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I am reading Jech's "Set Theory". First, he states that the existence of the empty set follows from the axiom of infinity. The empty set is defined, using the separation schema as $$\emptyset = \{ u \in X\mid u \neq u \},$$ which the presupposes the existence of some set X. The existence of some set X, in his argument, follows from the existence of an inductive set. Then, he uses the empty set to define an inductive set in the axiom of infinity: $$\exists S [ \emptyset \in S \wedge (\forall x \in S) [x \cup \{x\} \in S]].$$ This argument seems to be circular to me. Am I wrong?

Asaf Karagila
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user130888
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1 Answers1

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The argument is not circular. You can write the axiom of infinity in a much longer form,

$$\exists S(\exists u(\forall z(z\notin u)\land u\in S)\land\forall x(x\in S\rightarrow\exists v(v\in S\land\forall w(w\in v\leftrightarrow w\in x\lor w=x))))$$ No reference to the empty set there.

Of course from this assertion we can prove the existence of a set which has no elements.

Asaf Karagila
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  • Or you can slip in the empty set as ${y\in S: y\neq y}$ and all the required ontological content is still contained in the axiom. – Malice Vidrine Feb 23 '14 at 08:25
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    Thanks for your answers. So the existence of the empty set follows from the fact that the inductive set contains it as an element? – user130888 Feb 23 '14 at 08:32
  • @user130888: Exactly. – Asaf Karagila Feb 23 '14 at 08:36
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    @user130888: Actually, it's not just that. It's also by the fact that from the existence of any set separation proves the existence of the empty set. – Asaf Karagila Feb 23 '14 at 09:24
  • Oh, yes, that's right – user130888 Feb 23 '14 at 10:22
  • @AsafKaragila Thank you for pointing me to this post indirectly; your comments here were the answer I needed. – Trixie Wolf Oct 31 '22 at 17:48
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    @TrixieWolf: You're very welcome! – Asaf Karagila Oct 31 '22 at 17:54
  • @AsafKaragila I do have one question about Malice's comment on the same topic, and perhaps I should pose it to them instead but you clearly have the knowledge to confirm. Is it well-defined to use Replacement on the contents of S within the definition of S, as Malice attempts? That feels impredicative to me, and I suspect you could cause a contradiction in ZF by allowing a use of Replacement in that way, but I'd need to hash it out a bit to see. – Trixie Wolf Oct 31 '22 at 17:59
  • @TrixieWolf: How is it impredicative? The formula $x\neq x$ is almost as simple as it gets. – Asaf Karagila Oct 31 '22 at 21:07