Countably compact space

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. (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. • 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 [0,1] 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. More generally, the same holds for sequential spaces. • For metrizable spaces, countable compactness, sequential compactness, limit point compactness and compactness are all equivalent. The same holds for second countable Hausdorff spaces. • 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. • 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. More generally, every countably compact metacompact 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. • The product of two countably compact spaces need not be countably compact. ==See also==