# Prove that (if $R$ is set-like, then its transitive closure is set-like) implies the axiom of replacement

This is exercise I.9.6 from Kunen's Set Theory (2011). $$R$$ is set-like on the class $$A$$ iff $$y \in A$$ implies $$\{ x \in A: xRy \}$$ is a set. I am not so sure about this, but it is implied that this can be done without using the axiom of choice.

The obvious choice of $$R$$ is $$xRy$$ iff $$x= F(y)$$, however what I am most unsure of is how to pick the class such that it cointains all of $$A$$ and all of $$F(A)$$. I think the simple $$A \cup F(A)$$ doesn't work because for an $$x \in A$$, $$F(F(x))$$ may not be a set and hence $$R$$ wouldn't be set-like. This reasoning seems to apply to any set $$B$$, since $$A \cup F(A)$$ must be cointained in $$B$$ in order for the transitive closure to include $$F(A)$$.

As an specific example, consider the set $$\mathbb{N}$$ of natural numbers and the formula $$\phi(x,y)$$ "$$y$$ is $$\mathbb{N}$$ if $$x$$ is a natural number and, $$y$$ is $$\{w: w=w\}$$ if $$x$$ is $$\mathbb{N}$$". Clearly while $$R$$ is set-like in $$\mathbb{N}$$, it is not set-like in $$\mathbb{N} \cup F(\mathbb{N})$$, and in general I can't modify $$\phi(x,y)$$ as to exclude the $$x$$s in $$F(A)$$, as they may be part of $$A$$, too.

So, how can we prove it?