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Math Help - Topologically complete spaces

  1. #1
    Member ModusPonens's Avatar
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    Topologically complete spaces

    Hello.

    I'm having trouble doing exercise 7.43.6.c of Munkres' Topology. It says:

    Show that an open subspace of a topologically complete space is topologically complete.

    Now, why isn't the open interval (0,1), considered as a subspace of R, a counterexample to what the question is afirming? (since the sequence 1/n is a Cauchy sequence that converges to 0, a point which isn't in (0,1))

    Thanks.
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  2. #2
    Super Member girdav's Avatar
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    Re: Topologically complete spaces

    Hello,
    (X,d) is topologically complete means that we can find a metric d' which gives the same topology as d such that (X,d') is complete. Here, with X=(0,1) it doesn't work with the usual metric but with d'(x,y)=\left|\tan \frac{\pi x}2-\tan \frac{\pi y}2\right| it should.
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  3. #3
    Member ModusPonens's Avatar
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    Re: Topologically complete spaces

    I think I get the idea. That metric is to avoid sequences that converge to the frontier of (0,1), right? By the way, if that's the idea, that metric doesn't avoid that sequences converge to 0. Shouldn't it be  d'(x,y)=|tan(\pi (x-1/2))-tan(\pi (y-1/2))|
    Last edited by ModusPonens; April 29th 2012 at 11:30 AM.
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  4. #4
    Super Member girdav's Avatar
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    Re: Topologically complete spaces

    It's possible, I didn't make the computations.
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