1873_solutions

Of x must be closed solution suppose that x is a

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online