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Unformatted text preview: COMPACTNESS VS. SEQUENTIAL COMPACTNESS The aim of this handout is to provide a detailed proof of the equivalence between the two definitions of compactness: existence of a finite subcover of any open cover, and existence of a limit point of any infinite subset. Definition 1. K is compact if every open cover of K contains a finite subcover. K is sequentially compact if every infinite subset of K has a limit point in K . Theorem 1. K is compact K is sequentially compact. The first half of this statement (compact = sequentially compact) is Theorem 2.37 in Rudin and is proved there. Our aim is to prove the converse implication (sequentially compact = compact), following the lines of Exercises 23, 24 and 26 in Rudin Chapter 2. The proof requires the introduction of two auxiliary notions: Definition 2. A space X is separable if it admits a countable dense subset. For example R is separable ( Q is countable, and it is dense since every real number is a limit of rationals); for the same reason R k is separable (consider all points with only rational coordinates). Definition 3. A collection { V } of open subsets of X is said to be a base for X if the following is true: for every x X and for every open set G X such that x G , there exists such that x V G . In other words, every open subset of X decomposes as a union of a subcollection of the V s the V s generate all open subsets. The family { V } almost always contains infinitely many members (the only exception is if X is finite). However, if X happens to be separable, then countably many open subsets are enough to form a base (the converse statement is also true and is an easy exercise): Lemma 1. Every separable metric space has a countable base....
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 Fall '10
 Prof.KatrinWehrheim

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