204Lecture62006 - Economics 204 Lecture 6Monday, August 7,...

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Unformatted text preview: Economics 204 Lecture 6Monday, August 7, 2006 Revised 8/8/06, Revisions indicated by ** Section 2.8, Compactness Definition 1 A collection of sets U = { U : } in a metric space ( X, d ) is an open cover of A if U is open for all and U A ( may be finite, countably infinite, or uncountable.) A set A in a metric space is compact if every open cover of A contains a finite subcover of A . In other words, if { U : } is an open cover of A , there exist n N and 1 , , n such that A U 1 U n It is important to understand what this definition does not say. In particular, it does not say A has a finite open cover; note that every set is contained in X , and X is open, so every set has a cover consisting of exactly one open set. Like the - definition of continuity, in which you are given an arbitrary > and are challenged to specify an appropriate , here you are given an arbitrary open cover and challenged to specify a finite subcover of the given open cover. Example: (0 , 1] is not compact in E 1 . To see this, let U = U m = 1 m , 2 : m N Then m N U m = (0 , 2) (0 , 1] 1 Given any finite subset { U m 1 , . . . , U m n } of U , let m = max { m 1 , . . . , m n } Then n i =1 U m i = U m = 1 m , 2 6 (0 , 1] so (0 , 1] is not compact. Note that this argument does not work for [0 , 1]. Given an open cover { U : } , there must be some such that 0 U , and therefore U [0 , ) for some > 0, and a finite number of the U m s we used to cover (0 , 1] would cover the interval ( , 1]. This is not a proof that [0 , 1] is compact, since we need to show that every open cover has a finite subcover, but it is suggestive, and we will soon see that [0 , 1] is indeed compact....
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This note was uploaded on 03/06/2009 for the course ECON 0204 taught by Professor Staff during the Summer '08 term at University of California, Berkeley.

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204Lecture62006 - Economics 204 Lecture 6Monday, August 7,...

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