|$. $X_1'$ may not be transitive. One can try to let $X_1$ be the transitive closure; however, then $|X_1|$ may not be $\alpha$.

If one could pick Skolem functions such that every Skolem function restricted to $X_0$ has range lying in $H(\alpha^+)$, then $X_1$ define above would still have cardinality $\alpha$. Is this possible to find such Skolem function where the rank is bounded. Equivalently for any formula, when the other parameters lie in $V_\beta$, can quantification over $V_\kappa$ be replace by quantification over the sufficiently small $V_\gamma$ for $\gamma < \kappa$. If this is possible, I believe iterating the Skolem Hull construction and transitive closure, $\omega$-times and taking the union in the end will produce the desired elementary substructure.

Can this be done? Is there an alternative way to prove this transitive Lowenheim Skolem type result? Is this result even true? Thanks for any clarifications.

Answer 1

I don't think you can get such an $X$, even if there is no $R$ to worry about. Suppose you had an $X$ as in the question. Since $X$ properly includes $V_\alpha$, it has elements of rank $\geq\alpha$. Since the rank function is absolute for transitive models, $X$ contains the rank of such a set; in particular it contains an ordinal $\geq\alpha$, and, being transitive, it contains $\alpha$. Being an elementary submodel of $V_\kappa$, $X$ must contain all elements of $V_\kappa$ that are definable in $V_\kappa$ from $\alpha$. In particular, $X$ must contain $\alpha^+$ (and $\alpha^{++}$, etc.); but then to be transitive, $X$ must have cardinality at least $\alpha^+$ (and more), contrary to the requirement that $|X|=|V_\alpha|=\alpha$.

Answer 2

The following edited answer (hopefully) gives an $X$ of size $\alpha^{+}$.

Define $\langle X_i : i < \alpha^{+} \rangle$ as follows. $X_0 = V_{\alpha} \cup \{\alpha\}$. For $0 < i < \alpha^{+}$, $X_i$ is the transitive collapse of an elementary substructure $Y_i$ of $(V_{\kappa}, \in , R)$ of size $\alpha$ containing $\bigcup \{X_j : j < i\}$. Put $X = \bigcup \{X_i : i < \alpha^{+}\}$. Note that $X_i$'s are increasing and for a club no. of stages $i < \alpha^{+}$, we have $(X_i, \in, R \cap X_i)$ is an elementary substructure of $(X, \in, R \cap X)$. Now fix any $a_1, \dots, a_n \in X$, and suppose $(X, \in, R \cap X) \vDash \phi(a_1, \dots, a_n)$. Let $i < \alpha^{+}$ be large enough to include all $a_1, \dots, a_n$. Let $j > i$ be a limit stage such that $(X_j, \in, R \cap X_j)$ is elementary substructure of $(X, \in, R \cap X)$. Then $(X_j, \in, R \cap X_j) \vDash \phi(a_1, \dots, a_n)$. Let $\pi: Y_j \rightarrow X_j$ be the collapsing map used at stage $j$. Since $\pi$ fixes the transitive kernel of $Y_j$, it fixes $a_1, \dots a_n$. So $(Y_j, \in, R \cap Y_j) \vDash \phi(a_1, \dots a_n)$ hence also $(V_{\kappa}, \in, R) \vDash \phi(a_1, \dots, a_n)$.

More edit: Now for some $i < \alpha^{+}$, $(X_i, \in, R \cap X_i) \prec (X, \in, R \cap X) \prec (V_{\kappa}, \in, R)$ and $|X_i| = \alpha$ so that we are done.