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Math Help - Hausdorff metric

  1. #1
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    Hausdorff metric

    Let (X,d) be a metric space. Let {An} be a nested family of non empty compact subsets of X. Let A=Intersection of all An.
    We have that A is non empty and compact.

    We show An converges to A in the Hausdorff metric (D).

    I know D(A,B)= Inf {t>or eq.0 : A C B_t and B C A_t} Where A_t is the t-parallel body of A meaning A_t={x in X: d(x,A) < or eq.t}.

    But I am not sure how to proceed.
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  2. #2
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    Quote Originally Posted by math8 View Post
    Let (X,d) be a metric space. Let {An} be a nested family of non empty compact subsets of X. Let A=Intersection of all An.
    We have that A is non empty and compact.

    We show An converges to A in the Hausdorff metric (D).

    I know D(A,B)= Inf {t>or eq.0 : A C B_t and B C A_t} Where A_t is the t-parallel body of A meaning A_t={x in X: d(x,A) < or eq.t}.

    But I am not sure how to proceed.
    Clearly A\subseteq (A_n)_t for all t>0 (because in fact A\subseteq A_n). So we need to show that given t>0 there exists n such that A_n\subseteq A_t. This seems to be quite tricky to prove. One way that works (although it's not very elegant) is to suppose that the result is false, and get a contradiction.

    So suppose that there exists t>0 such that for all n there is an element x_n\in A_n with d(x,A)\geqslant t. Use compactness to conclude that there is a subsequence of (x_n) that converges, to x_0 say. Then get a contradiction by showing that x_0\in A but also d(x_0,A)\geqslant t.
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