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Math Help - Compactness and metric space

  1. #1
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    Compactness and metric space

    Let  (X,d) be a compact metric space, which metric is  d.

    Let be p and q two points of X.

    Can I say that, since X is compact, the distance between p and q is finite and reachable?

    Thank you.
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  2. #2
    Super Member Gamma's Avatar
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    The distance map d: X\times X \rightarrow \mathbb{R}^{nonneg} is a continuous map from a compact space (compact because X is, and continuity requires some proof if you haven't already proved this).

    But the continuous image of a compact set is compact which means closed and bounded in \mathbb{R} so the distance between the two points is definitely finite. I am not sure what you mean by reachable.
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  3. #3
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    By reachable I mean that your map d from R*R reaches is maximum and its minimum where the function is defined.

    For exemple the function sin(x) defined on [0,pi] has a maximum and a minimum and it reaches its maximum at Pi/2
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  4. #4
    Super Member Gamma's Avatar
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    Yeah you are just talking about the extreme value theorem then I think.

    Basically here is the deal. The image of the space is compact in \mathbb{R} which is complete. So we know the image is closed an bounded. In particular the inf and sup of this set would be a limit point, but it is closed, so it contains all of its limit points. Thus is achieves both its max and min? Sound good?
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