123 Theorem 5.2.1 Let D be a nonempty compact subset of R and let f : D → R be a continuous function. Then f (D) is a compact subset of R. In particular, f (D) is closed and bounded. Proof: Take any sequence {yn} in f (D). Then for each n, there exists an ∈ D such that yn = f (an). Since D is compact, there exists a subsequence {ank } of {an} and a point a ∈ D such that lim ank = a ∈ D. k→∞ It now follows from Theorem 3.3.2 that lim ynk = lim f (ank )= f (a) ∈ f (D). k→∞ k→∞ Therefore, f (D) is compact. The fnal conclusion follows from Theorem 5.1.5 □ We now prove the Extreme Value Theorem in our more general context. Theorem 5.2.2 — Extreme Value Theorem. Suppose f : D → R is continuous and D is a compact set. Then f has an absolute minimum and an absolute maximum on D. Proof: Since D is compact, A = f (D) is closed and bounded (see Theorem 5.1.5). Let m = infA = inf f (x). x∈D In particular, m ∈ R. For every n ∈ N, there exists an ∈ A such that m ≤ an < m + 1/n. For each n, since an ∈ A = f (D), there exists xn ∈ D such that an = f (xn) and, hence, m ≤ f (xn) < m + 1/n. By the compactness of D, there exists an element x0 ∈ D and a subsequence {xnk } that converges to x0 ∈ D as k → ∞. Because 1 m ≤ f (xnk ) < m+ for every k, nk by the squeeze theorem (Theorem 2.1.3) we conclude limk →∞ f (xnk )= m. On the other hand, by continuity we have limk →∞ f (xnk )= f (x0). We conclude that f (x0)= m ≤ f (x) for every x ∈ D. Thus, f has an absolute minimum at x0. The proof is similar for the case of absolute maximum. □ Remark 5.2.1 The proof of Theorem 5.2.2 can be shortened by applying Theorem 5.1.4. However, we have provided a direct proof instead. The version of the Extreme Value Theorem presented in Chapter 3 is now a simple corollary of the more general version. Corollary 5.2.3 If f : [a,b] → R is continuous, then it has an absolute minimum and an absolute maximum on [a,b]. Corollary 5.2.3 follows immediately from Theorem 5.2.2, and the fact that the interval [a,b] is compact (see Example 5.1.4). The following theorem shows one important case in which continuity implies uniform continuity.
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