[SOLVED] Closet Point/Sequentially compact subspace

Printable View

August 11th 2008, 05:06 PM
Paperwings

[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.
August 12th 2008, 04:44 AM
Plato

Quote:

Originally Posted by Paperwings

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 $\left| {f(q) - f(r)} \right| = \left| {d(p,q) - d(p,r)} \right| \le d(q,r)$ .
August 12th 2008, 09:34 AM
Paperwings

Quote:

Originally Posted by Plato

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.
August 12th 2008, 10:23 AM
Opalg

Quote:

Originally Posted by Paperwings

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.
August 12th 2008, 10:33 AM
Plato

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
August 12th 2008, 11:13 AM
Paperwings

Thank you Opalg and Plato.