1873_solutions

# X such that k b s yj j1 2 for each j in the event

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Unformatted text preview: t every nonempty open set must contain a rational number. Suppose that U is open and nonempty. Choose x U and choose   0 such that x , x   U. Since the interval x , x   must contain rational numbers, so must the set U. 5. Given that H is a closed subset of R 2 and that Q Q H, prove that H  R 2 . We need to show that the open set R 2 H is empty. To obtain a contradiction, suppose that and choose a point a, b R 2 H. Using the theorem on open subsets of R k we now R2 H choose   0 such that whenever a   x  a   and b   y  b   we have x, y R 2 H. Using the theorem on densesess of the set of rational numbers we choose rational numbers x and y such that a   x  a   and b   y  b   and we observe that x, y Q Q H, contradiction our assumption that QQ H. 6. Given that U is an open subset of a metric space X and that H is closed, prove that the set U that the set H U is closed. Since U HU XH which is the intersection of two open sets, the set U H is open. Since H UH XU which is the intersection of two closed sets, the set H U is closed. H is open and 7. Prove that every nonempty closed set of real numbers that is bounded below must have a least member. Suppose that H is a nonempty closed set of real numbers and that H is bounded below. We need to show that inf H H. To obtain a contradiction, suppose that inf H belongs to the open set R H. Choose   0 such that inf H , inf H   R H. Since no member of H can be less than inf H and no member of H can lie in the interval inf H , inf H   we see that the number inf H   is a lower bound of H, contradicting the fact that inf H is the greatest lower bound of H. 8. If A and B are subsets of the space R k for a given positive integer k, then the sum A  B of A and B is defined, by analogy with an earlier exercise. Prove that if A is any subset of R k and U is an open subset of R k then the set A  U is 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 B u,  U. We shall show that x is an interior point of A  U by showing that B x,  A  U. Suppose that t B x,  . Thus t au   which gives us t a u  and so t a B u,  U. Since 118 t  a t we see that t a A  U and so B x,  AU as promised. 9. Prove that the interval 1, 3 is not an open subset of the metric space R but that this interval is an open subset of the metric space 0, 3 which is a subspace of R. Given any positive number , the ball in the metric space R with center 3 and radius  will contain all of the numbers between 3 and 3   and will therefore fail to be included in the interval 1, 3 . Thus, in the metric space R, the point 3 fails to be an interior point of the interval 1, 3 and therefore this interval fails to be open in the metric space R. However, if 0    1 then, in the metric space 0, 3 , the ball center 3 with radius  is the interv...
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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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