# Every transitive $\in$-linearly ordered set is $\in$-well ordered without axiom of foundation

I try to prove that every ordinal number $(\alpha,\in)$ is well-ordered, where an ordinal number is defined as a transitive $\in$-linearly ordered set.

So all I have to show is, that every non-empty subset of $\alpha$ has an $\in$-minimal element. That follows quite simple by axiom of foundation, but how can I prove this without foundation?

• Don't you need foundation for this definition of ordinal to work? – Henno Brandsma Jun 17 '16 at 16:34
• See en.wikipedia.org/wiki/…, under "other definitions", which assumes regularity (foundation) for the equivalence. – Henno Brandsma Jun 17 '16 at 16:36
• @Henno: I guess I do need foundation to receive an equivalent definition to yours. Just consider it to be a transitive and linearly ordered set then without calling it ordinal. – E. Noether Jun 17 '16 at 16:41

There's a good reason you're having trouble - this is not provable without Foundation: it is consistent with $ZFC^-$ (that is, $ZFC$ without Foundation) that there is a sequence of sets $a_i$ ($i\in\omega$) such that $a_i=\{a_j: j>i\}$. Then $a_0$ is transitive and linearly ordered by "$\in$," but is clearly not well-ordered by "$\in$".
I suspect that AFA also provides counterexamples to your claim. However, this is a bit trickier; in fact, AFA disproves the existence of the counterexample I describe above! (In AFA we have $a_0=a_1=...$, so the result is a singleton - a (in fact, the) Quine atom.)