TITLE: Equivalence between different forms of the Axiom of Infinity
QUESTION [6 upvotes]: In Zermelo-Frankel set theory, the Axiom of Infinity is often stated as

"There exists a set $X$ such that $\emptyset \in X$ and such that if $y\in X$ then $S^{+}_1(y)\in X$",

where we take the successor function as $S^{+}_1(y) = y\cup \{y\}$. I believe the original formulation for the Axiom of Infinity by Zermelo had a different form, where the successor function was instead $S^{+}_2(y) = \{y\}$.
Let's call the form of the Axiom of Infinity with $S^{+}_1$ "Infinity 1" and the form with $S^{+}_2$ "Infinity 2". Changing from "Infinity 2" to "Infinity 1" made the axiom easier to work with, but the two formulations should be equivalent.
How can we prove that "Infinity 1" and "Infinity 2" are equivalent under the remaining ZF axioms?

REPLY [3 votes]: If you're allowed to use Replacement, you can recursively define a map from an inductive set of the second kind to an inductive set of the first kind by just setting $f(\varnothing)=\varnothing, f(S_2^+(y))=S_1^+(f(y))$, and then the image, which is a set by Replacement, will be an inductive set of the first kind.