psAnalysis2Ans - Fall 2007 ARE211 Problem Set #02 Second...

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Unformatted text preview: Fall 2007 ARE211 Problem Set #02 Second Analysis Problem Set Due date: Sep 18 Problem 1 For the following problem, consider an arbitrary universe X, together with an arbitrary metric d defined on X X . a) Prove that the union of arbitrarily many open sets is an open set. b) Prove that the intersection of finitely many open sets is an open set. c) Identify what part of your proof in part (b) would no longer hold if you would consider infinitely many sets. d) Prove that the intersection of arbitrarily many closed sets is closed. (This is another very short proof, you just have to come up with a little trick. Hint: use DeMorgans formula i I ( X \ A i ) = X \ ( i I A i ) ) (Note: The union of two sets A,B is: A B = { x X | x A x B } ). The intersection of two sets A,B is: A B = { x X | x A x B } ). Ans: Recall the definition of an open set: A set S X is said to be open in X w.r.t. a metric d if s S epsilon1 > such that B d ( s,epsilon1 ) = { x X | d ( x,s ) < epsilon1 } S . a) Show that the union S = i I S i of arbitrarily many open sets S i is an open set. (1) For any element s S we know by the definition of an union of sets that it must belong to at least one set S k , i.e., k I with s S k (2) Since the set S k is open, there is an epsilon1-ball around s that lies totally within the set S k , i.e., epsilon1 > such that B d ( s,epsilon1 ) = { x X | d ( x,s ) < epsilon1 } S k . (3) Since any set S k that is part of the union S is a subset of the union S we know that the epsilon1-ball from (2) also lies within S . In mathematical terms: By the definition of a union: S k S and hence from (2) epsilon1 > such that B d ( s,epsilon1 ) S k S . (4) Hence we can fit an epsilon1-ball around any element in the set and the ball lies within the set. The definition for an open set is thus fullfilled. b) Show that the intersection of finitely many open sets S i is an open set. Without loss of generality we can consider sets S 1 ,S 2 ,...,S N with N N and S = N i =1 S i . (1) For any element s S we know by the definition of the intersection of sets that it must belong to all sets S i i = 1 ...N . (2) Since all sets S i are open, there is an epsilon1-ball around s that lies totally within the set S i for all sets S i , i.e., epsilon1 i > such that B ( i ) d ( s,epsilon1 i ) = { x X | d ( x,s ) < epsilon1 i } S i for i = 1 ...N . (3) Let epsilon1 = min { epsilon1 i ,i = 1 ...N } > . (4) By the construction of epsilon1 we know that B d ( s,epsilon1 ) B i d ( s,epsilon1 i ) for i = 1 ...N . (The ball with radius epsilon1 is contained in all the balls with the larger radius epsilon1 i )....
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psAnalysis2Ans - Fall 2007 ARE211 Problem Set #02 Second...

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