Josh Zucker writes:
I'm not sure exactly what you're taking as your axioms here, so I'm not sure if I'm on the right track at all.
They include all the usual first-order axioms governing N, Z, Q, and R, but no second-order axioms other than the axiom of completeness.
This last line isn't making sense to me. x is a lower bound for E, and it is an integer, so x is in E, and thus not in S.
I fumbled the end of that proof. Plus, the whole thing was too terse to be readable. I should have written: Let S be some set of positive integers that contains 1 and has the property that for all n in S, n+1 is also in S. Suppose there exists some positive integer that is not in S. Let x be the greatest lower bound of the set E of all such exceptional positive integers. If x is not an integer, then we can find x' > x with no integers between x and x', and x' is also a lower bound on E, contradicting the choice of x as the greatest lower bound. Hence x is an integer. It follows that x-1 and x+1 are integers, and we'll make the assumption that there are no integers between x-1 and x, or between x and x+1. Now since x is a lower bound for E (remember, it's the greatest lower bound of E) and since x+1 isn't a lower bound for E (since it's greater than the greatest lower bound of E), there must be some element of E in the interval [x,x+1). But there's only one integer in this interval, namely x, so we must have x in E. That is, the integer x is not in S. Since the set S is inductive, it follows that x-1 is not in S. Hence x-1 is in E. But then x is not a lower bound on E: contradiction. (I grant that it seems a bit odd to derive a fundamental property of Z from a property of R, especially if you accept Kronecker's ontology of number!) Jim Propp