28 1.6 Applications of the Completeness Axiom Proof: Let us assume by contradiction that Nis bounded above. Since Nis nonempty, by the Completeness Axiom the supremum of Aexists and it is a real number. Say α=supN By Proposition 1.5.1 (2’) (withε =1), there exists n∈Nsuch that α−1<n≤α. But thenn+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 anyx ∈R, there exists n∈Nsuch that x <n. (ii) For anyε >0, there exists n∈Nsuch that 1/n<ε. (iii) For any x >0 and for anyy ∈R, there exists n∈Nsuch that y <nx. (iv) For any x ∈R, there exists m∈Zsuch that m−1≤x <m. Proof: (i) Fix any x ∈R. Since Nis not bounded above, x cannot be an upper bound of N. Thus, there exists n∈Nsuch that x <n. (ii) Fix any ε >0. Then 1/ε is a real number. By (i), there exists n∈Nsuch 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. Define the set A={n∈N: x <n}. From part (i), Ais nonempty. Since Ais a subset of N, by the Well-Ordering Property of the natural numbers, Ahas a smallest element ℓ. In particular, x < ℓ andℓ−1 is not in A. Since ℓ ∈N, either ℓ−1 ∈Nor ℓ−1 =0. If ℓ−1 ∈N, since ℓ−1̸ ∈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 withm=ℓ. In the case x ≤0, by part (i), there exists N∈Nsuch 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. □
RkJQdWJsaXNoZXIy NTc4NTAz