relationship among different kinds of compactness
The goal of this article is to prove
Proof. Choose an subset , and suppose has no limit points. Then contains its (vacuous set of) limit points and is therefore closed. But closed subsets of compact spaces are compact, so is compact. Since has no limit points, we may choose a neighborhood of each such that intersects only in . But this cover clearly has a finite subcover only if is finite. So any set without limit points is finite, and thus any infinite set has a limit point. This concludes the proof.
If is first countable, , and limit point compact, then is sequentially compact.
Proof. Let be any sequence of points in , and assume that takes infinitely many values (otherwise it obviously has a convergent subsequence). Choose a limit point for the sequence; we may assume wlog that is equal to for only finitely many (otherwise again the result holds trivially). So by ignoring a finite number of leading terms of the sequence, we may assume that for every . Since is first countable, choose a countable basis at ; by replacing with , we may assume that for all .
Now, choose such that . Inductively, assume we have chosen with . Since is , we may choose a neighborhood of that is disjoint from ; choose to be any point in . Then inductively the form a subsequence with , and clearly the converge to . This concludes the proof.
Note that every metric space and every second countable space is also first countable and .
Any sequentially compact metric space is second countable.
Claim first that for , the set of -balls in has a finite subcover. Suppose this is false for some particular . Let be any point, and construct inductively points with . Since is sequentially compact, we may replace the by a convergent subsequence, which we also call , with . But convergent sequences are Cauchy, so for large enough, we have , which contradicts the construction of the . This proves the claim.
If is second countable or is a metric space, and sequentially compact, then is compact.
Proof. Assume first that is second countable. Choose any open cover of ; it has a countable subcover . We use an argument very similar to that used in the above proposition. Suppose no finite subset of the covers , and choose . Since is sequentially compact, the have a convergent subsequence converging to . But for some ; since the converge to , all for large enough. But this is a contradiction to the construction of the , so that a finite subset of the cover and is compact.
Since any sequentially compact metric space is second countable by the above proposition, we are done.
The main theorem follows trivially from the above. Note that we have in fact proven the following set of implications:
Compact limit point compact for general topological spaces;
Limit point compact sequentially compact for first countable spaces;
Sequentially compact compact for second countable or metrizable spaces.
|Title||relationship among different kinds of compactness|
|Date of creation||2013-03-22 18:00:57|
|Last modified on||2013-03-22 18:00:57|
|Last modified by||rm50 (10146)|