Axiom of infinity and empty set

Question

The axiom of infinity is formulated as

$$\exists S ( \varnothing \in S \wedge (\forall x \in S) x \cup \{x\} \in S)$$

Can someone explain why the use of $\varnothing$ in the axiom of infinity makes sense, when the very existence of $\varnothing$ is predicated on it?

Answer 1

The Axiom of Existence states that the empty set exists. If you don't accept the Axiom of Existence as axiomatic, the Axiom of Infinity implies the existence of $\varnothing$, though you need another axiom to "extract" it from $S$.

Answer 2

The existence of the empty set can be proved from predicate calculus, the Rule of Generalization, and the Axiom of Separation; you don't need the Axiom of Infinity. You can find a formalized proof here:

http://us.metamath.org/mpegif/axnul.html

Of course in many treatments it is simply taken as an axiom.

Answer 3

You don't need the axiom of the empty set for the axiom of infinity.

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

The axiom states that there exists $S$ such that there is an element of $S$ which has no members, and $S$ is closed under successorship.