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Pitt MATH 1530 - HOMEWORK

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Chapter 4 15 Let f1,…,fN be functions from A ⊂ n to . Let mi be the maximum of fi, that is, mi = sup(fi(A)). Let f = S fi and m = sup(f(A)). Show that m ≤ S mi. Give an example where the equality fails. ∀x ∈ A fi(x) ≤ mi, i = 1,…,N Sfi(c) ≤ Smi f(x) ≤ Smi ⇒ m = sup(f(A)) ≤ Smi This fails for f1(x) = x, f2(x) = 1 – x on [0,1] 18 Let A ⊂ M be connected and let f: A →  be continuous with f(x) ∫ 0 for all x ∈ A. Show that f(x) > 0 for all x ∈ A or else f(x) < 0 for all x ∈ A. Suppose there exist points a,b ∈ A with f(a) < 0 and f(b) >0. Then by the intermediate value theorem there exists c ∈ (a,b) such that f(c) = 0. However, this contradicts f(x) ∫ 0 for all x ∈ A. Hence, either f(x) > 0 or f(x) < 0 for all x ∈ A. 24 Let f: A ⊂ M → N a. Prove that f is uniformly continuous on A iff for every pair of sequences xk,yk of A such that d(xk,yk) → 0, we have ρ(f(xk),f(yk)) → 0. “⇒” Since f is uniformly continuous on A, we have that ∀ε > 0 ∃δ > 0 such that if d(x,y) < δ then ρ(f(x),f(y)) < ε. If x and y are sequences xn and yn such that d(xn,yn) → 0, then ρ(f(xn),f(yn)) → 0. “›” If d(xk,yk) → 0 then we have x = xn and y = yn such that x = y. Then ρ(f(xk),f(yk)) < ε, ∀ε > 0 b. Let f be uniformly continuous, and let xk be a Cauchy sequence of A. Show that f(xk) is a Cauchy sequence. Let δ > 0 be such that if d(x,y) < δ for x,y ∈ A then ρ(f(x), f(y)) < ε. Since xk is a Cauchy sequence, there exists H(δ) such that d(xn,xm) < δ for all n,m > H(δ). Then for n,m > H(δ), we have ρ(f(xn),f(xm)) < ε. Therefore, f(xk) is a Cauchy sequence.c. Let f be uniformly continuous and N be complete. Show that f has a unique extension to a continuous function on cl(A). Let a be an accumulation point of A that is not in A. There exists a sequence xn in A that converges to a. This sequence is Cauchy, and its image in N is Cauchy, with an accumulation point b. Let f(a) = b. For any other sequence yn in A, we have that d(xn,yn) → 0. Then by the uniform continuity of f we have that every two image sequences represent the same point in N. If a is in cl(A), but not in A, the points of A within g of a map to points within ½ε of b. Let c be in cl(A), with f(c) = d. The distance from a to c is strictly less than g, so there is some r in A, that is within g of c. The image of r is within ε of both b and d, hence b and c are within ε of each


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Pitt MATH 1530 - HOMEWORK

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