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Thread: [SOLVED] Metric Space- Distance Function

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    Exclamation [SOLVED] Metric Space- Distance Function

    Hey guys, I have a problem from my Topology course that I can't seem to figure out. Any help would be appreciated. Here it is:

    The distance functions D'=D/(1+D) and D''= min{D,1} are each metrics on M. Let x$\displaystyle \in$M and let $\displaystyle \epsilon \in$ R>0.
    Let:

    S$\displaystyle \epsilon$'(x)={y$\displaystyle \in$M|D'(x,y)<$\displaystyle \epsilon$}
    S$\displaystyle \epsilon$''(x)={y$\displaystyle \in$M|D''(x,y)<$\displaystyle \epsilon$}

    a) Prove that S$\displaystyle \epsilon$''(x)$\displaystyle \subseteq$ S$\displaystyle \epsilon]$'(x).
    b)Prove that S$\displaystyle (\epsilon/(1+\epsilon))$'(x)$\displaystyle \subseteq$ S$\displaystyle \epsilon$''(x).

    It almost seems to me like the containment should be switched in part a at least, but I don't know.
    Thanks in Advance.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tidus89 View Post
    Hey guys, I have a problem from my Topology course that I can't seem to figure out. Any help would be appreciated. Here it is:

    The distance functions D'=D/(1+D) and D''= min{D,1} are each metrics on M. Let x$\displaystyle \in$M and let $\displaystyle \epsilon \in$ R>0.
    Let:

    S$\displaystyle \epsilon$'(x)={y$\displaystyle \in$M|D'(x,y)<$\displaystyle \epsilon$}
    S$\displaystyle \epsilon$''(x)={y$\displaystyle \in$M|D''(x,y)<$\displaystyle \epsilon$}

    a) Prove that S$\displaystyle \epsilon$''(x)$\displaystyle \subseteq$ S$\displaystyle \epsilon]$'(x).
    b)Prove that S$\displaystyle (\epsilon/(1+\epsilon))$'(x)$\displaystyle \subseteq$ S$\displaystyle \epsilon$''(x).

    It almost seems to me like the containment should be switched in part a at least, but I don't know.
    Thanks in Advance.
    This notation is hard to understand (at least for me), but I would tend to agree. If $\displaystyle D>1$ then $\displaystyle \frac{D}{1+D}=1-\frac{1}{1+D}<1=\min\left(D,1\right)$
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