Equivalence of Formulations of Axiom of Infinity for Zermelo Universe

Theorem
The following formulations of the  in the context of a Zermelo universe are equivalent:

Proof
Let the, in each of its formulations, be applied to a basic universe $V$ separately, as follows.

Formulation $1$ implies Formulation $2$
Let formulation 1 be taken as an axiom.

Then the class $\omega$ as so defined is a set.

But, by definition, $\omega$ is itself an inductive set.

Hence an inductive set exists.

Thus formulation 2 holds.

Formulation $2$ implies Formulation $1$
Let formulation 2 be taken as an axiom.

That is, there exists an inductive set, which we shall call $b$.

By the inductive set definition of the natural numbers:
 * $\forall n \in \omega: n \in b$

where $\omega$ denotes the set of natural numbers.

That is:
 * $\omega \subseteq b$

We have that:
 * $b$ is a set

and
 * $\omega$ is a subclass of $b$.

By the, $\omega$ is a set.

Thus formulation 1 holds.

Formulation $1$ implies Formulation $3$
Let formulation 1 be taken as an axiom.

In the basic universe $V$, we are able to create the set:
 * $S = \set {\O, \set \O}$

which is the natural number $2$.

By the, its power set is also a set:
 * $\powerset S = \set {\O, \set \O, \set {\set \O}, \set {\set \O, \set \O} }$

But $\powerset S$ is not a natural number.

Thus formulation 3 holds.

Formulation $3$ implies Formulation $2$
Let formulation 3 be taken as an axiom.

formulation 2 did not hold.

Then there would be no inductive sets.

But then vacuously every set is an element of all inductive sets.

Thus, by the inductive set definition of the natural numbers:
 * Every set is a natural number.

Thus it is seen that each of these formulations of the directly or indirectly implies each of the other two.

Hence the result.