204Lecture62006

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

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Economics 204 Lecture 6–Monday, August 7, 2006 Revised 8/8/06, Revisions indicated by ** Section 2.8, Compactness Definition 1 A collection of sets U = { U λ : λ Λ } in a metric space ( X, d ) is an open cover of A if U λ is open for all λ Λ and λ Λ U λ A (Λ 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 λ : λ Λ } is an open cover of A , there exist n N and λ 1 , · · · , λ n Λ such that A U λ 1 ∪ · · · ∪ U λ 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 ε - δ definition of continuity, in which you are given an arbitrary ε > 0 and are challenged to specify an appropriate δ , 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

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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 (0 , 1] so (0 , 1] is not compact. Note that this argument does not work for [0 , 1]. Given an open cover { U λ : λ Λ } , there must be some λ Λ such that 0 U λ , and therefore U λ [0 , ε ) for some ε > 0, and a finite number of the U m ’s we used to cover (0 , 1] would cover the interval ( ε, 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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