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Unformatted text preview: FUNCTIONAL ANALYSIS LECTURE NOTES: COMPACT SETS AND FINITEDIMENSIONAL SPACES CHRISTOPHER HEIL 1. Compact Sets Definition 1.1 (Compact and Totally Bounded Sets) . Let X be a metric space, and let E ⊆ X be given. (a) We say that E is compact if every open cover of E contains a finite subcover. That is, E is compact if whenever { U α } α ∈ I is a collection of open sets whose union contains E , then there exist finitely many α 1 , . . . , α N such that E ⊆ U α 1 ∪ ···∪ U α N . (b) We say that E is sequentially compact if every sequence { f n } n ∈ N of points of E contains a convergent subsequence { f n k } k ∈ N whose limit belongs to E . (c) We say that E is totally bounded if for every ε > 0 there exist finitely many points f 1 , . . . , f N ∈ E such that E ⊆ N S k =1 B ε ( f k ) , where B ε ( f k ) is the open ball of radius ε centered at f k . That is, E is totally bounded if and only there exist finitely many points f 1 , . . . , f N ∈ E such that every element of E is within ε of some f k . (d) We say that E is complete if every Cauchy sequence of points of E converges to a point in E . Exercise 1.2. Suppose that X is a complete metric space. Show that, in this case, E is complete ⇐⇒ E is closed . In a finite dimensionsional normed space, a set is compact if and only if it is closed and bounded. In infinite dimensional normed spaces, it is true all compact sets are closed and bounded, but the converse fails in general. We have the following equivalent formulations of compactness for sets in metric spaces. Theorem 1.3. Let E be a subset of a Banach space X . Then the following statements are equivalent. (a) E is compact. (b) E is sequentially compact. 1 2 COMPACT SETS AND FINITEDIMENSIONAL SPACES (c) E is complete and totally bounded. Proof. (a) ⇒ (c). Exercise. (c) ⇒ (b). Assume that E is complete and totally bounded, and let { f n } n ∈ N be any sequence of points in E . Since E is covered by finitely many balls of radius 1 2 , one of those balls must contain infinitely many f n , say { f (1) n } n ∈ N . Then we have ∀ m, n ∈ N , d( f (1) m , f (1) n ) < 1 . Since E is covered by finitely many balls of radius 1 4 , we can find a subsequence { f (2) n } n ∈ N of { f (1) n } n ∈ N such that ∀ m, n ∈ N , d( f (1) m , f (1) n ) < 1 2 . By induction we keep constructing subsequences { f ( k ) n } n ∈ N such that d( f ( k ) m , f ( k ) n ) < 1 k for all m , n ∈ N . Now consider the “diagonal subsequence” { f ( n ) n } n ∈ N . Given ε > 0, let N be large enough that 1 N < ε . If m ≥ n > N , then f ( m ) m is one element of the sequence { f ( n ) k } k ∈ N , say f ( m ) m = f ( n ) k . Then d( f ( m ) m , f ( n ) n ) = d( f ( n ) k , f ( n ) n ) < 1 n < ε....
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This note was uploaded on 08/25/2011 for the course MATH 7338 taught by Professor Heil during the Fall '09 term at Georgia Tech.
 Fall '09
 HEIL
 Sets

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