I'm currently reading about recursion, working on various exercises since I haven't formally studied the material before. The main book that I'm following is Kunen's Set Theory book
There's an exercise which asks to derive the Infinity Axiom using a lemma. This lemma is as follows:
Lemma: For a relation $R$ and a class A, if R is set-like on $A$, then $R^*$ is set-like on A.
His book defines $R^*$ as the transitive closure of $R$ on $A$, where $R$ is a relation and $A$ is a class. $R$ is set-like on $A$ iff $\{ x \in A : xRy\}$ is a set for all $y \in A$. The transitive closure $R^*$ is defined by $xR^*y$ iff there exists a path in $A$ from $x$ to $y$.
For the exercise, he provides a hint: let $A$ be the (possibly proper) class of all natural numbers and let $xRy$ iff $x = y + 1$.
My Attempt
I defined $A$ and $R$ as in the hint. Then, I considered $\{x \in A : xRy \} = \{x \in A : x = y + 1\}$. Since this is a set, $R$ is set-like. By the Lemma, $R^*$ is set-like, meaning that $\{x \in A : xR^*y \}$ is a set. I think $R^*$ would be the relation $x > y$. I'm guessing this would somehow lead to the fact that $\omega$ is set, which is equivalent to the Axiom of Infinity, but I'm just not seeing it.
Any help/hints with this exercise would be greatly appreciated.
HINT:
Note that $R^*$ being set-like means that given $n$, $\{k\in\omega\mid k\mathrel{R^*}n\}$ is a set. What happens when you set $n=0$?