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Thread: [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 $\displaystyle K \subseteq X$ be a sequentially compact subspace. Let p be a point in X\K. Prove that there is a point $\displaystyle q_{0} $ that is closet to p of all points in K in the sense that

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

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

    Define the function $\displaystyle f:K \rightarrow R$ by $\displaystyle f(q)=d(q,p)$ 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 $\displaystyle K \subseteq X$ is sequentially compact subspace of the metric space X. Then K is a closed subset of X. Define the function $\displaystyle f:K \rightarrow R$ by $\displaystyle f(q)=d(q,p)$ 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 $\displaystyle \left| {f(q) - f(r)} \right| = \left| {d(p,q) - d(p,r)} \right| \le d(q,r)$.
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    Quote Originally Posted by Plato View Post
    For this point you know that $\displaystyle \left| {f(q) - f(r)} \right| = \left| {d(p,q) - d(p,r)} \right| \le d(q,r)$.
    Hello Plato, I'm curious as how do you derive this? Is this because of the Triangle Inequality?

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

    so from this

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

    but how would you get this

    $\displaystyle |d(p,q)-d(p,r)| \leq d(q,r)$?
    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 $\displaystyle K \subseteq X$ be a sequentially compact subspace. Let p be a point in X\K. Prove that there is a point $\displaystyle q_{0} $ that is closet to p of all points in K in the sense that

    $\displaystyle d(p,q_{0}) \leq d(p,q)$ for all points q in K.
    Is this point unique?
    For the (non)uniqueness, let $\displaystyle X=\mathbb{R}^2$ with the norm $\displaystyle \|(x_1,x_2)\| = \max\{|x_1|,|x_2|\}.$ Take K to be the closed unit ball, $\displaystyle K = \{(x_1,x_2):|x_1\leqslant1,|x_2|\leqslant1\}.$ 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; Aug 12th 2008 at 10:42 AM.
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  5. #5
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    First we have $\displaystyle \left\{ \begin{array}{l} d(p,q) \le d(p,r) + d(r,q) \\
    d(p,q) - d(p,r) \le d(r,q) \\ \end{array} \right.$
    Likewise we get $\displaystyle d(p,r) - d(p,q) \le d(q,r) = d(r,q)$.
    Putting these together we have $\displaystyle - d(r,q) \le d(p,r) - d(p,q) \le d(r,q)$.
    QED
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  6. #6
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    Thank you Opalg and Plato.
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