This question is a follow-up from my recent question, Classifying set theories whose standard models sharing the same ordinals are equal

Let's say that a (recursively axiomatizable) set theory $T$ extending ZF is "$P(Ord)$-categorical" if, whenever $M$ and $N$ are transitive models of $T$ sharing the same sets of ordinals, one has $M=N$. For example, if $T$ proves the Axiom of Choice, then $T$ is $P(Ord)$-categorical, by Theorem 13.28 of Jech's {\it Set Theory}. Is the converse true? That is, if $T$ is $P(Ord)$-categorical, then must $T$ prove the Axiom of Choice? Or, alternatively, perhaps somehow there is a consistent extension of ZF + Axiom of Determinacy, for example, that is $P(Ord)$-categorical?

If a theory's $T$ being $P(Ord)$-categorical is not equivalent to $T \vdash $ Axiom of Choice, is there an axiom or axiom schema $A$ such that $T$ is $P(Ord)$-categorical if and only if $T \vdash A$ for all recursively axiomatizable extensions $T$ of ZF?


The answer to the first question is no, a theory can be $P(\text{Ord})$-categorical but not imply the axiom of choice.

To see this, consider the theory $T=\text{ZF}+V=L(\mathbb{R})$. This theory is $P(\text{Ord})$-categorical, because if two transitive models of this theory have the same sets of ordinals, then in particular, they have both the same ordinals (and hence the same height) and the same reals, and so they will build $L(\mathbb{R})$ in precisely the same way. So they will be equal.

But the theory is known to be relatively consistent with the failure of the axiom of choice.

  • $\begingroup$ Hmmm. Perhaps a candidate for an axiom $A$ characterizing such theories then might be "there is a set $X$ of ordinals such that $V = L(X)$"? (It's sufficient, but is it necessary?) $\endgroup$ – Jesse Elliott Dec 18 '14 at 13:17
  • $\begingroup$ I think you might mean that $X$ is a set of sets of ordinals? I don't think this will be necessary, however, since it could be that $X$ is a proper class of sets of ordinals, and we have $V=L(X)$. But if we make your assertion where $X$ is a class, it seems no longer to be first-order expressible. $\endgroup$ – Joel David Hamkins Dec 18 '14 at 13:23
  • $\begingroup$ For example, the theory $ZF+V=L(P(\text{Ord}))$ is $P(\text{Ord})$-categorical, for the same reasons that $V=L(\mathbb{R})$ is. $\endgroup$ – Joel David Hamkins Dec 18 '14 at 13:47
  • $\begingroup$ If two transitive models agree on ordinals, then they need not agree on sets of ordinals. Why, then, if they agree on sets of ordinals, must they also agree on sets of sets of ordinals? After I think about it more, I'm not clear on why $ZF+V = L(\mathbb{R})$ is $P(\Omega)$-categorical. $\endgroup$ – Jesse Elliott Dec 18 '14 at 22:40
  • $\begingroup$ In general, models of ZF with the same sets of ordinals need not agree on sets of sets of ordinals. But if they satisfy V=L(P(Ord)), then they do. If two models of V=L(R) have the same sets of ordinals, then in particular, they have the same real numbers and the same ordinals. And one can prove by induction on $\alpha$ that $L_\alpha(\mathbb{R})$ is the same in each of them, and so they are equal. The same argument works with P(Ord) in place of $\mathbb{R}$. $\endgroup$ – Joel David Hamkins Dec 18 '14 at 23:34

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