What is the Closest Point to a Compact Set in a Metric Space?

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Homework Statement



Let X be a metric space and let K be any non-empty compact subset of X, and let x be an element of X. Prove that there is a point y is an element of K st d(x,y) leq d(x,k) for every k an element of K.

Homework Equations



triangle inequality


The Attempt at a Solution



i tried proof by contradiction with the triangle inequality, and it didn't get me anywhere
 
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Look at the set {d(x,k) : k in K}. Note in particular that it's bounded below.
 
its bounded below as k gets closer to x is that inf K?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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