I'm following a proof in Jech's book that every well ordered set is isomorphic to a unique ordinal and hitting a point where I'm not sure why a certain move is justified. He writes
Proof. The uniqueness follows from Lemma 2.7. Given a well-ordered set $W$, we find an isomorphic ordinal as follows: Define $F(x) = \alpha$ if $\alpha$ is isomorphic to the initial segment of $W$ given by $x$. If such an $\alpha$ exists, then it is unique. By the Replacement Axioms, $ F(W) $ is a set. For each $x \in W$, such an $\alpha$ exists (otherwise consider the least $x$ for which such an $\alpha$ does not exist). If $\gamma$ is the least $\gamma \not\in F(W)$ then $F(W) =\gamma$ and we have an isomorphism of $W$ onto $\gamma$.
I've filled in all the details of this proof for myself up to the point where he assumes the class of ordinals not in the image of the function has a least element. Since this isn't a set, I'm not sure what can justify this. I could try to prove that every nonempty class of ordinals has a least element but that seems so substantial that it's hard to imagine it was a detail intentionally omitted. Am I missing something simple?
[Edit: Since the question has been associated with another question, it wants me to edit this one to show why the questions are different. I take it as pretty obvious how the questions are different, since I'm asking about something totally different from what was asked in that linked question. They're about the same theorem and proof, but I'm asking about a part of the proof not explained in the linked question. So ... here's my edit. I think someone is just a little over-zealous with the close-question button.]
Suppose $C$ is a nonempty class of ordinals, and pick some $\alpha\in C$. Either $\alpha$ is the least element of $C$, or it isn't. If it is, we're done. Otherwise, think about $$\alpha\cap C.$$ This is a set of ordinals, and nonempty by assumption on $\alpha$, so it has a least element $\beta$. Now do you see how to argue that $\beta$ is in fact the least element of $C$?