Definitons. A poset $P$ is directed if every finite subset has an upper bound in $P$. It is directed-complete if every directed subset has a least upper bound in $P$. For any poset $P$ let $[P\to P]$ denote the set of all order-preserving maps $P\to P$, ordered pointwise.
Lemma. $[P\to P]$ is directed-complete if $P$ is.
Theorem. Suppose $P$ is directed-complete, $f:P\to P$ order-preserving and inflationary. Then there is a unique smallest set $C\subseteq [P\to P]$ satisfying
- $f\in C$
- $C$ is closed under composition
- $C$ is closed under joins of directed subsets
Attempt of proof. Let
- $C_0 = \{f\}$
- $C_{\alpha+1} = \{f\circ g : g\in C_\alpha\}$
- $C_\lambda = \sup \bigcup_{\gamma<\lambda}C_\gamma$
Take $C=\bigcup_{\alpha\in ON}C_\alpha$. Note that $C$ and every subset of $C$ is directed as $\sup\{f^{\alpha_1},\ldots,f^{\alpha_n}\}=\max\alpha_i$. This the supremum in 3. is always defined due to the lemma.
Problem. I literally only applied composition and sup so my $C$ can hardly be too big, can it? But don't I have $C=\{f^\alpha : \alpha\in ON\}$ now? But this is not a set because otherwise, by replacement, $ON$ was a set.
I think I'm a bit confused here. Can someone elucidate?
No, you don't have a problem. Nobody told you that doing these things will enlarge your set. At some point, your sequence is going to stabilize. Namely, there will be some $\alpha$ such that $C_\alpha=C_\beta$ for all $\beta>\alpha$.
Try and see what happens if $f=\operatorname{id}$. This understanding extends to any situation when $C_0$ is already closed under compositions and suprema.