1873_solutions

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 HU XH which is the intersection of two open sets, the set U H is open. Since H UH 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 au   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,  AU 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...
View Full Document

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.

Ask a homework question - tutors are online