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Unformatted text preview: Mathematics 250 Fall 2008 Selected solutions, Assignment #1 Assignment # 1 MVM, Exercise 2.2.1: Are the following sets open? Closed? a) A = { x : 0 < x 2 bfR } . This set is neither open nor closed. It does not contain an interval around the upper boundary point 2, which is in the set, so it is not open. And it does not contain its lower boundary point, 0, which is a limit of sequences from the set, so it is not closed. b) B = { x : x = 2 k for some positive integer k, or x = 0 } . This set is closed. Only finitely many of the elements of the set are outside of any interval around 0, so 0 is the only possible limit point, and it is in the set. The set is not open, in fact, it does not contain an interval around any of its points. c) C = x y : y > . This set is open. Given a point p = x y in the set, put r = y . Then the open ball B r ( x y ) is inside the set. d) D = x y : y . This is closed. If a sequence is in D and converges in R 2 , then the sequence of second coordinates will be a sequence of nonnegative numbers converging in R . We assume that it is clear that a limit of nonnegative numbers will be nonnegative. This means that the limit will also be in D , so the condition for closedness is satisfied. Alternatively, we can observe that R 2 D == x y : y < . This set is open, for the same reasons that set C is open. Therefore, D is the complement of an open set, and hence is closed. e) E = x y : y > x . This set is open. If p = x y is in E , then set r = y x 2 . It can be checked that B r ( p ) is completely inside E , so that E satisfies the condition for openness. f) F = x y : xy 6 = 0 . This is open. We have decided that the set C is open. For similar reasons, the set C = x y : y < is open. Therefore, the set X = x y : y = 0 (which is the xaxis) is closed, since it is the complement of the open set C C . The same kind of reasoning shows that the yaxis is closed. But the condition xy = 0 is equivalent to the pair of conditions x = 0 or y = 0. That is, the condition xy = 0 defines the union of the x and y axes, and therefore is closed (since a union of two closed sets is closed). But the set F is the complement of the set defined by the condition xy = 0, so it is open....
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 Fall '06
 RogerHowe
 Sets

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