Lebegue Covering Lemma

Lebegue Covering Lemma - covers X entirely So let δ = min...

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The Lebesgue Covering Lemma copied from http://mathblather.blogspot.com/ February 3, 2012 Theorem(Lebesgue Covering Lemma): Given that X is a compact metric space, let U be a open covering of X . Then there exist δ > 0 such that for all x X , the ball B δ ( x ) U for some U ⊂ U . Proof. Take any x X , then x U x for some U x ⊂ U , since U x is open, there is a ball B ± ( x ) ( x ) U x . Now the set of all balls of the form: { B ± ( x ) / 2 ( x ) : x X } is an open cover of X , so by compactness of X , there is a finite collection of balls: B ± ( x 1 ) / 2 ( x 1 ) ,...,B ± ( x n ) / 2 ( x n ), where x 1 ,...,x n X , that
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Unformatted text preview: covers X entirely. So let δ = min { ± ( x 1 ) / 2 ,...± ( x n ) / 2 } . Pick any x ∈ X , I then claim that B δ ( x ) ⊂ U for some U ⊂ U . Since x ∈ B ± ( x j ) / 2 ( x j ) for some x j , we take any y ∈ B δ ( x ), by triangular inequality: d ( y,x j ) ≤ d ( y,x ) + d ( x,x j ) < δ + d ( x,x j ) ≤ ± ( x j ). Hence y ∈ B ± ( x j ) ( x j ), this implies that B δ ( x ) ⊂ B ± ( x j ) ( x j ) ⊂ U x j 1...
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This note was uploaded on 03/05/2012 for the course MATH 522 taught by Professor Patirck during the Spring '12 term at University of Wisconsin.

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