Question

(For this part, view Z as a subset of R.) Let A be a non-empty subset of Z with an upper bound in R. By the Completeness Axiom, A has a least upper bound in R, which we shall denote by m. Show that m ∈ A. (Hint: As m - 1 < m, which means that m - 1 cannot be an upper bound in R for A, we find that m - 1 < k ≤ m, or, equivalently, k ≤ m < k + 1 for some k ∈ A. Assume that k < m (which means that k cannot be an upper bound in R for A) and then apply the last part of Problem 1 to derive a contradiction.)

Answer 1