AnalysisHW4 125A

AnalysisHW4 125A - MAT 125a HW4 Solutions 10 Suppose that...

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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 > 0 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 p n k is a subsequence of p n (pick p n 1 from { p 1 , p 2 , p 3 , ... } , pick p n 2 from { p n k +1 , p n k +2 , p n k +3 , ... } , and continue in this manner). Now we see that by construction ( p n k

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