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Heine-borel related proof

  1. Aug 18, 2012 #1
    I'm trying to understand that proof with little success. Particularly I don't understand why "S has x_0 as a limit point, and S has no other limit point in R^k". Please help

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  2. jcsd
  3. Aug 18, 2012 #2
    The conditions [itex]|x_n-x_0|<1/n[/itex] imply that [itex](x_n)_n[/itex] converges to [itex]x_0[/itex]. Thus [itex]x_0[/itex] is a limit point of the [itex](x_n)_n[/itex] (which is the set S). Since limits of sequences are unique in [itex]\mathbb{R}^k[/itex], the limit point [itex]x_0[/itex] is unique.
  4. Aug 18, 2012 #3
    Alright, but can you explain it with basic topology, like neighbourhoods and the definition of a limit point? I don't see why any other interior point of S couldn't be a also a limit point in that case. Rudin says that it's because it would violate the triangular inequality theorem, but still it's not clear enough...
  5. Aug 18, 2012 #4
    If y is a limit point of [itex](x_n)_n[/itex], then there exists a subsequence [itex](x_{k_n})_n[/itex] that converges to y. But since the original sequence [itex](x_n)_n[/itex] converges to x, the subsequence must also converge to x.

    So [itex](x_{k_n})_n[/itex] converges to both x and y. Thus x must equal y.
  6. Aug 19, 2012 #5


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    By basic properties of the (standard) real numbers, any two real numbers that are

    indefinitely-close to each other, e.g., d(x,y)<1/n for all n , then x=y by,e.g., the

    Archimedean Property. Use the triangle inequality to show that, in a metric space,

    if a sequence has two limits L1, L2, then L1 is indefinitely-close to L2.
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