I've been reading some set theory lately, and I wanted to revisit a question I had in my first-year algebra course. We had to prove that the set of Zermelo ordinals $\mathbb{S}=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},...\}$ was a valid set under the ZFC axioms. I originally did this by using the axiom schema of replacement and defining a function $f:\mathbb{N}\rightarrow\mathbb{S}$ such that for every $n\in\mathbb{N}$ there is exactly once $s\in\mathbb{S}$. However, I wanted to try and re-do this using the recursion theorem as I believe this would be a cleaner approach. The obvious way to do this is as follows:
Let $g(x)=\{x\}$, then by said theorem there is a unique sequence $(f_n)_{n\in\mathbb{N}}$ s.t. $f_0=\emptyset$ and $f_{n+1}=g(f_n)$. The only problem is, letting $\mathbb{U}$ be the class of all sets, we have that $g\subset\mathbb{U}\times\mathbb{U}$ so it is thus not known to be a set, invalidating it's being a function and thus the entire argument. Is there an easy way to fix this or an alternative method to prove that $\mathbb{O}$ is a set using the recursion theorem.
The recursion theorem:https://en.wikipedia.org/wiki/Recursion#The_recursion_theorem
Because of the axiom of infinity we know that there exists an inductive set $X$. This is, $X$ satisfies the following property: $(\emptyset\in X)\wedge\forall x\in X(x\cup\{x\}\in X)$.
Power set axiom guarantees that $A=\mathscr P(X)$ is a set. Now you can consider $g\subseteq X\times A$, and by induction theorem $\mathbb N\subseteq X$.