# Axiom of Infinity implies existence of empty set. Is this circular?

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?

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

$$\exists S(\exists u(\forall z(z\notin u)\land u\in S)\land\forall x(x\in S\rightarrow\exists v(v\in S\land\forall w(w\in v\leftrightarrow v\in x\lor v=x))))$$

No reference to the empty set there.

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

• 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 8:25
• 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 8:32
• @user130888: Exactly. – Asaf Karagila Feb 23 '14 at 8:36
• @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 9:24
• Oh, yes, that's right – user130888 Feb 23 '14 at 10:22