Axiom of Power Set

Question

What does the axiom of power set add to ZF or ZFC set theory?

https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory

More specifically, it seems that once the axiom of infinity and the axiom schema of specification are used, then all of the elements of the power set of integers, P(N), could exist. If that is true, then wouldn't the axioms of pairing and union ensure that P(N) itself also existed?

P(N) contains many elements that are not computable sets: https://en.wikipedia.org/wiki/Recursive_set

Perhaps the axiom schema of specification could not be relied upon to assert the existence of any element of P(N) that is not computable (because there is no specifying formula)?

So, what is the point of the axiom of power set? It must add something to the other axioms... What elements of P(N) would not exist without the axiom of power set?

Bonus question: Given the existence of P(N), could the axiom schema of specification be used to assert the existence of...:

$$\{ p \in P( \,\mathbb{N}) \, : p \text{ is a computable set} \}$$

...where the formula for determining whether a set is computable is itself non-computable (think halting problem) and may not exist even though it is easily defined? I'm wondering what the limitations are when it comes to formulas that can be used in conjunction with the axiom schema of specification, thinking maybe it allows formulas to be used that literally cannot exist...

Answer 1

Here is a fun exercise. If you assume all the axioms except power set, then you cannot prove the existence of any uncountable set.

You can do it the hard way, or the smart way, by finding out that there is a model of the theory "$\sf ZFC-Power +\textrm{ Every set is countable}$".

In particular, to answer your question, no, you cannot prove that $\mathcal P(\Bbb N)$ exists. Because even without the power set axiom, you can show that given a countable set of subsets of $\Bbb N$, there is a subset of $\Bbb N$ which is not there. Namely, Cantor's theorem does not require $\mathcal P(\Bbb N)$ to actually be a set, if you formulate it correctly.

Answer 2

The axiom of specification only produces smaller sets than the natural numbers, i.e. sets that are subsets of $\mathbb{N}$ and are therefore members of $\mathcal{P} (\mathbb{N}) $. On the other hand, the axiom of infinity only guarantees that some $\aleph_0$ or larger infinite set exists -- it does not help us construct one from a smaller set like the naturals. What the power set axiom guarantees is not that any particular subset of $\mathbb{N}$ exists, but that the set of all these subsets is a set. This is also the axiom which guarantees the existence of a set with cardinality of exactly $2^{\lvert \mathbb{N} \rvert} $.