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

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SUMMARY

The discussion focuses on proving that for any non-empty compact subset K of a metric space X, there exists a point y in K such that the distance d(x,y) is less than or equal to the distance d(x,k) for every k in K. The triangle inequality is a key tool in this proof. Participants explored using proof by contradiction and the concept of infimum to establish the existence of such a point y, emphasizing the bounded nature of the set {d(x,k) : k in K}.

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  • Understanding of metric spaces and their properties
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  • Knowledge of the triangle inequality in mathematics
  • Concept of infimum in set theory
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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?
 

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