Here's my purported proof of the principle of induction, modulo a couple of tacit pesky details that can all be settled by an appeal to the assumption that 1 is the smallest positive integer: 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 m 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. Now since x is a lower bound for E and x+1 isn't, we must have x in S and x+1 not in S, which is also a contradiction. Jim