1873_solutions

# Of x must be closed solution suppose that x is a

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Unformatted text preview: al 3 , 3 which is included in the interval 1, 3 . Since all other members of this interval are obviously interior points, this interval is open in the metric space 0, 3 . 10. Given that S is a nonempty subset of a metric space X, d and that U is an open subset of the metric space X, d and that U S, prove that U must be open in the metric space S, d . Suppose that x U. Using the fact that U is open in the space X, d , choose a number   0 such that B x,  U. In the metric space S, d , the ball center x with radius  is S B x,  and since S B x,  B x,  U we have shown that x is an interior point of the set U in the metric space S, d . 11. Suppose that S is a nonempty subset of a metric space X, d and that U conditions are equivalent: S. Prove that the following two a. The set U is open in the metric space S, d . b. There exists an open subset V of the metric space X, d such that U  S V. Solution: First we prove that condition b implies condition a. Suppose that condition b holds and choose an open subset V of the metric space X such that U  S V. To prove that the set U is open in the subspace S, suppose that x U. Using the fact that x V and the fact that V is open in the metric space X we now choose   0 such that B x,  V and we observe that y S d x, y   S V. Since the set y S d x, y   is the ball center x with radius  in the metric space S, we have shown that the set U is open in the metric space S. Now we want to prove that condition a implies conditio b. Suppose that condition a holds. In other words, suppose that the set U is open in the metric space S. We know that for every point x U it is possible to find a positive number  such that B x,  S U. We now define V to be the union of all the balls of the form B x,  for which x U and   0 and B x,  S U. The set V, being a union of balls in the space X, must be open in X. Finally, we need to explain why U  V S. U we can choose   0 such that B x,  SU B x,  we have x V. Now given any point x and since x Now suppose that x V S. Using the definition of the set V we choose y 119 U and   0 such that B y,  and x B y,  . Thus x B y,  S S U U. 12. Given that S is an open nonempty subset of a metric space X, d and that U is an open subset of the metric space S, d , prove that U must be open in the metric space X, d . Using Exercise 11 we choose a set V that is open in the metric space X, d such that U  V S. Since both of the sets S and V are open in the space X, d , so is their intersection, which is U. 13. Given that S is a nonempty subset of a metric space X, d and that H is a closed subset of the metric space X, d and that H S, prove that H must be closed in the metric space S, d . From Exercise 11 and the fact that the set X H is open in the metric space X, d we conclude that the set S X H is open in the metric space S, d . Since S X H S H we conclude that S H is open in the space S, d which tells us that H is closed in the space S, d . 14. Given that S is a closed nonempty subset...
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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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