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Math Help - metric space

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    metric space

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    Last edited by patricia-donnelly; February 10th 2008 at 09:47 AM.
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    Quote Originally Posted by patricia-donnelly View Post
    Let (X,d) be a metric space and say A is a subset of X. If x is an accumulation point of A, prove that every r-neighbourhood of x actually contains an infinite number of distinct points of A (where r>0).
    Suppose x is an accumulation point of a set A. Then there exists distinct x_n \in A such that x_n \to x. Let r>0 then by definition of convergence it means d(x,x_n) < r for n\geq N. Thus, x_N,x_{N+1},... are distinct elements which lie in B(x,r) and hence there are infinitely many.

    Using this, prove that any finite subset of X is closed.
    A set is closed if and only if it contains all its accumulation points. A finite set clearly has no accumulation points, hence the set is empty. And it contains the empty set.
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    You've been a great help. Thank you very much
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