1873_solutions

y solution we need to find a number 0 such that the

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Unformatted text preview: serve that inf H   inf H x  inf H   and so inf H , inf H   H and we have shown that inf H is close to H. 10. Prove that no open set can have a largest member. Suppose that U is open and that x U. We need to show that x can’t be the largest member of U. Using the fact that x must be an interior point of U we choose   0 such that x , x   U. The interval x, x   is non empty and all its members must belong to U. Therefore x is not the largest member of U. 11. Given a real number x and a set S of real numbers, prove that the following two conditions are equivalent: a. The number x is an interior point of S. 98 b. It is possible to find an open set U such that x U S. Proof that condition a implies condition b: We assume that x is an interior point of S. Choose a number   0 such that x , x   S and define U  x , x   . In this way we have found an open set U such that x U S. Proof that condition b implies condition a. We assume that condition b holds. Choose an open set U such that x U S. Using the fact that x is an interior point of U we choose   0 such that x , x   U. Since x , x   S we deduce that x is an interior point of S. 12. Prove that if S is any set of real numbers then the set of all interior points of S must be open. Suppose that S is a set of real numbers and define U to be the set of interior points of S. To show that U is open, suppose that x U. Choose   0 such that x , x   S. We deduce from the results obtained in Exercise 11 that every number in the interval must be an interior point of S. In other words, x , x   U. Therefore, since every member of U is an interior point of U the set U must be open. 13. This exercise refers to the sum of two sets as it was defined in this exercise. Prove that if A is any set of real numbers and U is an open set then the set A  U must be open. Solution: We need to show that every member of A  U is an interior point of A  U. Suppose that x A  U. Choose a member a of A and a member u of U such that x  a  u. Using the fact that u U, choose   0 such that u , u   U. We shall show that x is an interior point of A  U by showing that x , x   A  U. Suppose that t x , x   . Thus |t a  u |   which gives us | t a u|   and so ta u , u   U. Since t  a t a we see that t A  U and so x , x   AU as promised. Exercises on Closure 1. Suppose that S  0, 1 Þ 1, 2 . a. What is the set of interior points of S? 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. 0, 1 Þ 1, 2  0, 1 Þ 1, 2  S. The purpose of parts a and b is to exhibit a set S such that, if U is the set of interior points of S then U  S. c. Give an example of a set S of real numbers such that if U is the set of interior points of S then U S. We could take S to be a singleton like 3 or it could be the set of all integers. It could also be the set of all rational numbers between 0 and 1. 99 d. Give an example of a subset S of th...
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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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