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hmw3s - 18.100B Fall 2002 Solutions to Homework 3 Rudin(1...

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18.100B, Fall 2002, Solutions to Homework 3 Rudin: (1) Chapter 2, Problem 6 Done in class on Thursday September 26. Here E X is a subset of a metric space and E is the set of limit points, in X, of E. (a) Prove that E is closed. If p X is a limit point of E then for each r > 0 , B ( p, r ) E q is not empty. Since q is a limit point of E and r d ( p, q ) > 0 , B ( q, r d ( p, q )) E is an infinite set. By the triangle inequality, B ( q, r d ( p, q )) B ( p, r ) so B ( p, r ) E is also infinite and p is therefore a limit point of E, i.e. p E . Thus E contains each of its limit points and it is therefore closed. (b) Prove that E and E have the same limit points. If p is a limit point of E then it is a limit point of E since E E. If p 1 is a limit point of E then B ( p, n ) ( E \{ 0 } ) decreases with n ; either it is infinite for all n or it is empty for large n. We show that the second 1 case cannot occur. Indeed this woould imply that B ( p, n ) ( E \{ p } ) is infinite for all n and hence that p is a limit point of E ; by the preceding result it is then a limit point of E contradicting the assumption that it is not. Thus a limit point of E is a limit point of E.

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hmw3s - 18.100B Fall 2002 Solutions to Homework 3 Rudin(1...

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