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Math Help - inf and sup of a compact subset of R

  1. #1
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    inf and sup of a compact subset of R

    Let K be a compact set
    I have to show that for some a, b in K, we can show |a-p| = inf{|x-p|: x in K} and |b-p|=sup{|x-p|: x in K}

    My proof
    Since K is compact, K is closed and bounded. Since K is bounded, it is bounded above and below. that is, there is a B in R s.t. B>=x for all x in K. B is the least upper bound. Similarly, there is a A in R s.t. A <= x for all x in K. A is the greatest lower bound.

    Let p<k for all k in K. Then choose a = min {values in K}. Then a inf|x-p| = |a-p|

    Let p>k for all k in K. Then choose b=man {values in K}. Then sup|x-p|=|b-p|

    I don't think my last part is right. Please provide me with some feedback
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by inthequestofproofs View Post
    Let K be a compact set
    I have to show that for some a, b in K, we can show |a-p| = inf{|x-p|: x in K} and |b-p|=sup{|x-p|: x in K}

    My proof
    Since K is compact, K is closed and bounded. Since K is bounded, it is bounded above and below. that is, there is a B in R s.t. B>=x for all x in K. B is the least upper bound. Similarly, there is a A in R s.t. A <= x for all x in K. A is the greatest lower bound.

    Let p<k for all k in K. Then choose a = min {values in K}. Then a inf|x-p| = |a-p|

    Let p>k for all k in K. Then choose b=man {values in K}. Then sup|x-p|=|b-p|

    I don't think my last part is right. Please provide me with some feedback
    I have no idea what this means, but I would assume we're dealing with some metric space K is some compact metric space, and we have some x\in X-K. Show that d(x,K)=d(x,k) for some k\in K.


    Note that \rho:K\mapsto\mathbb{R} given by k\mapsto d(k,x) is a continuous mapping, and since K is compact we have that \rho assumes a minimum on K. The conclusion follows.
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  3. #3
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    compact sets

    Thank you! We haven't covered metric sets. This is a tough proof, but your insight might provide some clarity
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by inthequestofproofs View Post
    Thank you! We haven't covered metric sets. This is a tough proof, but your insight might provide some clarity
    Oh! I'm sorry, I didn't notice it said \mathbb{R}. The same concept applies, just replace the metric space with \mathbb{R}.
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