# What is known about size-restricted power set axioms?

What is known about ZF without powerset but with an axiom "every set has a set of all its countable subsets"?

This seems stronger than positing that the set of natural numbers has a powerset, though I do not know a proof that it is.

More generally, for any definable cardinal $\alpha$, what about the axiom "every set has a set of all smaller-than-$\alpha$ subsets"?

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## 1 Answer

Isn't $H(\mathfrak c^+)$, the collection of sets whose transitive closures have cardinality at most $\mathfrak c=2^{\aleph_0}$, a model of your theory (but not the power set axiom)? The point is that a set of cardinality $\mathfrak c$ has only $\mathfrak c$ countable subsets.

EDIT: Colin actually asked whether this theory is stronger than ZF minus power set plus "the set of natural numbers has a power set". To get a model where $\omega$ has a power set but some sets don't have a set-of-all-countable-subsets, replace $\mathfrak c^+$ in my original answer with $\kappa^+$ where $\kappa$ is a singular cardinal, greater than or equal to $\mathfrak c$, and of cofinality $\omega$. So the proposed model consists of all sets whose transitive closures have size at most $\kappa$. From $\kappa\geq\mathfrak c$, it follows that the model contains the actual power set of the (von Neumann) natural numbers. From the fact that $\kappa$ has cofinality $\omega$, it follows that a set of size $\kappa$ has strictly more than $\kappa$ countable subsets. Therefore, some sets in the model, for instance $\kappa$ itself, don't have a set-of-all-countable-subsets in the model. (Note that each individual countable subset of $\kappa$ is in the model, so nothing other than the genuine set-of-all-countable-subsets of $\kappa$ can serve as the set-of-all-countable-subsets of $\kappa$ in the sense of the model.)

(By the way, I'm aware that, in this situation, we actually have strict inequality $\kappa>\mathfrak c$ because $\mathfrak c$ can't have cofinality $\omega$. I ignored that fact above, because it's not needed for the argument.)

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Can you prove that the replacement axiom scheme is true in this model? –  Zsbán Ambrus Jul 29 '12 at 15:28
Yes, I know this theory is weaker than ZF. But is it stronger than ZF[1]? (That is ZF without powerset but with an axiom saying the natural numbers have a power set.) There are really two parts (at least) to that question. Does ZF[1] actually not prove every set has a set of all countable subsets? I expect it does not, but I do not know. And if my expectation is true, does this theory have strictly higher consistency strength than ZF[1]? –  Colin McLarty Jul 29 '12 at 15:42
@Zsban: I claim that not only the replacement scheme but full second-order replacement is true in this model. If $X$ is a set in the model and $Y$ is a subset of the model and is the "image" of $X$ under a "function", then $Y$ is in the model. The transitive closure of $Y$ consists of the elements of $Y$ (of which there are at most $\mathfrak c$ since $Y$ is an image of $X$) together with the union of the transitive closures of all these elements of $Y$. That's a union of at most $\mathfrak c$ sets, each of size at most $\mathfrak c$, so the union still has size at most $\mathfrak c$. –  Andreas Blass Jul 29 '12 at 21:47
@Colin: I think my answer can be modified to get a model where the set of natural numbers has a power set but not every set has a set-of-countable-subsets. I'll edit to add this to my answer (so as to make it a real answer). –  Andreas Blass Jul 29 '12 at 21:48