That's not an axiom we take; the study of real numbers takes some algebraic and order axioms, and a completeness axiom: Any non-empty set of real numbers with

That's not an axiom we take; the study of real numbers takes some algebraic and order axioms, and a completeness axiom: Any non-empty set of real numbers with an upper bound has a least upper bound. We can't prove your statement Given ε = 1/∞ then: ε = 0 because it is not well-defined, but we can prove this: If ε ≥ 0 and, for every natural number n, ε < 1/n, the n ε = 0. For suppose there exists an ε which is a counter-example, i.e. ε > 0 and ε < 1/n for every natural number n. Then the set S = { x : x a real number, x > 0 and x < 1/n for every natural n}, is non-empty, and has an upper bound (e.g. 1). So it has a least upper bound, say y. In particular, y is an upper bound, so 2y is not in S. It is > 0, so there must exist a natural number N for which 2y >= 1/N. But then y/2 > 1/4N, and 4N is also a natural number. So for any element x of S, x < 1/4N < y/2; thus y/2 is an upper bound which is less than y. This is a contradiction, so the claim is proved.