Introduction to Mathematical Analysis I - Second Edition

27 Theorem 1.6.1 — The Archimedean Property. The set of natural numbers is unbounded above. Proof: Let us assume by contradiction that N is bounded above. Since N is nonempty, α = sup N exists and is a real number. By Proposition 1.5.1 (with ε = 1), there exists n ∈ N such that α − 1 < n ≤ α . But then n + 1 > α . This is a contradiction since n + 1 is a natural number. The following theorem presents several immediate consequences. Theorem 1.6.2 The following hold: (a) For any x ∈ R , there exists n ∈ N such that x < n ; (b) For any ε > 0, there exists n ∈ N such that 1 / n < ε ; (c) For any x > 0 and for any y ∈ R , there exists n ∈ N such that y < nx ; (d) For any x ∈ R , there exists m ∈ Z such that m − 1 ≤ x < m . Proof: (a) Fix any x ∈ R . Since N is not bounded above, x cannot be an upper bound of N . Thus, there exists n ∈ N such that x < n . (b) Fix any ε > 0. Then 1 / ε is a real number. By (1), there exists n ∈ N such that 1 / ε < n . This implies 1 / n < ε . (c) We only need to apply (a) for the real number y / x . (d) First we consider the case where x > 0. Define the set A = { n ∈ N : x < n } . From part (a) , A is nonempty. Since A is a subset of N , by the Well-Ordering Property of the natural numbers, A has a smallest element ` . In particular, x < ` and ` − 1 is not in A . Since ` ∈ N , either ` − 1 ∈ N or ` − 1 = 0. If ` − 1 ∈ N , since ` − 1 6∈ A we get ` − 1 ≤ x . If ` − 1 = 0, we have ` − 1 = 0 < x . Therefore, in both cases we have ` − 1 ≤ x < ` and the conclusion follows with m = ` . In the case x ≤ 0, by part (1), there exists N ∈ N such that | x | < N . In this case, − N < x < N , so x + N > 0. Then, by the result just obtained for positive numbers, there exists a natural number k such that k − 1 ≤ x + N < k . This implies k − N − 1 ≤ x < k − N . Setting m = k − N , the conclusion follows. The proof is now complete.