124 5.2 Continuity and Compactness Theorem 5.2.4 Let f : D → R be a continuous function. Suppose D is compact. Then f is uniformly continuous on D. Proof: Suppose by contradiction that f is not uniformly continuous on D. Then there exists ε0 > 0 such that for any δ > 0, there exist u,v ∈ D with |u− v| < δ and | f (u) − f (v)|≥ ε0. Thus, for every n ∈ N, there exist un,vn ∈ D with |un − vn|≤ 1/n and | f (un) − f (vn)|≥ ε0. Since D is compact, there exist u0 ∈ D and a subsequence {unk } of {un} such that unk → u0 as k → ∞. Then 1 |unk − vnk |≤ , nk for all k and, hence, we also have vnk → u0 as k → ∞. By the continuity of f , f (unk ) → f (u0) and f (vnk ) → f (u0). (5.1) Therefore, { f (unk ) − f (vnk )} converges to zero, which is a contradiction. The proof is now complete. □ We conclude this section with a second proof of Theorem 3.4.5 that does not depend on Theorem 3.4.4, but, instead, relies on the Nested Intervals Theorem (Theorem 2.3.2). Second Proof of Theorem 3.4.5: We construct a sequence of nested intervals as follows. Set a1 = a, b1 = b, and let I1 =[a,b]. Let c1 =(a + b)/2. If f (c1)= γ, we are done. Otherwise, either f (c1) > γ or f (c1) < γ. In the frst case, set a2 = a1 and b1 = c1. In the second case, set a2 = c1 and b2 = b1. Now set I2 =[a2,b2]. Note that in either case, f (a2) < γ < f (b2). Set c2 =(a2 + b2)/2. If f (c2)= γ, again we are done. Otherwise, either f (c2) > γ or f (c2) < γ. In the frst case, set a3 = a2 and b3 = c2. In the second case, set a3 = c2 and b3 = b2. Now set I3 =[a3,b3]. Note that in either case, f (a3) < γ < f (b3). Proceeding in this way, either we fnd some cn0 such that f (cn0 )= γ and, hence, the proof is complete, or we construct a sequence of closed bounded intervals {In} with In =[an,bn] such that for all n, (i) In ⊃ In+1,
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