124 5.2 Continuity and Compactness Theorem 5.2.4 Let f : D→Rbe a continuous function. Suppose Dis 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 anyδ >0, there exist u,v ∈Dwith |u−v| <δ and| f (u)−f (v)| ≥ε0. Thus, for every n∈N, there exist un,vn ∈Dwith |un −vn| ≤1/nand| f (un)−f (vn)| ≥ε0. Since Dis compact, there exist u0 ∈Dand a subsequence {unk}of {un}such that unk →u0 as k →∞. Then |unk −vnk| ≤ 1 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 first 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 first 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 find somecn0 such that f (cn0) =γ and, hence, the proof is complete, or we construct a sequence of closed bounded intervals {In}withIn = [an,bn] such that for all n, (i) In ⊃In+1,
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