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Unformatted text preview: 4.2 Properties of continuity 67  a n x n  fl fl fl fl a n 1 a n x fl fl fl fl + fl fl fl fl a n 2 a n x 2 fl fl fl fl + + fl fl fl fl a a n x n fl fl fl fl  a n x n  , since the expression in parentheses is < 1 for x >> 1. NEXT: The continuous image of a compact set is compact. Theorem 4.2.7. Let f be continuous, where K is compact. Then f ( K ) is compact. Proof. Let { U } A be a covering of f ( K ) by open sets U . NTS: a finite subcover. Since f is continuous, every set f 1 ( U ) will be open in K . Since the U cover f ( K ), we have an open cover { f 1 ( U ) } of K . K is compact by hypothesis, so there must be some finite subcover { f 1 ( U i ) } n i =1 of K . K n [ i =1 f 1 ( U i ) = f ( K ) f n [ i =1 f 1 ( U i ) ! by S&M 4(a) = n [ i =1 f ( f 1 ( U i ) ) by S&M 4(b) n [ i =1 U i by S&M 1(b) , and we have f ( K ) contained in the finite subcover { U i } n i =1 . I.e., f ( K ) is compact....
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 Fall '11
 Wong
 Continuity

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