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I've got 2 questions concerning the axiom of infinity $AI$.

  1. Given the axiom of empty set $AE$, it's obviously not necessary that the set prescribed by $AI$ is unique. Can $AC$ or some another condition make this set to be unique?
  2. Does $ZF$ or $ZFC$ imply that $\mathbb{N}$ contains exactly $\emptyset,\{\emptyset\},\{\emptyset\,\{\emptyset\}\},...$, i.e., can $\mathbb{N}$ contain some element of another type? The another variant of this question. If $ZFC$ has a model, is it necessary that all elements of $\mathbb{N}$ are of the given type?

Addition 1. I'm using the following variant of $AI$: there is an inductive set $x$, i.e. $\emptyset\in x$ and $\forall(y\in x)\exists(z\in x):z=y\cup\{y\}$.

Addition 2. I also consider $\mathbb{N}$ to be the intersection of all inductive sets.

P.S. If the full answer was given before, then please can you send me the link? Here's something close, but it's not precisely my question: Could the natural numbers be unique after all?

Bertrand Haskell
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    I don't understand what you are trying to get at in question 1--what is the relevance of the axiom of empty set? And by "the set prescribed by $AI$" do you just mean an inductive set? As for question 2, what exactly does $\dots$ mean? The usual way of making that precise is exactly to say "the minimal inductive set" and so your question becomes a tautology. – Eric Wofsey Jan 15 '20 at 18:24
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    I would also recommend including in your question the exact statement of $AI$ you are talking about, since there are several different equivalent formulations that are sometimes used. – Eric Wofsey Jan 15 '20 at 18:26
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    Regarding question 2, this seems to be addressed somewhat in the other question. The sentence “$\mathbb N$ contains exactly $\emptyset,$ ...” can’t even be expressed in the first order language of set theory, much less proven. – spaceisdarkgreen Jan 15 '20 at 20:14
  • Have you seen this? – Arturo Magidin Jan 15 '20 at 20:27
  • @spaceisdarkgreen thanks, seems the same quiestion is here https://math.stackexchange.com/questions/1626849/proof-that-the-minimal-inductive-set-does-not-contain-any-more-elements-than?rq=1 – Bertrand Haskell Jan 15 '20 at 20:42

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The set isn't unique. The natural numbers are usually defined to be the intersection of all sets satisfying $\emptyset \in S \wedge \forall x \in S, \{x\}\cup x \in S$. The axiom says such a set exists. The natural numbers will be some subset of the set shown to exist by AI.

Christopher Hughes
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  • Thanks. I'm following exactly this definition of the natural numbers. Is it true that it isn't unique both in $ZF$ and $ZFC$? – Bertrand Haskell Jan 15 '20 at 18:40
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    Any limit ordinal is also an inductive set. – Christopher Hughes Jan 15 '20 at 19:36
  • @Elmar Inductive sets aren’t unique. The intersection of all inductive sets is unique. – spaceisdarkgreen Jan 15 '20 at 20:08
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    Well, except that you can’t take the intersection over all such sets in ZF, because that family is a proper class. Instead, you take a particular inductive set $A$, then take the intersection of all subsets of $A$ that are inductive; and then you show that the resulting set is in fact a subset of any inductive set and so the construction does not depend on the choice of inductive set. See for example here . – Arturo Magidin Jan 15 '20 at 20:25
  • @ArturoMagidin thanks I've just been not too precise. I know the definition, but can't understand how all this lead us (or not) to $\mathbb{N}={0,1,2,3,...}$ for any given model (if there is) – Bertrand Haskell Jan 15 '20 at 20:34
  • No, @Arturo. You can take the intersection without any issues. – Andrés E. Caicedo Jan 16 '20 at 04:34
  • @AndrésE.Caicedo: My understanding is that you can define the intersection of a family, but a family must be a set of sets. The intersection is defined to be the subset of the union such that the elements are in each set, and the union is only defined for families. I guess there are other ways this can be done, at least in the case at hand? – Arturo Magidin Jan 16 '20 at 06:20
  • You need the family to be nonempty. If I take the empty intersection I would end up with the set of all sets, given the defining property of intersection. – Christopher Hughes Jan 16 '20 at 09:27
  • If the family is nonempty, then I can just pick a set and take the intersection by axiom of specification. – Christopher Hughes Jan 16 '20 at 09:29
  • On a tangential note, this does point to one small disadvantage of this definition. While our definition of the set of natural numbers can and should break in the absence of AI, our definition of the class of natural numbers shouldn’t... and this one does. – spaceisdarkgreen Jan 16 '20 at 15:40
  • @spaceisdarkgreen: I usually define a “relative-natural-number” in terms of a specific inductive set; then show that the sets obtained relative to possibly distinct inductive sets are equal, and define “the” natural numbers as the result of applying the construction to any inductive set. – Arturo Magidin Jan 16 '20 at 18:38
  • @Arturo But how does this work when there aren’t any inductive sets? – spaceisdarkgreen Jan 16 '20 at 18:52
  • @spaceisdarkgreen: If there aren’t any inductive set, then there is no set of natural numbers. Why would it “work”? – Arturo Magidin Jan 16 '20 at 19:48
  • @Arturo But there is still a class of natural numbers. That was my point, the definition breaks for defining the class when the set doesn’t exist. You can define the class just fine by other means: something is a natural number if it is an ordinal with no lesser limit ordinals. – spaceisdarkgreen Jan 16 '20 at 20:36