# Compactness and every infinite subset has a limit point

1. Jan 13, 2012

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.

2. Jan 14, 2012

### JG89

I believe it is only true in metric spaces. Try using the Lebesgue number lemme to prove it.

Last edited: Jan 14, 2012
3. Jan 14, 2012

### AxiomOfChoice

I am interested in where (chapter and section, since I'm not sure our page numbers would match) you think this result is used.

Anyway, proving this ain't easy, but I found a proof online over at http://www.math.umn.edu/~jodeit/course/CPTandLIMITpts.pdf.

4. Jan 14, 2012

### dylanbyte

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.

5. Jan 14, 2012

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.

6. Jan 14, 2012

### mathwonk

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: Jan 15, 2012
7. Jan 15, 2012

### lavinia

Try assuming that there is an infinite set without a limit point and find an open cover of C with no finite subcover.

8. Jan 15, 2012

### ebola1717

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.

9. Jan 17, 2012

### mathwonk

i thought the question was how to prove sequential compactness implies open cover compactness.