1873_solutions

Of r we need to show that h k is nonempty and that

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Unformatted text preview: the set Z of integers. 2. Prove that in the metric space R we have L Q  R. Suppose that x is any real number. To show that x is a limit point of Q, suppose that   0. Since there are rational numbers in the interval x, x   we know that the set x , x   Q x . 3. Prove that in the metric space R we have L 1 n Z  0 . n If x is any negative number then the interval Ý, 0 is a neighborhood of x that fails to contain any members of the set 1 n Z  . Thus a negative number can not be a limit point of n 1  nZ. n 128 If x is any positive number then the interval infinitely many members of the set 1 n n then the condition , Ý is a neighborhood of x which fails to contain Z  . To see why, note that if n is a positive integer x 2 1 x ,Ý n 2 can hold only if n  2 . Therefore no positive number can be a limit point of 1 n Z . x n 1  Finally, we need to explain why 0 must be a limit point of n n Z . Suppose that   0. Choose an integer k  1 and observe that  1 1 0 , 0   n Z 0 n k 1 0 must be nonempty. n Z from which we deduce that the set 0 , 0   n 4. a. Give an example of an infinite subset of R that has no limit point. As we saw in Exercise 1, the infinite set Z has no limit point. b. Give an example of a bounded subset of R that has no limit point. A finite set like 2 will not have any limit points. We could also look at the empty set . c. Give an example of an unbounded subset of R that has no limit point. As we saw in Exercise 1, the infinite set Z has no limit point. d. Give an example of an unbounded subset of R that has exactly one limit point. The unbounded set Z Þ 1 n Z  has only the limit point 0. n e. Give an example of an unbounded subset of R that has exactly two limit points. The set ZÞ 1 n Z Þ 1  1 n Z n n has the two limit points 0 and 1. We can see this directly or we can use the assertion proved in Exercise 6 below. 5. Prove that if A and B are subsets of a metric space X and if A B then L A LB. Suppose that A and B are subsets of a metric space X and that A B. Suppose that x is a limit point of A. We need to explain why x has to be a limit point of B. Suppose that   0. Since the set B x,  A x is nonempty and since B x,  A x B x,  B x we deduce that the set B x,  B x is nonempty. 6. Prove that if A and B are subsets of a metric space X then L AÞB  L A ÞL B . Solution: Since A LB L A Þ B . Thus A Þ B we know that L A L A Þ B and similarly we know that L A ÞL B L AÞB . Now suppose that a point x fails to belong to the set L A Þ L B . Choose a number  1  0 such that the ball B x,  1 contains only finitely many members of the set A. Choose a number  2  0 such that the ball B x,  2 contains only finitely many members of the set B. We now define  to be the smaller of the two numbers  1 and  2 and we observe that, although   0, the interval B x,  contains only finitely many members of the set A Þ B. Therefore no number that lies outside the set L A Þ L B can be a limit point of A Þ B and w...
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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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