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

Equivalent definitions:
A topological space X is 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 subset A of X has an ω-accumulation point in X (a point x such that every neighborhood of x contains infinitely many points of A).
(3) Every sequence in X has an accumulation point in X.
(4) Every countable family of closed subsets of X with empty intersection has a finite subfamily with empty intersection.

(1) ⇒ (2): Suppose (1) holds and let A be an infinite subset of X with no ω-accumulation point. By passing to a subset we may assume A is countable. For each x in X choose an open neighborhood U_x with U_x ∩ A finite (possibly empty), which exists because x is not an ω-accumulation point. For each finite subset F of A let V_F = {x ∈ X : U_x ∩ A = F}. Each V_F is open, the family {V_F} (indexed by the finite subsets F of A) covers X, and there are only countably many such F, so {V_F} is a countable open cover of X. But each V_F meets A in at most the finite set F, so no finite subfamily of the V_F can cover A (and hence cannot cover X), contradicting (1). Therefore (2) holds.

(2) ⇒ (3): Suppose (2) holds and let (x_n) be a sequence in X. If some value x occurs infinitely many times in the sequence, then x is an accumulation point of the sequence. Otherwise the set of distinct values of the sequence is infinite, so by (2) it has an ω-accumulation point, which is an accumulation point of the sequence. That x is then an accumulation point of the sequence is easily checked.

(3) ⇒ (1): Suppose (3) holds and {U_n} is a countable open cover without a finite subcover. Then for each n choose a point x_n not in U_1 ∪ … ∪ U_n. The sequence (x_n) has an accumulation point x, so x belongs to some U_k. But U_k is a neighborhood of x that contains no x_n with n ≥ k, so x is not an accumulation point of the sequence after all. This contradiction proves (1).

(4) ⇔ (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; the converse does not hold. For example, the product of continuum-many closed intervals with the product topology is compact and hence countably compact, but it is not sequentially compact. For first-countable spaces, countable compactness and sequential 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, σ-compactness, nor paracompactness implies countable compactness. Closed subspaces of a countably compact space are countably compact, and 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 countably compact Hausdorff first-countable space is regular. Every normal countably compact space is collectionwise normal. The product of a compact space and a countably compact space is countably compact, but the product of two countably compact spaces need not be countably compact.

See also: sequentially compact space; compact space; limit point compact; Lindelöf space; compactness (mathematics).