Real Analysis: Completeness Axiom

Question

Suppose that k > 0 and S is a subset of R is a bounded set. For any set S, we define k * S = {k * x: x is an element of S}. Show that sup (k * S) = k * sup (S).

a) Show that sup (k * S) <= k * sup S

b) Show that k * sup S <= sup (k * S)

Answer 1

If s is in S, then s ≤ sup S; as k > 0, then ks ≤ k • sup S. Thus k • sup S is an upper bound for k • S, which implies a).

Given e > 0, sup S - e/k is not an upper bound for S. So there is some s in S such that s > sup S - e. We obtain sup(k • S) ≥ ks > k • sup S - e. Since e was arbitrary, b) follows.