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Is this a homeomorphism that does not preserve metric completeness?

  1. Apr 12, 2012 #1
    I'm well aware and understand that homeomorphism do not need to preserve metric completeness, I'm just trying to work out a simple counterexample. I have tried searching around just for kicks, but only seem to find more complex ones. I'm wondering if the one I have works for it for sure?

    On the interval [-pi/2, pi/2], use the sequence {an} such that an = pi/2 - 1/n. Then {an} converges to pi/2. Use the functions f(x) = arctan(x) and and f-1(x) = tan (x) as the functions for the homeomorphism. Then the fist metric would be complete since {an} is cauchy and converges in the set, but the second metric would not complete since an would not converge in the set.

    Does that work? Am I off a little bit?
     
  2. jcsd
  3. Apr 12, 2012 #2

    micromass

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    What would [itex]f(\pi/2)[/itex] be?? A function must have an image for each point.
     
  4. Apr 12, 2012 #3
    f(pi/2) = arctan(pi/2) which is like 1.000?? something isn't it? Do you mean f-1? because tan(pi/2) would be undefined sadly...

    Making it an open interval wouldn't help would it? Because then an wouldn't be convergent since pi/2 is not in the set.
     
  5. Apr 12, 2012 #4

    micromass

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    Indeed, so that won't work. But perhaps we can adjust the idea a bit?

    We know that there is a homeomorphism between [itex]]-\pi/2,\pi/2[[/itex] and [itex]\mathbb{R}[/itex]. Indeed, the tangent you just listed.

    Now, [itex]\mathbb{R}[/itex] carries a metric (the absolute value function). So what if we "pull back that metric through the homeomorphism.

    That is, for [itex]x,y\in ]-\pi/2,\pi/2[[/itex], define [itex]d(x,y)=|\tan(x)-\tan(y)|[/itex]. Is the sequence in your OP a Cauchy sequence for your metric?? Is the space complete under this metric??

    Of course, we also equip [itex]]-\pi/2,\pi/2[[/itex] with the usual absolute value metric. Does the sequence in your OP converge for your metric?? Is the space complete under this metric??

    Does there exist a homeomorphism between these two spaces??
     
    Last edited: Apr 12, 2012
  6. Apr 12, 2012 #5
    I must say I'm a little confused my topology isn't great, so I'm trying to get there.

    The first metric would be Cauchy and complete I think? The second one wouldn't? Can you use the original functions to to create the homeomorphism? And weren't we trying to construct the homeomorphism between [-∏/2, ∏/2] and ℝ, which were the ones I listed?

    Sorry, I feel way off now..
     
  7. Apr 12, 2012 #6

    micromass

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    The sequence would not be Cauchy in the first metric. But the space would be complete.

    The sequence would be Cauchy in the second metric, so not convergence. So the space would not be complete.

    Yeah, but that won't work. So I'm trying to give you two metric spaces (one of which is complete and one of which is not) that are homeomorphic.
     
    Last edited: Apr 12, 2012
  8. Apr 12, 2012 #7
    I disagree. Homeomorphisms preserve, among other things, compactness. While [itex][-\pi/2,\pi/2] [/itex] is compact (by, say, Heine-Borel), [itex]\mathbb{R}[/itex] is not.
     
  9. Apr 12, 2012 #8

    micromass

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    I never said there is a homeomorphism between the two spaces you listed. I said there is one between the OPEN interval and [itex]\mathbb{R}[/itex].
     
  10. Apr 12, 2012 #9
    Possibly I'm having trouble with the formatting, but it looks an awful lot like you used the closed brackets.

    In any event, apologies for the misunderstanding.
     
  11. Apr 12, 2012 #10
    The definition I was given for complete is that a space is complete if every Cauchy sequence is convergent in the space. So for the first metric, how could the space couldn't be complete if the sequence was not Cauchy.. The absolute value of arctan should get closer and the functions does not sprout off to infinity like tan(x) would, so I don't quite get that.

    Can you use just a simple f:(-∏/2, ∏/2) defined by f(x)= x, just to keep it simple. Then ℝ would be complete but (-∏/2, ∏/2) wouldn't be with the metrics above?
     
  12. Apr 12, 2012 #11
    And just to help my confusion, in your post explaining each metric, you said the first metric was cauchy and was not cauchy
     
  13. Apr 12, 2012 #12

    micromass

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    To show that the first space is complete, then it of course isn't sufficient to show that that sequence is Cauchy. We need to prove more.

    However, there is an isometry between the open interval [itex]]-\pi/2,\pi/2[[/itex] with the first metric and [itex]\mathbb{R}[/itex] with the standard metric. And this isometry preserves completeness.

    Sorry, that were typos. I meant to say that the first metric is cauchy and the second metric is not.

    Furthermore, I made another typo. I want the first metric on [itex]]-\pi/2,\pi/2[[/itex] to be

    [tex]d(x,y)=|\tan(x)-\tan(y)|[/tex]

    I corrected it.
     
  14. Apr 12, 2012 #13
    Micro, a little off-topic, is this a modern trend to use ]0, 1[ to denote the open unit interval? I've seen it, but not very often. Is it gaining mindshare because people confuse (0,1) with a particular point in 2-space?
     
  15. Apr 12, 2012 #14

    micromass

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    Don't know that really. I have always learned to use ]0,1[ and not (0,1).
    This page: http://demonstrations.wolfram.com/IntervalNotation/ says that the difference in notation is between Europe and America. That would make sense.

    I don't know if it's gaining much mindshare.
     
  16. Apr 12, 2012 #15
    It's sufficient to say I'm lost...I think I'm exhausted from looking at top. all day..

    So we have a complete metric space and an incomplete metric space, they're both on the open interval from -pi/2 to pi/2 into the reals. So all you need are the continuous functions linking the two metrics? Still using the sequence an = pi/2 - 1/n
     
  17. Apr 12, 2012 #16

    micromass

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    All I'm saying is that if we equip the open interval [itex]]-\pi/2,\pi/2[[/itex] with the metric

    [tex]d(x,y)=|\tan(x)-\tan(y)|[/tex]

    then the space will be complete.
    If we equip the open interval with the metric

    [tex]|x-y|[/tex]

    then the space will be incomplete.

    However: [itex]T:]-\pi/2,\pi/2[\rightarrow ]-\pi/2,\pi/2[:x\rightarrow x[/itex] is a homeomorphism.

    Alternatively, you can show that both [itex]]-\pi/2,\pi/2[[/itex] and [itex]\mathbb{R}[/itex] with their standard metrics are homeomorphic. But one space is complete and the other one is not. That might be a little bit easier than my weird metrics.
     
    Last edited: Apr 12, 2012
  18. Apr 12, 2012 #17
    They both help, big time! Thank you!!! My biggest problem was that I've not seen a straight up example, just was told this fact, so was having trouble constructing and trying to include the metrics, and figuring out completeness since we didn't do a whole with that either..just all confused me
     
  19. Apr 12, 2012 #18

    mathwonk

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    send punctured circle to real line.
     
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