28 1.6 Applications of the Completeness Axiom Proof: Let us assume by contradiction that N is bounded above. Since N is nonempty, by the Completeness Axiom the supremum of A exists and it is a real number. Say α = supN By Proposition 1.5.1 (2’) (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: (i) For any x ∈ R, there exists n ∈ N such that x < n. (ii) For any ε > 0, there exists n ∈ N such that 1/n < ε. (iii) For any x > 0 and for any y ∈ R, there exists n ∈ N such that y < nx. (iv) For any x ∈ R, there exists m ∈ Z such that m− 1 ≤ x < m. Proof: (i) 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. (ii) Fix any ε > 0. Then 1/ε is a real number. By (i), there exists n ∈ N such that 1/ε < n. This implies 1/n < ε. (iii) We only need to apply (i) for the real number y/x. (iv) First we consider the case where x > 0. Defne the set A = {n ∈ N : x < n}. From part (i), A is nonempty. Since A is a subset of N, by the Well-Ordering Property of the natural numbers, A has a smallest element ℓ. In particular, x <ℓ and ℓ − 1 is not in A. Since ℓ ∈ N, either ℓ − 1 ∈ N or ℓ − 1 = 0. If ℓ − 1 ∈ N, since ℓ − 1 ̸ ∈ A we get ℓ − 1 ≤ x. If ℓ − 1 = 0, we have ℓ − 1 = 0 < x. Therefore, in both cases we have ℓ − 1 ≤ x <ℓ and the conclusion follows with m = ℓ. In the case x ≤ 0, by part (i), 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. □
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