Sequential Compactness

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In summary, the conversation discusses how to prove that every metric space that is sequentially compact and separable is compact. The participants suggest using the fact that a metric space is compact if and only if it is complete and totally bounded, and also discuss the possibility of using sequential compactness as a stepping stone. They also consider using the properties of dense subsets and infinite sequences to construct a sequence with no convergent subsequence, leading to a contradiction. Ultimately, it is agreed that the open sets must be nested in order to prove this statement.
  • #1
Dragonfall
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How do I prove that every metric space that is sequentially compact and separable is compact? I can't seem to use either hypotheses.
 
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  • #2
A metric space is compact iff it is complete and totally bounded. Also, I don't think the separable condition is needed.
 
  • #3
That's if you use the sequential definition of compactness. Here 'compactness' alone is defined a la topology; i.e., open covers, etc. Basically, I want to prove that if X is sequentially compact and separable (sequential => separable, but I can assume it without proof, since I proved in some earlier place), then for all open cover there exists a finite subcover for X.
 
  • #4
I thought if you couldn't just use that fact, it might at least be a good stepping stone. But actually it's probably easier to use sequential compactness as a stepping stone to get that fact.

If so, I would try assuming there is some open cover that has no finite subcover and use it to construct a sequence with no convergent subsequence. Specifically, construct a sequence of infinitely many disjoint open balls.
 
  • #5
Even if you construct such a sequence of open balls, how would you construct a sequence of points in X that has no convergent subsequence? X is sequentially compact.

The sequence of points in X must converge to something outside of X, but I can't think of what, since X could be complete.
 
  • #6
If X is complete and non-compact, it is not totally bounded. Take the real numbers for example. One open cover of them are the sets (n-1,n+1) as n ranges over the integers. One infinite sequence with no convergent subsequence is n, as n ranges over the integers. The connection won't be as straightforward in general, but you should be able to find one.
 
  • #7
X is sequentially compact and separable, and hence if an open cover A does not have a finite subcover, it has a countable subcover since we can map a dense subset into a countable subcover of A. Take an element that is in X\Ai for i=1 to infty and form a sequence, that sequence has a convergent subsequence that cannot converge to anything in the subcover, and hence outside X. Contradiction.

I don't know where you're getting at with completeness though.
 
  • #8
Dragonfall said:
X is sequentially compact and separable, and hence if an open cover A does not have a finite subcover, it has a countable subcover since we can map a dense subset into a countable subcover of A.

I'll agree with your conclusion, but I don't know what "since we can map a dense subset into a countable subcover of A" means.

Take an element that is in X\Ai for i=1 to infty and form a sequence, that sequence has a convergent subsequence that cannot converge to anything in the subcover, and hence outside X.

Why not? For example, if X is the set of real numbers and your cover is (n-1,n+1), take x_i=1 for the index corresponding to (-1,1) and x_i=0 for all other terms and you have a convergent sequence. I think you'll need your sets to be disjoint, and then pick an element in each set. I'm not 100% sure though. Sorry, I'm really busy now, but I'll look over this more carefully later tonight. If you need help sooner, maybe someone else has some ideas...
 
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  • #9
Ok, I've thought about it a bit more. Your idea in post 7 was on the right track, but you need the open sets to be nested. Pick x_i outside of A_i, then you can show that each A_i can only contain finitely many of the x_i, and use this to show there can be no convergent subsequence.
 

1. What is sequential compactness?

Sequential compactness is a property of a topological space, which means that every sequence in the space has a convergent subsequence.

2. How is sequential compactness different from compactness?

Sequential compactness only requires that every sequence has a convergent subsequence, while compactness requires that every open cover has a finite subcover. Therefore, sequential compactness is a weaker condition than compactness.

3. What are some examples of sequentially compact spaces?

Examples of sequentially compact spaces include closed and bounded intervals of real numbers, compact subsets of Euclidean spaces, and the Cantor set.

4. Is every compact space also sequentially compact?

Yes, every compact space is also sequentially compact. This is a consequence of the Bolzano-Weierstrass theorem, which states that every bounded sequence in Euclidean space has a convergent subsequence.

5. What is the significance of sequential compactness in mathematics?

Sequential compactness is an important concept in mathematics, particularly in the study of topological spaces and analysis. It is used to prove many important theorems, such as the Heine-Borel theorem, and is also a key property in the definition of a complete metric space.

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