Norm replaced by the norm c prove that in the metric

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Unformatted text preview: ded. See the solution of Exercise 3. Ý S such that E fails to be totally 3. True or false? If x is a point in a metric space X then there exists a number  0 such that the ball B x, is totally bounded. We take an infinite set S and define X  Ý S . Suppose that  0. For each x S we define the function f x : S R by the equation fx t  0 2 if t S if t  x x . The set E of all these functions f x fails to be totally bounded because no ball with radius 2 can contain more than one member of E. Since E is included in the ball B O, where O is the constant function 0 we conclude that the assertion made in Exercise 3 is false. 4. Prove that if S is an infinite totally bounded subset of a metric space X and if  0 then it is possible to find two different members x and y of the set S such that d x, y  . Suppose that S is an infinite totally bounded subset of a metric space X and that  0. Choose finitely many sets with diameter less than whose union includes S. At least one of these sets must contain more than one member of S. 5. True or false? If X is a metric space then the following two conditions are equivalent: a. No infinite subset of X is totally bounded. b. There exists a number  0 such that for every pair x and y of different points of X we have d x, y . If you decide that these two conditions are equivalent, prove that each implies the other. If you decide that one of these conditions is sufficient but not necessary for the other, supply a proof and a counter example. If you decide that neither of these statements implies the other, supply two counter examples. The fact that condition b implies condition a follows at once from Exercise 4. Condition a does not imply condition b. We define X  Z Þ n  1 n Z . n An infinite subset of this set X has to be unbounded but, in spite of this, condition b does not hold. Exercises on Open Sets and Closed Sets 1. Given a subset H of a metric space X, prove that the following conditions are equivalent: a. The set H is closed. H there exists a number   0 such that B x,  H . Condition b says that for every x X H there exists a number   0 such that B x,  and this is the condition for the set X H to be open. b. For every point x X X H 2. Prove that if X is any metric space then the singleton x is closed. Then use the fact that every finite set is a finite union of singletons to show that every finite subset of X must be closed. Solution: Suppose that x is a point in a metric space X. We need to show that the set X Suppose that y X x . We define   d x, y and we observe that   0 and that B y,  X x. 3. Give an example of a sequence U n of open subsets of the metric space R such that the set 117 x is open. Ý  Un n1 fails to be open. We have observed that Ý  n1 1, 1 nn  0. 4. Given that H is a closed subset of R and that every rational number belongs to H, prove that H  R. Solution: The set R H is open in R and contains no rational number. To prove that R H must be empty we shall observe tha...
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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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