Countably compact space

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 setA 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.

Proof of equivalence

(1) ${\displaystyle \Rightarrow }$ (2): Suppose (1) holds and A is an infinite subset of X without ${\displaystyle \omega }$-accumulation point. By taking a subset of A if necessary, we can assume that A is countable. Every ${\displaystyle x\in X}$ has an open neighbourhood ${\displaystyle O_{x}}$ such that ${\displaystyle O_{x}\cap A}$ is finite (possibly empty), since x is not an ω-accumulation point. For every finite subset F of A define ${\displaystyle O_{F}=\cup \{O_{x}:O_{x}\cap A=F\}}$. Every ${\displaystyle O_{x}}$ is a subset of one of the ${\displaystyle O_{F}}$, so the ${\displaystyle O_{F}}$ cover X. Since there are countably many of them, the ${\displaystyle O_{F}}$ form a countable open cover of X. But every ${\displaystyle O_{F}}$ intersect A in a finite subset (namely F), so finitely many of them cannot cover A, let alone X. This contradiction proves (2). (2) ${\displaystyle \Rightarrow }$ (3): Suppose (2) holds, and let ${\displaystyle (x_{n})_{n}}$ 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 ${\displaystyle A=\{x_{n}:n\in \mathbb {N} \}}$ 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) ${\displaystyle \Rightarrow }$ (1): Suppose (3) holds and ${\displaystyle \{O_{n}:n\in \mathbb {N} \}}$ is a countable open cover without a finite subcover. Then for each ${\displaystyle n}$ we can choose a point ${\displaystyle x_{n}\in X}$ that is not in ${\displaystyle \cup _{i=1}^{n}O_{i}}$. The sequence ${\displaystyle (x_{n})_{n}}$ has an accumulation point x and that x is in some ${\displaystyle O_{k}}$. But then ${\displaystyle O_{k}}$ is a neighborhood of x that does not contain any of the ${\displaystyle x_{n}}$ with ${\displaystyle n>k}$, so x is not an accumulation point of the sequence after all. This contradiction proves (1). (4) ${\displaystyle \Leftrightarrow }$ (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. ^[3]

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. ^[4] The converse does not hold. For example, the product of continuum-many closed intervals ${\displaystyle [0,1]}$ with the product topology is compact and hence countably compact; but it is not sequentially compact. ^[5]
• For first-countable spaces, countable compactness and sequential compactness are equivalent. ^[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. ^[9]
• Every countably compact paracompact space is compact. ^[9]
• 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

Notes

1. Steen & Seebach, p. 19
2. "General topology - Does sequential compactness imply countable compactness?".
3. Steen & Seebach 1995, example 42, p. 68.
4. Steen & Seebach, p. 20
5. Steen & Seebach, Example 105, p, 125
6. Willard, problem 17G, p. 125
7. Willard, problem 17F, p. 125
8. Willard, problem 17F, p. 125
9. "General topology - Countably compact paracompact space is compact".
10. Steen & Seebach, Figure 7, p. 25
11. "General topology - Prove that a countably compact, first countable T₂ space is regular".
12. Willard, problem 17F, p. 125
13. "General topology - is the Product of a Compact Space and a Countably Compact Space Countably Compact?".
14. Engelking, example 3.10.19, p. 205

References

• Engelking, Ryszard (1989). General Topology. Heldermann Verlag, Berlin. ISBN   3-88538-006-4.
• James Munkres (1999). Topology (2nd ed.). Prentice Hall. ISBN   0-13-181629-2.
• Steen, Lynn Arthur; Seebach, J. Arthur Jr. (1995) [1978]. Counterexamples in Topology (Dover reprint of 1978 ed.). Berlin, New York: Springer-Verlag. ISBN   978-0-486-68735-3.
• Willard, Stephen (2004) [1970], General Topology (Dover reprint of 1970 ed.), Addison-Wesley