I know the WOP is treated like axiom of a natural number, but I was curious if I can prove WOP defined for the set of natural numbers N by following:
Suppose A is a subset of N, which then obeys all axioms from Peano postulates, and let us assume that A does not have a least element. Then, 1 in A must be a successor to another natural number, which is 0. However, 0 is neither in A nor N. (If 0 is in N, then we argue that 0 must be a successor to -1, which is not part of N). Also, 1 in A being a successor contradicts with the Peano axiom (1 cannot be a successor). Hence, our assumption that A having no least element is not true. Therefore, A has a least element. Without out loss o generality, every subset of the N has a least element.
I understand this is silly proof, but I wanted to make sure that I have reasoning for WOP in N before moving on.