Let (X, d) be a metric space and suppose r> 0 and (Pn) is a sequence...
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Let (X, d) be a metric space and suppose r> 0 and (Pn) is a sequence from X such that d(pn, Pm) ≥r for nm. Show E = {pn: ne N} is closed. chatgpt to scott: To show that E is closed, we must prove that its complement is open. Let a be a point in the complement of E, so x is not equal to any pn. By the assumption that d(pn. Pm) ≥r for nm, we know that the distance between any two distinct points in the set E is at least r. Thus, we can find an open ball of radius r/2 centered at r that does not intersect the set E. This shows that the complement of E is open, so E is closed. (A) Point our a serious (mathematical) error in chat's proof. Give a (correct) proof of the statement. (B) Let (X, d) be a metric space and suppose r > 0 and (pr) is a sequence from X such that d(pn, Pm) ≥r for nm. Show, directly from the open cover definition of compactness, that E = {pn: ne N} is not compact. (C) Let fn: [0, 1] → R, for n E N, denote the sequence of functions described as follows: Given n E N, let mn denote the midpoint of the interval (1). The function fn is 0) outside the interval (1) and between 1 and mn its graph (in R2) is the line segment connecting (1,0) and (mn, 1); and between mn and its graph is the line segment connecting (mn, 1) to (,0). n+1 n In the metric space (C([0, 1], R), doo), the set B = {fe C([0, 1], R): do(f,0) = ||f||0 ≤ 1} is closed and bounded (proofs not required). Show, using the sequence (f) (or otherwise), B is not compact. Let (X, d) be a metric space and suppose r> 0 and (Pn) is a sequence from X such that d(pn, Pm) ≥r for nm. Show E = {pn: ne N} is closed. chatgpt to scott: To show that E is closed, we must prove that its complement is open. Let a be a point in the complement of E, so x is not equal to any pn. By the assumption that d(pn. Pm) ≥r for nm, we know that the distance between any two distinct points in the set E is at least r. Thus, we can find an open ball of radius r/2 centered at r that does not intersect the set E. This shows that the complement of E is open, so E is closed. (A) Point our a serious (mathematical) error in chat's proof. Give a (correct) proof of the statement. (B) Let (X, d) be a metric space and suppose r > 0 and (pr) is a sequence from X such that d(pn, Pm) ≥r for nm. Show, directly from the open cover definition of compactness, that E = {pn: ne N} is not compact. (C) Let fn: [0, 1] → R, for n E N, denote the sequence of functions described as follows: Given n E N, let mn denote the midpoint of the interval (1). The function fn is 0) outside the interval (1) and between 1 and mn its graph (in R2) is the line segment connecting (1,0) and (mn, 1); and between mn and its graph is the line segment connecting (mn, 1) to (,0). n+1 n In the metric space (C([0, 1], R), doo), the set B = {fe C([0, 1], R): do(f,0) = ||f||0 ≤ 1} is closed and bounded (proofs not required). Show, using the sequence (f) (or otherwise), B is not compact.
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Related Book For
Introduction to Algorithms
ISBN: 978-0262033848
3rd edition
Authors: Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest
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