1873_solutions

# But from the fact that p n we see that p must be

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Unformatted text preview: rval x , 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 we conclude that L AÞB  L A ÞL B . 7. Is it true that if A and B are sets of real numbers then LA B LA LB? What if A and B are closed? What if A and B are open? What if A and B are intervals? The answers are no, no, no and no. Look at the following example: A  0, 1 and B  1, 2 These two sets are closed and LA B L 1  while LA L B  0, 1 1, 2  1 . Now look at the following example: A  0, 1 and B  1, 2 . In this case LA B L  and LA L B  0, 1 1, 2  1 . 8. Is it true that if D  R then L D  R? Hint: Yes. Suppose that D  R. We know that whenever a and b are real numbers and a  b there must be 105 members of D lying between a and b. Now suppose that x is a real number. To show that x is a limit point of D, suppose that   0. Since there must be members of D in the interval x, x   we conclude that the set x , x   D x is nonempty. 9. Given that a set S of real numbers is nonempty and bounded above but that S does not have a largest member, prove that sup S must be a limit point of S. State and prove a similar result about inf S. To show that sup S is a limit point of S, suppose that   0. Since sup S   sup S and since sup S is the least upper bound of S the number sup S  fails to be an upper bound of S. Choose a member x of S such that sup S   x. Since x sup S and since sup S does not belong to S we have x  sup S. We conclude that sup S , sup S   S sup S . 10. Given any set S of real numbers, prove that the set L S must be closed. Solution: We shall show that any number that fails to belong to L S must fail to belong to L S . Suppose that x R L S . Choose a number   0 such that the interval x , x   contains only finitely many members of S. Given any number t in the interval x , x   , it follows from the fact that x , x   is a neighborhood of t and the fact that x , x   contains only finitely many members of S that t is not a limit point of S. Thus x , x   LS  and we have shown, as promised, that x does not belong to L S . 11. Prove that if a set U is open then L U  U. Of course L U U. Now suppose that x U. To show that x L U , suppose that   0. Using the fact that x U, choose a number y in the set U x , x   . Using the fact that the set U x , x   is open, choose  0 such that y ,y  U x , x   . We have now found more than one member of U that belongs to the interval x , x   and so we know that U x , x   x . 12. Suppose that S is a set of real numbers, that L S numbers x and y in S such that |x y |  . , and that   0. Prove that there exist two different Solution: Choose a limit point t of the set S. Using the fact that the interval t /2, t  /2 contains infinitely many members of S, choose two different members x and y of S that lie in the interval t /2, t  /2 . We observe that |x y |  . Alternative 6: The Topology of Metri...
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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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