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Unformatted text preview: Math 104  Lecture 1  Problem Set 6 Solutions 2.6 (3 points) (a) T is the minimum closed set containing T , and T contains S , so T contains S . Since overlineS is the minimum closed set containing S , and T is such a set, it must be that T S . (b) Since S T , ( X \ T ) ( X \ S ) . So X \ T X \ T . That gives us that: int ( S ) = X \ X \ S X \ X \ T = int ( T ) 2.30a (3 points) If h x n i is increasing and bounded, put L = sup { x i : i N } . We will prove that L is the limit of h x n i . Now for any > , L < L . Since L is the least upper bound, this means L is not an upper bound. So for some i , x i > L . But then for every n > i , also x n x i > L . So by taking N = i , we see that the sequence does converge to L . If instead, h x n i is decreasing, apply the same argument to h x n i . 2.84 (5 points) (a) Since the numerator is nonnegative and the denominator is positive, ( p,q ) always exists and is nonnegative. Also p = q d ( p,q ) = 0 ( p,q ) , so is positive definite. Since d is symmetric, is symmetric. To show the triangle inequality, let a,b,c M be given. There are two cases. Ifbe given....
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 Summer '08
 RIEMAN
 Math

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