Compact sets and sup/inf

  • #1
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Homework Statement



Show if K contained in R is compact, then supK and inf K both exist and are elements of K.

Homework Equations





The Attempt at a Solution



Ok we proved a theorem stating that if K is compact that means it is bounded and closed.

So if K is bounded that means that for every element k in K there exists an M in R so that,
|k|<M
which is equiv to:
-M<k<M
meaning for all k in K
k>-M and k<M and therefore K is bounded above AND below.
According to the axiom of completeness, that means that the supK and the infK exists.

Alright, this is where I get stuck, I know that the supK and infK are the limit points of K, and since we know that K is closed, they would be contained in K, but I don't know how to show that supK and infK are the limit points. Any help would be great!
 

Answers and Replies

  • #2
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Ok, first of all, if K is finite, then it doesn't have a limit point; however, it's trivial to prove the theorem for this.

Let s be the sup. Then by definition every open ball centered at s contains a point in K. Can you construct a sequence of points in K converging to s from this?
 
  • #3
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so could I just say as epsilon approaches zero the epsilon-neighborhood (ball) around s gets smaller and smaller and thus s is a limit point of K, and since K is closed, it contains s? Then i can make a similar argument for the infK?
 
  • #4
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I think you have the right idea and just need that final step in order to make it precise. To do that, however, requires that you know exactly what definitions you are using for each concept. What definition are you using for the limit point?
 
  • #5
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well we have one definition and one theorem for the limit points. The first is x is a limit point of A if every epsilon neighborhood Ve(x) intersects the set A in some point other than x. The theorem states a point x is a limit point of A if and only if x=lim(a_n) for some sequence (a_n) contained in A satisfying a_n is unequal to x for all n.

Couldn't I just use the definition of the supremum and say that since supK is the LEAST upper bound, that would mean for any epsilon greater than zero, s-epsilon is an element in K (because then if it wasn't then s-epsilon would be the supremum). Which would mean that every epsilon neighborhood Ve(s) intersects the set K in some point other than s, which would make s a limit point of K. And since K is closed, s would be contained in K.
 
  • #6
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Yes. That's the reason I asked you about your definition of limit point. If that's what you're using, then you can just simply say that.
 
  • #7
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thanks so much for the help!
 

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