1873_solutions

Prove that the interval 1 3 is not an open subset of

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Unformatted text preview: of a metric space X, d and that H is a closed subset of the metric space S, d , prove that H must be closed in the metric space X, d . Using Exercise 11 and the fact that S H is open in the space S, d we choose a set V that is open in the space X such that S HS V and we observe that HS X V. Thus H, being the intersection of two sets that are both closed in the space X, d , must be closed in the space X, d . 15. Skip this exercise if you are not familiar with the concept of an uncountable set. Prove that every nonempty open subset of R k is uncountable. The proof follows at once from an earlier theorem about equivalence to R of subsets of R that include intervals. 16. Suppose that X, d is a given metric space and that we have defined  x, y  d x, y if d x, y 1 1 if d x, y  1 . a. Prove that the function  is also a metric on the set X. b. Prove that the metric spaces X, d and X,  have exactly the same open sets. We saw in an earlier exercise that the function  is a metric. Part b follows from the fact that whenever x X and 0  r  1, the ball with center x and radius r n the metric space X, d is exactly the same as the ball with center x and radius r in the metric space X,  . 17. This exercise refers to the optional topic of a convex hull that appears in the reading material on convexity. Prove that if U is an open subset of R k then the convex hull of U is open. We assume that U is an open subset of R k . Suppose that x is any member of the convex hull co U n of U and choose a positive integer n and numbers r j 0, 1 for j  1, 2, , n such that j1 r j  1 and n x rjxj. j1 We may assume that the coefficient r 1 is not zero. Now using the fact that U is open and that x 1 U we choose   0 such that B x 1 ,  U. We see that if y is any member of the ball B x 1 , r 1  , since 120 n yy x rjxj j1  r1 y x n r1  x1   x1 B x1,  rjxj j2 and since y x r1 we see that y U co U . 18. The purpose of this exercise is to exhibit an example of a closed subset H of R 2 whose convex hull fails to be closed. We define 1 1 x 1 1x 1 fx  1 if x 0 if x  0 0.8 0.6 0.4 0.2 -4 -2 0 2 x 4 and we define H  x, y 0 y fx . Prove that the set H has the desired properties. Given any positive integer n and points A j  x j , y j in the set H for j  1, 2, , n and nonnegative n numbers r j for which j1 r j  1 we have n n n rj xj, yj  rjAj  j1 j1 n rjxj, j1 rjyj j1 n and since j1 r j y j  1 we know that the point 0, 1 does not belong to co H . On the other hand, whenever p  0 the point p2 1 1 0, 2  1 p, 1  1 p, 1 2 1p 2 1p p 1 must belong to co H and so the point 0, 1 is close to co H . Therefore co H is not closed. We can, of course, do much better. In fact co H  0, 0 Þ x, y R2 0  y  1 . Exercises on Closure 1. Suppose that S  0, 1 Þ 1, 2 . a. What is the set of interior points of S in the metric space R? The set of interior points of S is 0, 1 Þ 1, 2 . b. Given that U is the set of interior points of S, evaluate U. 121 0, 1 Þ 1, 2  0, 2...
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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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