Compactness and every infinite subset has a limit point

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Discussion Overview

The discussion revolves around the statement from Rudin's Mathematical Analysis regarding compactness: if every infinite subset of a set C has a limit point in C, then C is compact. Participants explore the implications of this statement, its validity in different contexts, and the challenges in proving it, particularly in relation to metric spaces and the concepts of limit points and open covers.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to prove the statement and questions its truth, noting the need for a finite collection of open sets to cover C.
  • Another participant suggests that the statement may only hold in metric spaces and proposes using the Lebesgue number to assist in the proof.
  • A participant references a proof found online and seeks clarification on where the result is used in Rudin's text.
  • Discussion includes references to specific chapters and theorems in Rudin, particularly Theorem 2.41, which connects closed and bounded sets to compactness and limit points.
  • One participant recalls that in metric spaces, compactness is equivalent to completeness and total boundedness, suggesting these concepts might aid in proving the statement.
  • Another participant proposes assuming the existence of an infinite set without a limit point to find an open cover of C with no finite subcover as a potential approach to the proof.
  • A later reply discusses the relationship between limit point compactness and sequential compactness, outlining a sketch of a proof that connects subsequences and limit points.
  • One participant questions whether the original inquiry was about proving the equivalence of sequential compactness and open cover compactness.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the statement in various contexts, particularly regarding metric spaces. There is no consensus on the proof or the implications of the statement, and multiple competing approaches are presented.

Contextual Notes

Some participants note the difficulty of proving the statement and the reliance on specific properties of metric spaces, such as total boundedness and completeness. The discussion also highlights the need for clarity on definitions and assumptions related to compactness and limit points.

faradayslaw
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In studying Rudin's Mathematical Analysis, it seems that he uses the following statement:

If every infinite subset of a set C, has a limit point in C, then C is compact.

I can prove the converse of this statement, and its converse is also proven in the text, but I am unsure on how to prove the above statement, since I must provide a collection of finitely many open sets, whose union contains C, but the only thing I know is that there is a point in C complement s.t. every neighborhood of that point contains at least one point of C. The neighborhood of such a point is indeed open, but I don't know how to construct a finite collection of open sets whose union contains C. So, I was wondering if the above statement is actually true, and if so, what is the general reasoning behind it.
 
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I believe it is only true in metric spaces. Try using the Lebesgue number let me to prove it.
 
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I think you will find the proof in Rudin as a series of questions at the end of the chapter, if I recall correctly.

It begins with the notions of separability and of a base.
 
Thanks, there is a series of questions which leads to the result. Also, for those wondering, I was referring to Theorem 2.41 (ch. 2) in which it states if E subset of R^k has one of the following three properties, then it has the other two:

E is closed and bounded
E is compact
Every infinite subset of E has a limit point in E

Rudin does not mention anything about showing how 3 implies 2. He shows 1 implies 2 and 3, and how 3 implies 1. 2 implies 1 and 3 are shown previously.
 
as i recall, in a metric space, compact is equivalent to complete and totally bounded. i.e. every sequence has a cauchy subsequence and every cauchy sequence has a limit. maybe those concepts will allow you to prove it.i recall it being difficult for me to work out all this stuff last time i taught this course about 10-20 years ago. i recommend starting by proving that a closed bounded subset of R^2 is compact., then try to generalize by using totally boundedness as a version of the subdivision property of R^2 by small squares.
 
Last edited:
faradayslaw said:
In studying Rudin's Mathematical Analysis, it seems that he uses the following statement:

If every infinite subset of a set C, has a limit point in C, then C is compact.

I can prove the converse of this statement, and its converse is also proven in the text, but I am unsure on how to prove the above statement, since I must provide a collection of finitely many open sets, whose union contains C, but the only thing I know is that there is a point in C complement s.t. every neighborhood of that point contains at least one point of C. The neighborhood of such a point is indeed open, but I don't know how to construct a finite collection of open sets whose union contains C. So, I was wondering if the above statement is actually true, and if so, what is the general reasoning behind it.

Try assuming that there is an infinite set without a limit point and find an open cover of C with no finite subcover.
 
If you try to prove limit point compactness is equivalent to sequential compactness, it's actually rather natural. I sketched the proof below, so don't read it if you want to figure it out for yourself.



The main idea is that we can go back and forth between subsequences and infinite subsets of the space. Take a sequence, and ignore that it's a sequence. Then you have a subset, which must have a limit point. Pick such a limit point. Now you can show using cardinality arguments that there must be a subsequence converging to the limit point by only accepting points into your subsequence if they're closer to your limit point than the last point you accepted (and further along in the sequence of course). And the converse requires almost no work with sequential compactness.
 
i thought the question was how to prove sequential compactness implies open cover compactness.
 

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