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) 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.

Proof of equivalence

(1) $\Rightarrow$ (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) $\Rightarrow$ (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 $A=\{x_{n}:n\in {\mathbb {N}}\}$ 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) $\Rightarrow$ (1): Suppose (3) holds and $\{O_{n}:n\in {\mathbb {N}}\}$ 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 $\cup _{i=1}^{n}O_{i}$. 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.
• A countably compact space is always limit point compact.
• For T1 spaces, countable compactness and limit point compactness are equivalent.
• 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.
• The continuous image of a countably compact space is countably compact.
• 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.
• Every normal countably compact space is collectionwise normal.

See also

• Sequentially compact space
• Compact space
• Limit point compact

References

• 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.