In article <email@example.com>, Dan Christensen <Dan_Christensen@sympatico.ca> wrote:
> Assuming the implicit successor functions in each case, these sets are all > isomorphic to one another, i.e. you are simply using different symbols for > the various elements, but the successor relation is preserved, as is the > induction principle. As I said, such structures is embedded in EVERY infinite > set. So, you are on fairly safe ground to assume the existence of one such > structure at the beginning of a proof.
If one does not have something like an axiom of choice, there is no guarantee that any infinite set need have any well-orderable subsets.
So in ZFC you are correct but not necessarily so in ZF. --