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Absolutely Closed Metric Spaces.

  1. Aug 30, 2009 #1
    Hi.

    An absolutely closed metric space M is such that: If N is a meric space containing M, then M is closed in N.

    I would like to show that an absolutely closed metric space is complete, how do I do this? I know the proof of the converse but that's no help obviously.
    I know intuitively that an absolutely closed space should be complete: if you "imbed" the space in its completion, it will be a closed subset, and hence complete itself. But! We only know that any space is isometric to a subset of its completion, not that a completion containing the space exists (am I making sense?).
    The problem seems to me that for a given metric space M, you do not know if there are any other metric spaces containing it.
    I tried to show that the image of an absolutely closed metric space under an isometric function is itself absolutely closed, but I couldn't. If I could show that, then the image of M would be a closed subset of M's completion, and hence complete, thus M would be complete.

    Any help?
     
  2. jcsd
  3. Aug 30, 2009 #2

    quasar987

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    But isn't that isometry is also a bijection? Call a bijective isometry an isomorphism of metric spaces.

    If two metrics spaces are isomorphic, and one of them is complete, then surely the other is too, since Cauchy sequences are preserved by isomorphisms.
     
  4. Aug 30, 2009 #3
    Yes, but then we would have to show that the image of the absolutely closed space is also absolutely closed, or at least a closed subset of its completion. But it is not possible to show that the image of M as a subset of its completion is closed.

    Lets say our isometry from M to a subset of its completion is f. We want to show f(M) is a closed subset of the completion of M, let's call it C.
    Take a sequence in f(M) converging to some point in C...how do we show this limit is in f(M)?
     
  5. Aug 30, 2009 #4

    quasar987

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    I see...

    But suppose we have constructed the completion (M',d') of (M,d) and h:M-->M' is an injective isometry. Can't we define M'' as [itex]M\cup (M'-h(M))[/itex] with the obvious metric, so that M'' is a complete metric space effectively containing M?
     
  6. Aug 30, 2009 #5
    Yes, that looks like it will work.

    The metric we will use on M'' takes on the values of the corresponding points in M'. We must first check M'' is complete: obvious. If x_n is Cauchy in M'', the corresponding sequence is Cauchy in M' and hence converges in M', so x_n converges to the corresponding point in M''. M is absolutely closed, so is a closed subset of M'', hence it is complete. Is this correct?

    Thanks for your help quasar987, your solution is so obvious, yet I've been thinking about this problem for a few days.
     
  7. Aug 30, 2009 #6

    quasar987

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    I'm glad I could help!
     
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