Existence of subsets in ZFC without the Axiom of Power Set

Question

Does the elements of $\mathcal{P}(\mathbb{N})$ exist in ZFC without the Axiom of Power Set?

Or, more generally, for which sets $S$, all the elements of $\mathcal{P}(S)$ exist in ZFC without the Axiom of Power Set?

Answer 1

The question is what do you mean by "existence". In the formal sense of the word, it means that the object is an element of the universe. So even without any axiom, if $\Bbb N$ exists, there are some subsets of $\Bbb N$ which exist, and others which might not.

If your theory is strong enough, it will prove there is no bijection between the natural numbers and the set of natural numbers. For this some basic set theory is needed, but certainly not the power set axiom.

On the other hand, if $S$ exists and we have the separation schema to our disposal, then we can indeed prove that there is no definable surjection from the $S$ onto the subsets of $S$ in the universe. Since the existence of a function would imply the existence of the range of the function, and Cantor's theorem only uses separation.

Naively, however, you might think about some universe of set theory, and ask if $M$ is a model of set theory without power set, can we prove that all the subsets of $\Bbb N$ are in that model. The answer is negative, for nearly obvious reasons. First of all, $H(\omega_1)$, the set of all hereditarily countable sets, is a model of all the axioms of set theory except power set (and it does include all the subsets of $\Bbb N$). Using the Lowenheim-Skolem theorem, it has a countable submodel. That countable model by definition does not know about all the subsets of $\Bbb N$, since the universe knows there are uncountably many of them, and the countable model is countable.

In fact, the same naive argument fails even if we do assume the power set axiom. Granted, now we need to assume the existence of a model of $\sf ZFC$, but other than that, the same argument follows. If there is a model, there is a countable model, which will not know about all the subsets of $\Bbb N$. To learn more, read about Skolem's paradox.