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Thread: Proof relative to metric spaces.

  1. #1
    Junior Member RaisinBread's Avatar
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    Proof relative to metric spaces.

    I'm trying to do a proof on some metric spaces, I think I've almost completed it but I am stuck on the last part of it.

    The two metric spaces I am working with are $\displaystyle (R^+,d_1)$ $\displaystyle (R^+,d_2)$ where $\displaystyle d_1(x,y)=|x-y|$ and $\displaystyle d_2(x,y)=|ln(y/x)|$

    So far I can show that any Cauchy sequence in d2 is also a cauchy sequence in d1, in short because if $\displaystyle |ln(y/x)|$ converges to zero then (y/x) has to converge to one, which means that the difference between x and y has to converge to zero.

    I can similarly show that whenever a sequence converges to some limit $\displaystyle l\in R^+$ (notice it cant be zero) in d1, then it converges to the same limit in d2.

    Now the only thing that I need to prove next is that if a sequence is Cauchy in both metrics, it cannot converge to zero in d1, or equivalently if a sequence converges to zero in d1, it will not be Cauchy in d2.

    I don't really know how to approach this; I see that if x and y are really close to zero, ln(x) and ln(y) will be very large, but I can't manage to turn this into something useful...
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Re: Proof relative to metric spaces.

    Notice that $\displaystyle \bigl|\ln(y/x)\bigr| = \bigl|\ln y -\ln x\bigr|$. It follows that if the sequence $\displaystyle (x_n)$ is Cauchy for d2, then the sequence $\displaystyle (\ln(x_n))$ is Cauchy for the usual metric d1 and therefore converges to a finite limit. So $\displaystyle x_n$ must be bounded away from 0.
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  3. #3
    Junior Member RaisinBread's Avatar
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    Re: Proof relative to metric spaces.

    Eureka, thank you very much.
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