Compact sets are covered by finitely many epsilon disks

[redone632]

Homework Statement

Let K be a compact sebset of a metric space (X, d) and let $\epsilon$ greater than 0.
Prove that there exists finitely many points $x_1 x_2, ... x_n \in K$ such that K is a subset of the union of the $\epsilon$ neighborhoods about $x_i$

Homework Equations

N/A

The Attempt at a Solution

I think all I need to show is that all of the neighborhood epsilon disks about all the x's cover K and since K is is a compact set then finitely many of these neighborhoods cover K. If that is the correct approach then I'm just not sure how to start it. We didn't do any examples of proving that a certain set covers another.

 

[VeeEight]

Yes, that is the right approach. A space is compact if every open cover has a finite subcover. Take a cover of K by $\epsilon$ balls. If you assume there does not exist fintely many $x_1 x_2, ... x_n \in K$ such that K is a subset of the union of the $\epsilon$ neighborhoods about $x_i$, then this is an open cover without a finite subcover.

 

[redone632]

Alright, but how do I go about defining my epsilons to be?

 

[lanedance]

the proof should be sound for any epsilon

 

[redone632]

the proof should be sound for any epsilon

That's what I thought. We've just been dealing with defining epsilon so much I want to be sure. I'll go over my notes to see if I can come up with something. Thanks!