# Naive set theory really need axiom of power?

given the axioms of extension, pairing, specification, unions, unordered pair, as stated in naive set theory, this do not ensure the existence for each set of a collection of sets containing among its elements all subsets of the given set?

Because if exists a set we can by the axioms of specification create all subsets of the given set, also with help of axiom of unordered pairs we can create a singleton of each set and with axiom of union collect all of these singletons in one set whose existence is guaranteed by the axiom of unions.

Thank you.

This breaks down at two points:

• First, specification (also called separation) only gives us the subsets which are definable (from parameters); but lots of subsets probably won't be.

• Second, how exactly do you propose to use the axiom of unions in your last line? Remember you need an "indexing set", which you don't seem to have here.

Note that these obstacles are hard to see if you use the informal versions of the axioms: i.e. Specification = "subsets exist" and Union = "we can take unions of families." In each case, there's a subtlety: for specification, it's definability (with parameters) of the subset, and for union it's the existence of the family of sets whose union you want to take, to begin with.

• Fine, except that I don't see why specification even has anything to do with it. As I understand it (but please correct me if I'm all wet), the power set axiom tells us that $\{X:X\subseteq A\}$ is is a set and not a proiper class, i.e., that a set does not have too many subsets. That being the case, I don't see why the fact that specification doesn't give us as many subsets as we might wish is part of the reason we need the power set axiom. – bof May 13 '16 at 2:47
• @bof Because in certain circumstances, you might be able to prove "$A$ has a subset with property ----", but not be able to define any such subset. Then the "powerset" you get via separation would not satisfy, "I have an element with property ----." For example, if we let $\mathcal{X}$ be the set of subsets of $\mathbb{R}$ which are definable from real parameters, then $\mathcal{X}$ (under certain hypotheses) does not contain a (set coding a) well-ordering of $\mathbb{R}$, although if AC is an axiom, it needs to in order to be the powerset. – Noah Schweber May 13 '16 at 3:03
• Thanks. I don't see what your reply has to do with my question, but I'm sure the problem is at my end: either my inability to understand or my inability to express myself. – bof May 13 '16 at 3:30
• How are we to construct the reals without powerset? – André Nicolas May 13 '16 at 4:15
• @AndréNicolas We aren't - I was outlining how you could have a model of ZFC minus powerset, in which some set exists but does not have a powerset (in this example, it would be $\mathbb{R}$). – Noah Schweber May 13 '16 at 4:40