In ZF, the Power Set Axiom (POW) says that given any set $A$, there exists a set $\mathcal{P}(A)$ such that $$ \forall a(a\in\mathcal{P}(A)\leftrightarrow a\subseteq A)\tag{1} $$


  1. On many occasions, I've seen POW stated like this: "For any set $A$ there is a set consisting of all subsets of $A$."

    Would a more precise formulation be: "For any set $A$ which is known to exist there exists a set containing all subsets of $A$ which are known to exist."?

  2. An instance of the Axiom Schema of Separation (SEP) essentially gives the existence of a subset of a given set. So, why doesn't POW imply SEP? Is it because POW doesn't give the existence of the individual subsets but only the set containing them all?

  3. Conversely, does SEP and the remaining ZF axioms imply POW?

  • 2
    $\begingroup$ What would a $\textit{non-existing}$ set be? $\endgroup$ Commented Sep 22, 2015 at 13:49
  • $\begingroup$ How does the axiom schema of separation give the existence of all elements of $\mathcal{P}(A)$, given a set $A$? $\endgroup$
    – Servaes
    Commented Sep 22, 2015 at 13:51
  • $\begingroup$ @T_M A non-existing set would be a set whose existence can't be proved from the ZF axioms, for example the "set" of all sets? $\endgroup$
    – Guest
    Commented Sep 22, 2015 at 13:53
  • $\begingroup$ @Guest In formulation of POW only sets are involved. Not "sets". $\endgroup$
    – drhab
    Commented Sep 22, 2015 at 13:55
  • $\begingroup$ But the set of all sets isn't a set. This doesn't mean that it doesn't exist. So the formulation of SEP is ok, "for all $\textit{sets}$ there exists...". Existence of the set of all sets isn't the issue here, the issue is that it's not an actual set. $\endgroup$ Commented Sep 22, 2015 at 13:55

1 Answer 1


1: Axioms do not have any sense of time or knowledge. They can't speak about whether sets or other elements exist prior to (or as a consequence of) other sets -- at least not then in standard first-order logic that axiomatic set theory is usually done within.

It may be intuitively instructive to imagine all of the sets of set theory to come into being in some particular order sequence, guided by "when" axioms say they ought to exist, but that is not what the axioms say.

What the axioms say is just "Let's imagine we have a universe of sets, and such-and-such claims happen to be true about the universe of sets we have". The axiom of power sets does not create a power set -- it simply asserts that there is a power set of whatever you hold in your hand, and this power set has existed the whole time. It comes together with all of the other sets in your universe, and the entire totality of sets magically happens to make your axioms true.

2: What the power set axiom says is only that for every set $A$ there's a set whose members are those sets in the universe that happen to be subsets of $A$. It doesn't say anything about how many such things exist -- all it tells you that there's a set $\mathcal P(A)$ such that if you manage to find some set $B$ and discover that $B\subseteq A$, the set you have found will also be an element of $\mathcal P(A)$.

On the other hand, the axiom of separation can, in most variants of the formulations, be derived from the axiom of replacement (together with the axiom of the empty set).

3: The axiom of separation lets you know that any subcollection of a set that you can define (in the language of set theory) will exist as a set inside your universe of sets. It does not in itself guarantee that there's a set that contain all of those subsets; for that you need a separate axiom.

For example, if you take a model of ZFC and restrict it to only the hereditarily finite-or-countable sets, you will have something that satisfies all axioms of ZFC except for the power set axiom. Thus, the other axioms cannot (if they are consistent) imply the power set axiom.

  • $\begingroup$ Thank you. Your answer raises more questions in my mind, though. (a) If I understand, we have 'the' intuitive class of all sets in mind and the main reason why we use the axioms is to make sure we have a set with the desired properties in our class (for example, we use the axioms to make sure that in any model of ZF there is a set behaving like $\mathbb{R}$)? (b) Since there are many models of ZF, when we talk about sets, do we talk about the (intuitive) union of all these models, if that makes any sense? $\endgroup$
    – Guest
    Commented Sep 22, 2015 at 15:13
  • $\begingroup$ (c) Also, what if in my model I have the set $A$ of all cats in Florida. Cats are not set, so for every set $B$ we have $B\not\in A$ and $A$ acts like an empty set? But the empty set should be unique according to ZF, and the set $C$ of all cats in Egypt also acts like the empty set in my model and clearly $A\neq C$... $\endgroup$
    – Guest
    Commented Sep 22, 2015 at 15:13
  • 1
    $\begingroup$ (a) Formally we don't "use the axioms to make sure we have a set". We use the axioms to build proofs. The axioms are what we already (think we) know about the model, and then proofs allow us to know more things about it. (b) Usually we leave it implicitly that we're working in some model of ZFC, and leave it to the reader to decide which one he wants to think of. As long as the model he chooses to think of satisfies the axioms that we work from, he can be sure that what we say is true about it. $\endgroup$ Commented Sep 22, 2015 at 15:17
  • 1
    $\begingroup$ @Guest, (c) Something that contains a set of all cats in Florida cannot be a model of ZF, for the reason you sketch. ZF is designed to talk about universes where everything we ever want to speak formally about sets. If you want to speak directly about things that are not sets, ZF is not the axiom system you should be using. $\endgroup$ Commented Sep 22, 2015 at 15:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .