1873_solutions

Space r now we repeat the same argument in r k we

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Unformatted text preview: S E is open in the metric space S. Using the fact that E choose a point x E. Since V is a neighborhood of x and x S we know that S V . This shows that S E . The same argument applied to the set S E shows that S E is a nonempty open subset of the metric space S. Therefore the space S fails to be connected, as we promised. 22. True or false? If S is a bounded subset of a metric space X then the set S is also bounded. This statement is true. Suppose that S is a bounded subset of a metric space X and choose a number p such that the inequality d x, y p holds whenever x and y belong to S. We shall now observe that the ame inequality holds for all members x and y of S. In other words, diam S p. 125 To obtain a contradiction, suppose that x and y belong to S and that d x, y  p. We define   d x, y p. Using the fact that x and y are close to S we choose members u and v of S such that u B x, and v B y,  . We observe that 2 d x, u  d u, v  d v, y    d u, v    d u, v    d u, v  d x, y 2 2 which implies that d u, v  p, contradicting the way in which p was chosen.  2 d x, y p 23. True or false? If S is a totally bounded subset of a metric space X then the set S is also totally bounded. This statement is true. Suppose that S is a totally bounded subset of a metric space X and that  0. Using the fact that S is totally bounded we choose a positive integer n and points x 1 , x 2 , , x n such that n B S xj, j1 . 2 We shall now show that n  B xj, S . j1 Suppose that y S. Choose a member x of S such that d x, y  x B x j , 2 . Since d y, x j we see that y d y, x  d x, x j  2  2 . Choose j such that 2 B xj, . 24. Prove that the closure of a convex subset of R k is also convex. 25. Suppose that S is a convex subset of R k , that x is an interior point of S, that y that the point 1 t xy is an interior point of S. S and that 0  t  1. Prove 26. This exercise refers to the notion of a subgroup of R that was introduced in an earlier exercise. That exercise should be completed before you start this one. a. Given that H and K are subgroups of R, prove that the set H  K defined in the sense of an earlier exercise is also a subgroup of R. To prove that H  K is a subgroup of R we need to show that H  K is nonempty and that the sum and difference of any members of H  K must always belong to H  K. To show that H  K is nonempty we use the fact that H and K are nonempty to choose x H and y K. Since x  y H  K we have H  K . Now suppose that w 1 and w 2 are any members of the set H  K. Choose members x 1 and x 2 of H and members y 1 and y 2 of K such that w 1  x 1  y 1 and w 2  x 2  y 2 . Since the numbers x 1  x 1 and x 1 x 2 belong to H and the numbers y 1  y 2 and y 1 y 2 belong to K, and since w1  w2  x1  x2  y1  y2 and w1 w2  x1 x2  y1 y2 we see at once that w 1  w 2 and w 1 w 2 belong to H  K. b. Prove that if a, b and c are integers and if a 2  b 3 c then a  b  c  0. 126 Solution: From the equation a 2  b 3 c we see...
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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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