In this post we discuss several notions of compactness, which are presented in the context of the real line. This is part of a series of posts on the topology of the real line. Our goal is to use this series of posts on the real line to motivate the topological notions being discussed. For basic definitions of open intervals, open sets and closed sets of the real line, refer to these previous posts:
This is a post on introductory topology. Concepts discussed here are basic notions found in beginning topology courses and elementary analysis courses. For the readers who are taking math courses that transition their math career from calculation to proving of theorems, we encourage such readers to think about the concepts and theorems discussed here before reading the proofs.
Let be the set of all real numbers. Let be a collection of subsets of . The set denotes the union of all sets in and is the set of all real numbers which belong to at least one set .
Let . By an open cover of , we mean a collection of open sets of such that , that is, every point in belongs to some open set in . The subset of is said to be compact if for each open cover of , there is a finite subcollection of such that .
The real line is not compact since is an open cover of such that no finite subcollection of which is a cover of the real line.
Any convergent sequence along with its limit is a compact set. For example, the set is compact since any open interval containing contains all but finitely many points of the convergent sequence .
Any subset of the real line that is not closed can never be compact since every compact subset of the real line is closed. Thus we have the following theorem.
Let . If is compact, then is closed in .
Suppose is not closed. Then there is a limit point of such that . Let be the set of all positive integers. Then the following is an open cover of such that it has no finite subcollection covering :
Therefore, if is compact, then must be closed.
Recall from the previous post (The Euclidean topology of the real line (2)) that open sets and closed sets are relative to the space in which they are embedded. For example, take . The singleton set is surely not an open subset of the whole real line. However, it is open relative to the space . The property of being closed sets is also relative. The interval is a closed set relative to the space . However, is not a closed set in the whole real line since both endpoints are limit points that are not in the .
The notion of compactness is not a relative concept. Let . Let’s call the definition of being a compact set as defined above the property of being compact relative to . We say that is compact relative to if in the same definition of compactness above, the open cover consists of open sets relative to . It turns out that these two notions are equivalent. We have the following theorem.
Let . Then is compact relative to if and only if it is compact relative to .
Let be compact relative to . We show that is compact relative to . To this end, let be an open cover of . Here consists of open sets of . Then the following is an open cover of consisting of open sets relative to .
Since is compact relative to , has a finite subcollection that is also an open cover. The following lists out this finite subcollection:
Then and covers .
Let be compact relative to . We show that is compact relative to itself. To this end, let be an open cover of consisting of open sets relative to . Note that each set in is of the form where is an open subset of . So consider the following:
Note that is an open cover of . Since is compact relative to , we have a finite subcollection such that is a cover of . Consider the following:
Note that is a cover of and is a finite subcollection of .
The above theorem shows that compactness does not depend on the space in which the compact set is embedded. A subset of the real line is compact in its own right as well as compact relative to the real line. This same thinking applies in topological spaces in general.
Let . Suppose that is a compact set. Then if is a collection of compact subsets of such that the intersection of every finite subcollection of is nonempty, then .
Suppose is a collection of compact subsets of with the property indicated in the theorem (called the finite intersection property). We want to show that . Suppose . Then the following is an open cover of the compact set .
Since is compact, there exists such that covers . This means , contradicting that the collection has the finite intersection property.
There are other notions of compactness that are equivalent in Euclidean spaces (the real line in particular) and some in metric spaces in general. Let . The set is said to be countably compact if every countable open cover of has a finite subcollection that is also a cover of , i.e. there exists some positive integer such that . The set is said to have the Bolzano-Weierstrass property if is closed relative to and every infinite subset of has a limit point. The set is said to be sequentially compact if is closed relative to and every sequence of points of has a subsequence that converges.
Let be a collection of open subsets of . Then there is a countable subcollection such that .
Proof. Note that there are only countably many open intervals with rational endpoints. Thus we can enumerate these open intervals in a sequence:
Let . For each , there is an open set such that . Furthermore, there is an open interval such that the endpoints are rational numbers and . Each is associated with at least one such that . Thus for each , we choose one such that . Then is the desired subcollection.
By definition, any compact subset of the real line is also countably compact (this is also true in any topological spaces). The above lemma indicates that in the real line, the notion of compactness is equivalent to the notion of countably compactness. Outside of Euclidean spaces, there are classes of spaces in which these two notions coincide. But the topological structure along does not guarantee that these two notions of compactness coincide.
There are also other topological notions at work in this lemma, namely the Lindelof property and the property of having a countable base. A topological space is Lindelof if every open cover of the space has a countable subcollection that is also an open cover. The proof for lemma 4 essentially shows that any space whose topology is generated by a countable base is Lindelof. Furthermore, the Lindelof property is hereditary (any subspace is also Lindelof).
Let . The following are equivalent:
- is compact.
- is countably compact.
- is closed and bounded.
- has the Bolzano-Weierstrass property.
- is sequentially compact.
This follows from definition and lemma 3. See the remark right below lemma 3.
Suppose that is countably compact. The fact that it is a closed set of follows from the same proof in Theorem 1. We only need to show is bounded. Suppose not. Then no open interval contains where is positive integer. So is an open cover of such that no finite subset of can cover , implying that is not countably compact. So if is countably compact, it must be bounded.
Suppose that is closed and bounded. The Bolzano-Weierstrass property follows from condition 4 of the main theorem in the previous post (The Euclidean topology of the real line (3) – Completeness).
Suppose that has the Bolzano-Weierstrass property. We claim that is a bounded set. Once we show that it is bounded, the property that is sequentially compact follows from the direction of in the main theorem in the previous post (The Euclidean topology of the real line (3) – Completeness).
Suppose that is not bounded. Then for each positive integer , there is some such that . The set is an infinite set. Note that is unbounded and every infinite subset of is also unbounded. By the Bolzano-Weierstrass property, has a limit point . Then every open interval containing contains infinitely many points of , contradicting the fact that every infinite subset of is unbounded.
Suppose that is sequentially compact. Suppose that is not countably compact. Then there is an open cover such that no finite subset of can cover . Then for each , choose such that .
By the assumption of sequential compactness, the sequence has a convergent subsequence where is an infinite subset of . Any convergent sequence of real numbers is bounded, contradicting the fact that the orginal sequence and any infinite subsequence of it are unbounded.
As consequence of Theorem 3 and theorem 5, we have the following corollary.
Suppose we have closed intervals such that .
In Theorem 5, the direction is the Heine-Borel theorem. Theorem 5 completely characterizes the compact subsets of the real line, namely the subsets that are closed and bounded (closed relative to the real line). By the theorem, closed intervals and closed subsets of are the only compact subsets of the real line. The theorem also holds in higher Euclidean spaces .
Of the properties 2 through 5 listed in Theorem 5, it is a great topology exercise to find out which of the conditions are equivalent to compactness when considering spaces other than Euclidean spaces. What if we venture outside of Euclidean spaces? For example, what if we just consider abstract metric spaces? What if we only consider topological spaces? In cases where countably compact does not imply compactness, what kind of additional conditions would make countably compactness equivalent to compactness?
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- Steen, L. A. and Seebach, J. A., Counterexamples in Topology, 1995, Dover Publications, Inc, New York.
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