Countably compact space explained

In mathematics a topological space is called countably compact if every countable open cover has a finite subcover.

Equivalent definitions

A topological space X is called countably compact if it satisfies any of the following equivalent conditions:^{[1] [2]}

(1) Every countable open cover of X has a finite subcover.

(2) Every infinite set A in X has an ω-accumulation point in X.

(3) Every sequence in X has an accumulation point in X.

(4) Every countable family of closed subsets of X with an empty intersection has a finite subfamily with an empty intersection.

(1)

⇒ (2): Suppose (1) holds and A is an infinite subset of X without \omega -accumulation point. By taking a subset of A if necessary, we can assume that A is countable.Every x\in X has an open neighbourhood O _{x} such that O _{x\cap} A is finite (possibly empty), since x is not an ω-accumulation point. For every finite subset F of A define O _{F} = \cup\{O _{x:} O _{x\cap} A = F\} . Every O _{x} is a subset of one of the O _{F} , so the O _{F} cover X. Since there are countably many of them, the O _{F} form a countable open cover of X. But every O _{F} intersect A in a finite subset (namely F), so finitely many of them cannot cover A, let alone X. This contradiction proves (2).

(2)

⇒ (3): Suppose (2) holds, and let (x _{n)} _{n} be a sequence in X. If the sequence has a value x that occurs infinitely many times, that value is an accumulation point of the sequence. Otherwise, every value in the sequence occurs only finitely many times and the set A = \{x _{n:} n\in N\} is infinite and so has an ω-accumulation point x. That x is then an accumulation point of the sequence, as is easily checked.

(3)

⇒ (1): Suppose (3) holds and \{O _{n:} n\in N\} is a countable open cover without a finite subcover. Then for each n we can choose a point x _{n\in} X that is not in \cup ^{n} _{i =1} O _{i} . The sequence (x _{n)} _{n} has an accumulation point x and that x is in some O _{k} . But then O _{k} is a neighborhood of x that does not contain any of the x _{n} with n>k , so x is not an accumulation point of the sequence after all. This contradiction proves (1).

(4)

\Leftrightarrow (1): Conditions (1) and (4) are easily seen to be equivalent by taking complements.

Examples

• The first uncountable ordinal (with the order topology) is an example of a countably compact space that is not compact.

Properties

• Every compact space is countably compact.
• A countably compact space is compact if and only if it is Lindelöf.
• Every countably compact space is limit point compact.
• For T1 spaces, countable compactness and limit point compactness are equivalent.
• Every sequentially compact space is countably compact.^[3] The converse does not hold. For example, the product of continuum-many closed intervals

[ 0 , 1 ] with the product topology is compact and hence countably compact; but it is not sequentially compact.^[4]

• For first-countable spaces, countable compactness and sequential compactness are equivalent.^[5] More generally, the same holds for sequential spaces.^[6]
• For metrizable spaces, countable compactness, sequential compactness, limit point compactness and compactness are all equivalent.
• The example of the set of all real numbers with the standard topology shows that neither local compactness nor σ-compactness nor paracompactness imply countable compactness.
• Closed subspaces of a countably compact space are countably compact.^[7]
• The continuous image of a countably compact space is countably compact.^[8]
• Every countably compact space is pseudocompact.
• In a countably compact space, every locally finite family of nonempty subsets is finite.
• Every countably compact paracompact space is compact.^[9] More generally, every countably compact metacompact space is compact.
• Every countably compact Hausdorff first-countable space is regular.^{[10] [11]}
• Every normal countably compact space is collectionwise normal.
• The product of a compact space and a countably compact space is countably compact.^{[12] [13]}
• The product of two countably compact spaces need not be countably compact.^[14]

See also

• Sequentially compact space
• Compact space
• Limit point compact
• Lindelöf space

References

• Book: Engelking, Ryszard. Ryszard Engelking. General Topology. Heldermann Verlag, Berlin. 1989. 3-88538-006-4.
• Book: James Munkres . James Munkres . 1999 . Topology . 2nd . . 0-13-181629-2.
• Book: Steen . Lynn Arthur . Lynn Arthur Steen . Seebach . J. Arthur Jr. . J. Arthur Seebach, Jr. . . 1978 . . Berlin, New York . Dover reprint of 1978 . 978-0-486-68735-3 . 1995.

Notes and References

1. Steen & Seebach, p. 19
2. Web site: General topology - Does sequential compactness imply countable compactness?.
3. Steen & Seebach, p. 20
4. Steen & Seebach, Example 105, p, 125
5. Willard, problem 17G, p. 125
6. , Theorem 1.20
7. Willard, problem 17F, p. 125
8. Willard, problem 17F, p. 125
9. Web site: Countably compact paracompact space is compact.
10. Steen & Seebach, Figure 7, p. 25
11. Web site: Prove that a countably compact, first countable T₂ space is regular.
12. Willard, problem 17F, p. 125
13. Web site: Is the Product of a Compact Space and a Countably Compact Space Countably Compact?.
14. Engelking, example 3.10.19