Is defining a successor operation equivalent to defining a well-ordering operation?

Question

I've read that a well-ordering operation on a set implies a successor operation on that set. Is the converse true? That is, if, given some well-ordering operation $\leq$ on $A$, $succ\; a$ is the successor of $a \in A$, is defining $succ$ equivalent to defining $\leq$?

Answer 1

Your question is not entirely well-defined, but there are two general reasons why something like this would fail.

1. The ordering $(\mathbb{Z}, \leq)$ also has a successor operation, but is clearly not well-ordered.
2. If you consider the ordinal $\omega + \omega$, and only look at the successor relation, then how are you supposed to tell whether $0 < \omega$ or $\omega < 0$? Both are not the successor of any element, but that is about the only property we can formulate with successors.