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Unformatted text preview: 245B, Winter 2009, Assignment 7: Notes and selected model answers (Model solutions follow question numbers in bold .) Folland Chapter 4 3. Suppose that X is a metric space. Then it is clear that individual points are closed, so X is T 1 [technically this is part of the definition of normality and as such it needs to be checked, although its not the heart of this question and wasnt worth any marks on the homework]. Now suppose that A,B X are closed and disjoint. We need to find disjoint open sets U A and V B . Let us take U := { x X : ( x,A ) < ( x,B ) } and V := { x : X : ( x,A ) > ( x,B ) } and show that these will do. Firstly, they are manifestly disjoint, since any x U V would have to satisfy two contradictory inequalities. Secondly, any x A has ( x,A ) = 0 , but since x 6 B (because A B = ) and B is closed , x actually lies in the interior of X \ B and therefore lies at the centre of some open ball disjoint from B , and so in particular we have ( x,B ) > . Therefore x U , and so since x A was arbitrary we have A U . An exactly analogous argument gives B V . Finally, suppose that x U , so r := ( x,B ) ( x,A ) > . Then if y B ( x,r/ 3) the triangle inequality gives for any b B , ( y,b...
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 Spring '10
 tao/analysis

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