More general Axiom of Infinity

Question

I was reading https://en.wikipedia.org/wiki/Axiom_of_infinity, in relation to Zermelo Set theory and was wondering why the axiom is posed as $$ \exists Z(\varnothing \in Z \wedge \forall x(x\in Z \to x\cup \{x\}\in Z)) $$ or an alternative mentioned here $$ \exists Z(\varnothing \in Z \wedge \forall x(x\in Z \to \{x\}\in Z)). $$ My question is that, is there a reason for these presentations that specify exactly what kinds of sets are in the infinite set $Z$? Would it be insufficient to instead take $$ \exists Z(\varnothing \in Z \wedge \forall x(x\in Z\to \exists y(y\in Z \wedge x\in y))) $$ to be the axiom?

Answer 1

The version you suggest will work in the standard framework of $\sf ZF$, or at least in the presence of Regularity, as we can prove that $\in$ does not contain any cycles, so the set you describe will necessarily be infinite.

In fact, we don't even care if $\varnothing$ is a member or not at this point. All we need is that $Z$ is not empty.

However, it is consistent without Regularity that $A=\{\varnothing, A\}$ exists. In that case, $A=Z$ will satisfy your requirements and is certainly finite. This can be extended to produce "longer cycles" as well.

The key point in the standard definition is that it will ensure $Z$ must be infinite even if Regularity fails, as it allows us to prove there is a smallest such $Z$, which must not contain any $\in$-cycles.