1873_solutions

R j for which j1 r j 1 we have n n n rj xj yj rjaj

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Unformatted text preview: e that if A and B are sets of real numbers and ABR then A B  R? The answer is no. Look at A  Q and B  R Q. 9. Prove that a subset S of a metric space X is dense in X if and only if we have SU whenever U is a nonempty open subset of X. Suppose that S is dense in X and that U is a nonempty open set. Choose x U. Since x S and U is a neighborhood of x we know that U S . Now suppose that the condition U S holds for every nonempty open subset U of X. To show that S  X, suppose that x X and that  0. Since the ball B x, is a nonempty open set we must have B x, S . 10. Prove that if U and V are open dense subsets of a metric space X then the set U if only one of the two sets U and V is open? V is also dense in X. What Solution: All we need to know is that at least one of the sets U and V is open. Suppose that A and B are sets of real numbers, that UVX and that the set U is open. To prove that U V  X, suppose that x X and that   0. Since x U we know that the set B x,  U is nonempty and we also know that this set is open. Therefore, since V  X we know that B x,  UV . We have therefore shown that every point of the space X must belong to U V. 11. Skip this exercise if you are not familiar with the concept of a countable set. Find a sequence U n of dense open subsets of the metric space Q such that Ý  Un  n1 123 . Using the fact that the set Q of rational numbers is countable we express Q in the form Q  r 1 , r 2 , r 3 , , r n ,  . We now define Un  rj j n for each positive integer n. Since the set Q U n is finite for each n we know that each set U n is open in the metric space Q. Finally, since every neighborhood of a point in Q must contain infiitely many rational numbers, each neighborhood of a point of Q must intersect with each of the sets Q U n and so each of the sets Q U n must be dense. 12. Skip this exercise if you are not familiar with the concept of a countable set. Prove that if S is a countable subset of the metric space R then R S is dense in R. Extend this fact to the metric space R k for an arbitrary positive integer k. Suppose that x is a real number and that   0. Since the interval x , x   is uncountable we know that x , x   S , in other words x , x   RS . Therefore R S is dense in the metric space R. Now we repeat the same argument in R k . We assume that k is a positive integer and that S is a countable subset of R k . Suppose that x R k and that   0. Since the ball B x,  is uncountable we know that B x,  S , in other words . B x,  Rk S Therefore R k S is dense in the metric space R k . 13. Suppose that D is a dense subset of a metric space X and that U is a neighborhood of a point x X. Prove that there exists a point y D and a rational number r  0 such that x B y, r U. Choose   0 such that B x,  U. Using the fact that D is dense we now choose a member y of the set D such that d x, y   . Finally we choose a positive integer n  3 which makes 1   . n  3 3 z y B y, x 1 n B x,  We shall now observe tha...
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