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The axioms of ZF define what a set is by:

  1. $\omega$ is a set
  2. If $x$ and $y$ are sets, then $\{x, y\}$ is a set
  3. If $x$ is a set, then $\bigcup x$ is a set
  4. If $x$ is a set, then $\mathcal{P}(x)$ is a set
  5. If $x, y_1, ..., y_n$ are sets and $P$ is a statement free variables $x, y, z_1, ..., z_n$, then $\{z \in x : P\}$ is a set

and also defines equality between sets by the axiom of extensionality.

But I don't understand why we can't just add "nothing else is a set" as an axiom to the list, as we usually do when defining a certain mathematical object.

Wouldn't adding that axiom solve the problem of some statements being independent? If we can't prove a set exists, then it means it can't be produced using the rules listed above, so it's not a set.

  • This would be akin to a minimal, countable model of ZF. It wouldn't solve incompleteness as Godel's theorem would still apply. Being able to prove that a set exists is a different matter than a set existing. –  Jan 18 '19 at 19:56
  • Proving that a statement is true merely means it holds in every model of $ZF$. – J.F Jan 18 '19 at 19:58
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    You forgot collection/replacement – Not Mike Jan 18 '19 at 20:36
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    For example, there is a set called $\mathcal P(\omega)$, but it has only countably many elements (only countably many can be proved from these axioms, and "nothing else is a set"). – GEdgar Jan 18 '19 at 20:40
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    See also https://en.wikipedia.org/wiki/Constructible_universe – Not Mike Jan 18 '19 at 20:55
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    "The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times... – Jonathan Schilhan Jan 18 '19 at 22:30
  • @Jonathan: What do you think that the axioms of ZF describe, then? – Asaf Karagila Jan 19 '19 at 08:22
  • @Jonathan: It is a matter of mathematical history and culture. ZFC seems to be an elegant first-order theory that allows mathematicians to construct what they think are well-defined abstract collections, without being so silly as to be obviously inconsistent. The longer it stands, the more it is taught, and sooner or later (if not already) all mathematics students will affirm that sets are described by ZFC even if they don't really know what that means. – user21820 Jan 19 '19 at 10:27
  • @Stefan: You're making the basic error of conflating truth with provability. You can have a first-order theory with a predicate for is-a-set, and you can have an axiom that says that every set is constructible from your operations, but the problem is how to express "constructible". If you appeal to sets to encode the construction, then you might get something like V=L. V=L does decide a lot of questions including GCH. If you don't want to invoke sets in this, you would have to sort of go beyond first-order logic. – user21820 Jan 19 '19 at 10:40
  • I say "sort of" because you can use second-order logic with Henkin axioms and semantics, where you can quantify over subclasses of the universe and hence can say "no proper subclass of the universe is closed under the operations". But this may be even worse, because together with "every set is a subclass of the universe" you would be able to prove "there is no (set) model of Z". Will that shake your belief in Z set theory? – user21820 Jan 19 '19 at 10:51
  • By the way, it should be very obvious from the incompleteness theorems that no foundational system can ever escape independence. So it is utterly futile to try to "solve the problem of some statements being independent". – user21820 Jan 19 '19 at 11:03
  • @user21820 But the suggested system is no longer first-order... – Andrés E. Caicedo Jan 19 '19 at 13:02
  • @AndrésE.Caicedo: As I said, with Henkin axioms and semantics, it is "sort of" first-order, since it is equivalent to some 2-sorted first-order theory, or some 1-sorted first-order theory with an extra predicate-symbol to capture the "set" sort, as in MK set theory. But, you know this, right? – user21820 Jan 19 '19 at 13:08
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    @user21820 I do. But I didn't think it was quite made explicit in your post that the suggested reduction in your post had as a consequence that the resulting system was still subject to incompleteness. – Andrés E. Caicedo Jan 19 '19 at 13:26
  • @user21820 Anyway, my "real" point is that it is not clear that it is futile to try to escape the incompleteness phenomenon, as long as we are willing to forgo the first-order requirement for our foundational system. – Andrés E. Caicedo Jan 19 '19 at 13:28
  • @AndrésE.Caicedo: Well, I usually say "foundational system" when I'm lazy to spell out the precise conditions (i.e. has a program proof verifier and can reason about finite program runs). I insist that no other kind of system can be considered foundational. And, as you can see from my linked post, it doesn't matter whether the foundational system is based on first-order logic or not; it still will be incomplete. – user21820 Jan 19 '19 at 13:37
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    @AsafKaragila As any other first-order theory they describe properties of a structure. The ZF axioms don't distinguish "sets" from "non sets" (which the term "defining" would imply). I thought this was at the heart of OP's misunderstanding. – Jonathan Schilhan Jan 19 '19 at 21:08
  • I deleted my not very helpful answer to this question. Noah Schweber kindly provided this very relevant link in a comment on my deleted answer. – Rob Arthan Jan 19 '19 at 22:53
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    @Jonathan: Yes, but if you describe the structure of a vector space, then you could say that you're talking about properties of vectors (e.g., closed under addition and scalar multiplication). So in that sense, why wouldn't it be correct to say that the axioms of ZF describe the properties of sets (e.g., if $x$ is a set, then it has a power set, and subclasses of $x$ are sets, etc.)? – Asaf Karagila Jan 20 '19 at 00:45

2 Answers2

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It's not clear why we need to. One reason we "can't" has to do with a technicality. ZF is a theory in "first-order logic". Saying that nothing is a set except that which the axioms imply is a set is not a "first-order" statement, so if we tried to add it we'd no longer have a first-order theory. That would be too bad, because first-order logic works out a lot nicer than higher-order logics.

There's no way to say "nothing is a set except things that the axioms imply are sets" using just $\forall$, $\exists$ and $\in$.

(In fact, the axioms don't quite say what you say they say! There's no mention of "is a set" in the actual axioms. What you say is maybe how one thinks off what the axioms mean, but in fact, for example the actual axiom (2) is $$\forall x\forall y\exists z(\forall t(t\in z\iff (t= x\lor t= y))).$$The axioms talk about sets, not about what is or is not a set; adding "is a set" to the formalism makes it a totally different sort of thing.)

David C. Ullrich
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Your proposed rule would lead to contradictions. Take the continuum hypothesis, for example. We can't prove (in ZFC, assuming ZFC is consistent) that there is a set of cardinality between that of $\mathbb{N}$ and $\mathbb{R}$, so the rule says there isn't one. So, given the rule, the continuum hypothesis is true.

But wait! We also can't prove that there is a bijection between $\omega_1$ and $\mathbb{R}$, so the rule says there isn't one, and the continuum hypothesis is false.

Eric M. Schmidt
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