-1

B is a finite non-empty set which belongs to $\mathbb{R}^n$. I have to prove that it is compact. The two necessary conditions for a set to be compact are that it should be bounded and closed.

B is bounded. Since it has a finite number of elements, an infimum and supremum exist. This means that it is bounded since both a supremum and infimum exist.

However, I don't know how to prove that B is closed. Would it be sufficient proof to say that no limit points exist in B since it is finite and it is closed because B contains all the limit points that exist?

Thank you so much!

J.-E. Pin
  • 40,163
realanalysisbeta
  • 1
  • 1
    Hint: show a single point is closed. Then a finite set is a finite union of closed sets. – Thomas Andrews Oct 02 '21 at 23:53
  • 1
    Or you can show the complement is open pretty easily. – Thomas Andrews Oct 02 '21 at 23:54
  • It is actually incredibly easy to prove the compactness using the topological definition of compact space/subset. Take a covering of B made of open sets, then for each point choose one set that contains it, and you obtain a finite covering. – Lorenzo Pompili Oct 03 '21 at 00:00
  • @ThomasAndrews Okay thank you so much for your response. Can you please let me know if I am correct with my proof? C is a subset of B. C = {1}. C is closed because no limit points exist as all points in a finite set are isolated points. The same is true for all finite points in B as B is just a union of all finite points. Since no limit points exist, an empty set exists for the set of all limit points. The empty set is both closed and open thus satisfying the condition for being compact. – realanalysisbeta Oct 03 '21 at 00:18

2 Answers2

1

I would use the “open cover” definition of compactness. You have a set of n points that you cover with open sets. At worst, you only need n open sets, one for each point. That’s a finite subcover, so it’s compact.

Rob Dukes
  • 69
  • 6
1

In the metric space case, your definition is probably: "from every sequence you can extract a convergent subsequence". If your set is finite, any sequence will have some element repeated infinitely many times. That subsequence is clearly convergent.

GReyes
  • 16,446
  • 11
  • 16