:Theorem 4.4.5 (Sequential Criterion for Absence of Uniform Conti-nuity). A function f AR fails to be uniformly continuous on A if andonly if there exists a particular eo > 0 and two sequences (xn) and (yn) in Asatisfyingxn - Yn → 0but|f(xn) — f(yn)|≥ €0. Exercise 4.4.1. (a) Show that f(x) = x³ is continuous on all of R.(b) Argue, using Theorem 4.4.5, that f is not uniformly continuous on R.(c) Show that f is uniformly continuous on any bounded subset of R.

Answered: Image /qna-images/answer/97c52df9-e0ea-4811-9ba4-fc41933ab89b.jpg

Answered: : Theorem 4.4.5 (Sequential Criterion…

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter1: Fundamentals

Section1.5: Permutations And Inverses

Problem 5E: Let f:AA, where A is nonempty. Prove that f a has right inverse if and only if f(f1(T))=T for every...

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Q1 Theorem 18.1 states that "Let f be a continuous real-valued function on a closed interval [a, b]. Then f is a bounded function.". Read the proof of Theorem 18.1 with [a, b] replaced by (a, b). Where does it break down? Explain.
2. Is the proof correct? Either state that it is, or circle the first error and explain what is incorrect about it. If the proof is not correct, can it be fixed to prove the claim true? Claim: If f : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, and f(1) = g(1), then the function h : (0, 2) → R, defined by h(æ) = { F(x) for x € (0, 1] | 9(x) for x e [1, 2) is uniformly continuous on (0, 2). Proof: Let e > 0. Since f is uniformly continuous on (0, 1], there exists d1 > 0 such that if x, y E (0, 1] and |æ – y| 0 such that if æ, y E [1, 2) and |r – y| < d2, then |9(x) – 9(y)| < €/2. Let 8 = min{d1, &2}. Now suppose r, Y E (0, 2) with x < y and |x – y| < 8. If x, y E (0, 1], then |x – y| < 8 < 81 and so |h(x) – h(y)| = |f (x) – f(y)|< e/2 < e. If x, y € (1,2), then |æ – y| < 8 < d2 and so |h(x) – h(y)| = |9(x) – g(y)| < e/2 < e. If x € (0,1) and y E (1,2), then |x – 1| < |x – y| < 8 < dị and |1 – y| < |æ – y| < 8 < 82 so that |h(x) – h(y)| = |f (x) – f(1) + g(1) –…
Is the proof correct? Either state that it is, or circle the first error and explain what isincorrect about it. If the proof is not correct, can it be fixed to prove the claim true?Claim:If f : (0, 1] → R and g : [1, 2) → R are uniformly continuous on their domains, andf(1) = g(1), then the function h : (0, 2) → R, defined by h(x) = f(x) for x ∈ (0, 1] g(x) for x ∈ [1, 2),is uniformly continuous on (0, 2).Proof:Let e > 0.Since f is uniformly continuous on (0, 1], there exists δ1 > 0 such that if x, y ∈ (0, 1] and|x − y| < δ1, then |f(x) − f(y)| < e/2.Since g is uniformly continuous on [1, 2), there exists δ2 > 0 such that if x, y ∈ [1, 2) and|x − y| < δ2, then |g(x) − g(y)| < e/2.Let δ = min{δ1, δ2}.Now suppose x, y ∈ (0, 2) with x < y and |x − y| < δ.If x, y ∈ (0, 1], then |x − y| < δ ≤ δ1 and so |h(x) − h(y)| = |f(x) − f(y)| < e/2 < e.If x, y ∈ [1, 2), then |x − y| < δ ≤ δ2 and so |h(x) − h(y)| = |g(x) − g(y)| < e/2 < e.If x ∈…
Exercise 3.3.10: Suppose f: [0, 1] [0, 1] is continuous. Show that f has a fixed point, in other words, show that there exists an x = [0, 1] such that f(x): = X.
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AP1) Prove that f : E R; x → x is uniformly continuous if E is bounded. AP2) Prove that f: R R; X→ x*is not uniformly continuous.
Please do exercise 3.2.10 with detailed explanations

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