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Prove that (-1,1) is homeomorphic to R (real numbers)

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that (-1,1) is homeomorphic to R (real numbers), with the topology given by the usual metric.


    2. Relevant equations

    None.


    3. The attempt at a solution

    I constructed the function f(x) = [1/(1-x) - 1/(1+x)]/2 = x/[(1+x)(1-x)] which is continuous and maps (-1,1) to R. Next I need to show that f has a continuous inverse. I'm stuck here.
     
  2. jcsd
  3. Oct 9, 2009 #2

    LCKurtz

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    Re: Homeomorphism

    Can you use theorems such as the formula for the derivative if the inverse of an increasing function?

    Or if not that, there is a similar looking trig function that might be easier to work with.
     
  4. Oct 9, 2009 #3

    HallsofIvy

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    Re: Homeomorphism

    It's not all that difficult to explicitely find the inverse function using the quadratic formula.
    Then the only "technical difficulty" is if y= 0.
     
  5. Oct 9, 2009 #4
    Re: Homeomorphism

    I'm still not sure. This is a topology course so no difficult analysis type theorems should be involved I think. To prove that the function (not its inverse) is continuous is it sufficient to say that it is a rational function who's denominator is non-zero on the interval?
    I can't think of any simpler trig function to use either.
     
  6. Oct 9, 2009 #5

    LCKurtz

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    Re: Homeomorphism

    That is "whose", not "who's". Think about tan(x) appropriately modified. Hard for us to know what theorems your instructor expects you to know and use. At some level, your statement about rational function would be sufficient to claim continuity.
     
  7. Oct 9, 2009 #6
    Re: Homeomorphism

    Hehe thanks for the spelling tip. Is it really so simple as f(x) = tan(pi*x/2) and
    x = (2/pi)arctan(y)?
    And hopefully since these are well known functions I don't have to prove that they're continuous right?
     
  8. Oct 9, 2009 #7

    LCKurtz

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    Re: Homeomorphism

    Hopefully :tongue2:
     
  9. Oct 10, 2009 #8
    Re: Homeomorphism

    Hi,I am doing a question very similar to this, but still don't understand... can you possibly explain further?
     
  10. Oct 10, 2009 #9
    Re: Homeomorphism

    What don't you understand?
     
  11. Oct 10, 2009 #10
    Re: Homeomorphism

    sorry, I think I get it now, but is it just enough to say that since f(x) as above is continuous and bijective, and x is continuous, therefore (-1,1) homeomorphic to the reals?
     
  12. Oct 10, 2009 #11

    Hurkyl

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    Re: Homeomorphism

    The inverse* of a continuous bijection need not be continuous.

    *: as defined when viewing it as a function on the underlying sets
     
  13. Oct 10, 2009 #12
    Re: Homeomorphism

    oh ok, it's just a definition that I have written down says a homeomorphic function must be continuous and invertible and the inverse function must also be continuous? Is that not necessarily true?

    so for madness' question above for example, would it suffice to say:

    Take f(x) as tan(pi*x/2) and f^-1(x) as (2/pi)arctan(y)

    this shows that (-1,1) is homeomorphic to the reals?
    That seems too easy? Does anything else need to be proved?
     
  14. Oct 10, 2009 #13

    Hurkyl

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    Re: Homeomorphism

    No, what you said here is true.

    It is also true that there exist continuous bijections whose inverse is not continuous.

    So when taken together, what do these imply?
     
  15. Oct 10, 2009 #14
    Re: Homeomorphism

    the inverse of a continuous bijection need not be continuous, but the inverse of a homeomorphic function must be continuous? ...hence the idea of a homeomorphism?
     
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