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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 $\leq$?

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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.
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