18.100pset3 - 18.100B/C Fall 2008 Solutions to Homework 3 1...

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18.100B/C: Fall 2008 Solutions to Homework 3 1. By condition (a), E is nonempty. By condition (b), E is open. By condition (c), every point that is a limit of a sequence of points in E is itself in E . Note that if e is a limit point of E , we can choose a sequence of points of E whose limit is e : choose x n to be a arbitrary point of E that lies in the ball B ( e, 1 n ) and then lim n →∞ x n = e . Thus, E contains all its limit points and so E is closed. By Homework 2, Problem 8, a nonempty open and closed subset of a connected metric space must be the entire space, so E = X as desired. 2.(a) Equivalent: Suppose x n p . Then for all r > 0 there is some M N such that n M implies d ( x n , p ) < r . Choose any N and define r = N - 1 , and with the corresponding M set ε = M - 1 , so M = ε - 1 . Then we have that for all n ε - 1 , d ( x n , p ) < N - 1 . Thus, for every N there is some ε > 0 such that for all n ε - 1 , d ( x n , p ) < N - 1 and so statement (a) is true. Conversely, suppose that statement (a) holds of a sequence x n and a point p . Then for all M N there is some r > 0 such that for all n r - 1 , d ( x n , p ) < M - 1 . Choose any ε > 0 and choose some M ε - 1 , so M - 1 ε . For this value of M these is an associated value of r , and pick some integer N r - 1 . Then for all n N we have d ( x n , p ) < M - 1 ε . Thus, for all ε > 0 there is some N N such that n N implies d ( x n , p ) < ε and so x n p . (The moral of this question: variables are just placeholders, and there’s nothing special about the letters N and ε .) (b) Not equivalent: consider X = R and x n = 1 n . Then lim n →∞ x n = 0. However, for any N N , with ε = 1 N +1 and n = N we have d ( x n , 0) = d ( x N , 0) = 1 N > 1 N +1 = ε , so the given statement is false. (In fact, statement (b) is equivalent to the statement “ x n is eventually constant.”) (Moral: quantifiers matter.) (c) Not equivalent: consider X = R and the sequence x n defined by x 2 n = 1 n , x 2 n - 1 = n . (The first few terms of this sequence are 1 , 1 , 2 , 1 2 , 3 , 1 3 , . . . .) The range of this sequence has a single limit point at 0, but the sequence x n has no limit. (Another counterexample is the sequence defined by x 2 n = 1 n , x 2 n - 1 = 1, since 1 is a limit of a subsequence, but an isolated point of the range.) (d) Not equivalent: consider X = R and x n = ( - 1) n , then the range {- 1 , 1 } is finite and hence B r ( p ) C ∩ {- 1 , 1 } is finite for all r > 0. However, the sequence is known to diverge. (On the other hand, the statement “For all r > 0 the set { n N | x n B r ( p ) C } is finite.” is equivalent to convergence; see Rudin 3.2(a).) 3. Suppose s n z . Then for all ε > 0 there is some N N such that for all n N we have d C ( s n , z ) < ε . But d C ( s n , z ) = | s n - z | ≥ || s n | - | z || = d R ( | s n | , | z | ), so for all ε > 0 there is some N N such that for all n N we have d R ( | s n | , | z | ) < ε , i.e., | s n | → | z | and so | s n | converges.
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