1873_solutions

# That x n is a sequence of real numbers that x is a

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Unformatted text preview: 1 n for each n then, although x n is frequently in the set 1 it has the partial limit 1 that is not close to 1 . 4. Prove that a subset U of a metric space X is open if and only if every sequence in X that converges to a 167 member of U must be eventually in U. We know from a recent theorem that a set H is closed if and only if no sequence that is frequently in H can have a limit that doesn’t belong to H. Therefore if U is a set of real numbers then the set R U is closed if and only if no sequence with a limit in U can fail to be eventually in U. Of course the exercise can also be done directly. 5. Given that S is a subset of a metric space X and that x X, prove that the following conditions are equivalent: a. The point x is a limit point of the set S. b. There exists a sequence x n in the set S x such that x n x as n Ý. Solution: The assertion in this exercise follows at once from the corresponding theorem about limits of sequences and closure of a set, and from the fact that x is a limit point of S if and only if x S x. 6. Prove that if x n is a sequence in a metric space X then the set of all partial limits of x n is closed. Suppose that x n is a sequence in a metric space X and write the set of partial limits of x n as H. In order to show that H is closed we shall show that X H is open. Suppose that x X H. In other words, suppose that x is not a partial limit of x n . Choose a number  0 such that the condition x n B x, holds for at most finitely many integers n. Given any number y B x, , it follows from the fact that B x, is a neighborhood of y and the fact that x n belongs to this neighborhood of y for at most finitely many integers n that y is not a partial limit of x n . In other words, B x, RH and we have shown, as promised, that the set X H is open. 7. (This exercise is more advanced than the others.) Prove that if X is a separable metric space then every closed subset of X is the set of partial limits of some sequence. 1. Solution: Suppose that S is a closed subset of a separable metric space X. Using the fact that the subspace S of X must also be separable, choose a countable dense subset of S. We express this countable set in the form x n n  1, 2,  . Now we look at the sequence yn  x1, x1, x2, x1, x2, x3, x1, x2, x3, x4, x1, x2, x3, x4, x5,  . A point of X lies in the closure of the set x n n  1, 2,  if and only if it is a partial limit of the sequence y n . 8. Suppose that A and B are nonempty sets of real numbers and that for every number x y B we have x  y. Prove that the following conditions are equivalent: A and every number a. We have sup A  inf B. b. There exists a sequence x n in the set A and a sequence y n in the set B such that y n n Ý. xn 0 as Solution: From the information given about A and B we know that sup A inf B. In the event that sup A  inf B we know that for every sequence x n in A and every sequence y n in B we have y n x n inf B sup A  0 for every n and so we can’t have y n x n 0 as n Ý. Therefor...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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