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Math Help - [SOLVED] Closet Point/Sequentially compact subspace

  1. #1
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    [SOLVED] Closet Point/Sequentially compact subspace

    Problem:
    Let X be a metric space and let K \subseteq X be a sequentially compact subspace. Let p be a point in X\K. Prove that there is a point q_{0} that is closet to p of all points in K in the sense that

    d(p,q_{0}) \leq d(p,q) for all points q in K.
    Is this point unique?

    ====================
    Suppose that K \subseteq X is sequentially compact subspace of the metric space X. Then K is a closed subset of X.

    Define the function f:K \rightarrow R by f(q)=d(q,p) for all q in K and p be a point in X\K. I want to show that this function f is continuous first, but I don't know how nor what theorem to use.

    Thank you for your time. Much appreciated.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    Suppose that K \subseteq X is sequentially compact subspace of the metric space X. Then K is a closed subset of X. Define the function f:K \rightarrow R by f(q)=d(q,p) for all q in K and p be a point in X\K. I want to show that this function f is continuous first, but I don't know how nor what theorem to use.
    For this point you know that \left| {f(q) - f(r)} \right| = \left| {d(p,q) - d(p,r)} \right| \le d(q,r).
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    Quote Originally Posted by Plato View Post
    For this point you know that \left| {f(q) - f(r)} \right| = \left| {d(p,q) - d(p,r)} \right| \le d(q,r).
    Hello Plato, I'm curious as how do you derive this? Is this because of the Triangle Inequality?

    d(p, q)= |p(x)-q(x)| \leq |p(x)-r(x)| + |r(x)-q(x)| \leq d(p,r) + d(q,r)

    so from this

    d(p,q) \leq d(p,r) + d(q,r)

    but how would you get this

    |d(p,q)-d(p,r)| \leq d(q,r)?
    Would you just subtract d(p,r) from both sides?

    Thank you.
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    Quote Originally Posted by Paperwings View Post
    Let X be a metric space and let K \subseteq X be a sequentially compact subspace. Let p be a point in X\K. Prove that there is a point q_{0} that is closet to p of all points in K in the sense that

    d(p,q_{0}) \leq d(p,q) for all points q in K.
    Is this point unique?
    For the (non)uniqueness, let X=\mathbb{R}^2 with the norm \|(x_1,x_2)\| = \max\{|x_1|,|x_2|\}. Take K to be the closed unit ball, K = \{(x_1,x_2):|x_1\leqslant1,|x_2|\leqslant1\}. So K is a square with vertices at the points (1,1). Let p be the point (2,0). The distance from p to K is 1, but d(p,q)=1 for all points q of the form q=(1,y) where -1≤y≤1.
    Last edited by Opalg; August 12th 2008 at 11:42 AM.
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    First we have \left\{ \begin{array}{l} d(p,q) \le d(p,r) + d(r,q) \\ <br />
 d(p,q) - d(p,r) \le d(r,q) \\  \end{array} \right.
    Likewise we get d(p,r) - d(p,q) \le d(q,r) = d(r,q).
    Putting these together we have  - d(r,q) \le d(p,r) - d(p,q) \le d(r,q).
    QED
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  6. #6
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    Thank you Opalg and Plato.
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