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Unformatted text preview: MAT 125a, HW4 Solutions 10 Suppose that lim n p n = p in a given metric space. The set of points S = { p,p 1 ,p 2 ... } is closed. lemma: If x 6 { a 1 ,a 2 ,a 3 ,... } and for all > B ( x ) { a 1 ,a 2 ,... } 6 = then > 0 and n N we have B ( x ) { a n +1 ,a n +2 ,a n +3 ... } 6 = . proof We will show the contrapositive. Suppose there exist > 0 and n N such that B ( x ) { a n +1 ,a n +2 ,... } = . Let * = min { ,d ( x,a 1 ) ,d ( x,a 2 ) ,...,d ( x,a n ) } / 2 (Why is this > 0?) and consider B * { a 1 ,a 2 ,... } . a i 6 B * { a 1 ,a 2 ,... } for i { 1 ,...,n } because * < d ( x,a i ) for i { 1 ,...,n } . Also for i { n + 1 ,n + 2 ,... } we know a i 6 B * { a 1 ,a 2 ,... } because * < and B ( x ) { a n +1 ,a n +2 ,... } = . So we have that B * ( x ) { a 1 ,a 2 ,... } = , proving our lemma. To show that S is closed, we will show its complement is open. Consider some point x S { . We wish to show there is a ball centered at x and contained entirely in the complement. For a contradiction, suppose there is no such ball, then for every k N , there is a point p n k such that p n k B 1 /k ( x ). Furthermore, by our lemma, we can pick the indices such that)....
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This note was uploaded on 10/23/2010 for the course MATH math 125A taught by Professor Mulase during the Fall '10 term at UC Davis.
 Fall '10
 Mulase

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