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Math Help - Complete metric space

  1. #1
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    Complete metric space

    Let X_{n+1}\subset X_n for each n\in\mathbb N. Let (M,d) be a metric space. Prove that the following assertions are equivalent:

    a) (M,d) is a complete metric space.
    b) For each X_{n+1}\subset X_n being closed and non-empty where \text{diam}(X_n)\to0 as n\to\infty, we have \bigcap_{n=1}^\infty X_n\ne\varnothing.

    I already did a) \implies b) but for the other implication, I have a long solution, is there a short way to do it?
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  2. #2
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    Re: Complete metric space

    let {xn} be a cauchy sequence in X. by definition for any ε > 0 there is some N such that d(xm,xn) < ε for all m,n > N.

    so pick εk = 1/2k, and set Uk = cl(B(xNkk)).

    define Xk = U1∩U2∩...∩Uk.

    it's clear that diam(Xk) ≤ 1/2k-1 and that each Xk is closed being the intersection of finitely many closed sets.

    furthermore since Nk < Nk+1 its clear that for any k:

    xNk is in Um, for all m ≤ k, so each Xk is non-empty.

    so, we have from (b) that ∩kXk ≠ .

    but this means the subsequence {xNk} of {xn} converges to some point x of X (the sole element of the infinite intersection, since if d(x,y) = 0, y = x, and for any y in ∩kXk,

    we have d(y,xNk) < 1/2k-1 for ALL k in N). hence {xn} also converges to x, so X is complete.

    (i was tempted to use just the 1/2k balls at first, but i realized that the 2nd ball could contain points OUTSIDE the first ball, but their intersection has no such defect. the center of the 3rd ball has to lie within the intersection of the first 2, and the center of the 4th ball has to lie within the intersection of the first 3, etc. the diameter of the intersections can be no larger than the diameter of the ball, which is twice its radius).
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  3. #3
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    Re: Complete metric space

    Okay, thank you!

    I was thinking the following: let x_n be a Cauchy sequence on X, then let X_n=\overline{\{x_m:m\ge n\}}, so X_n is closed and satisfies X_{n+1}\subset X_n, yet I don't see how to prove that \text{diam}(X_n)\to0.
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