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The open interval (0, 1)

  1. Feb 11, 2013 #1
    The open interval (0, 1) is isomorphic to ℝ. One can find many bijections, yet ℝ is complete but (0, 1) is not (because the Cauchy seq. 1/n has no limit point in (0, 1)?

    what am I missing? isomorphism means groups are and behave similarly?
     
  2. jcsd
  3. Feb 12, 2013 #2

    jbunniii

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    Completeness is a metric property, not a topological property, so it is not necessarily preserved by homeomorphisms.

    Another example: ##(0,\infty)## with the usual metric is homeomorphic to ##(0,\infty)## with the metric ##d(x,y) = |\log(x) - \log(y)|##, but the latter space is complete whereas the former is not. This is because with the log metric, it's not possible to get close to the missing point 0: no matter how small we choose ##x \in (0, \infty)##, we have ##d(x, x/2) = \log(2)##.
     
  4. Feb 12, 2013 #3
    Interesting. The more I learn the more I find out I know nothing.
     
  5. Feb 12, 2013 #4
    So I guess completeness depends on the metric and is independent of the metric space we choose.
     
  6. Feb 12, 2013 #5

    micromass

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    It is dependent of the metric space we choose! Completeness depends indeed crucially on the metric. It is not a topological concept since we can not define what complete is by using only open sets. This is the reason that it is not preserved by homeomorphisms. You need some metric space (or a bit more general: a uniform or approach space) to make sense of completeness.

    A related concept is that of "completely metrizable". A topological space is completely metrizable if there exists some metric on the space that makes it complete. Of course, that doesn't mean that every metric makes it complete. This notion of "completely metrizable" is preserved under homeomorphisms. So both [itex](0,1)[/itex] as [itex]\mathbb{R}[/itex] are completely metrizable, even though [itex](0,1)[/itex] is not complete under the usual metric.

    Other notions which need a metric are Cauchy sequence, totally bounded, uniform continuity. None of these can be expressed properly in topological context. They are thus not preserved under homeomorphisms.
     
  7. Mar 1, 2013 #6

    Bacle2

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    There is an obscure result that every Gδ subset of a complete metric space is topologically complete, and the proof is constructive, i.e., the actual metric is constructed in the proof. As an example, the irrationals are subset of the reals that are topologically complete. Sorry don't have much time now, do you want to know the actual argument and metric?
     
  8. Mar 1, 2013 #7
    By all means...Thank you.
     
  9. Mar 1, 2013 #8

    Bacle2

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    Glad to help. I'll do the proof when I have more time. Still, I need to think thru to see if this is an iff. result.
     
  10. Apr 1, 2013 #9

    Bacle2

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    Here is the proof:

    You use these facts: an open or closed subset of topologically-complete spaces, the countable product of
    topologically-complete spaces are all topologically-complete. You also use the fact that the diagonal in a Hausdorff
    space is closed ( we work with the diagonal in a product space). Let your Gδ set D be given as D= O1\capO2...

    And use the embedding d--> (d,d,d....) of D into O1xO2x......

    Then D is a closed subset of the Hausdorff space ( the product of the Oi's) , which is itself a
    Gδ , as a countable product of Gδ's .

    Let me work a bit more on the proof that gives you the actual metric.

    This is just a sketch from my notes, which were not too clear.
     
    Last edited: Apr 1, 2013
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