Countably compact space

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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.


Proof of equivalence

(1) (2): Suppose (1) holds and A is an infinite subset of X without ω-accumulation point. By taking a subset of A if necessary, we can assume that A is countable. Every xX has an open neighbourhood Ox such that OxA is finite (possibly empty), since x is not an ω-accumulation point. For every finite subset F of A define OF={Ox:OxA=F}. Every Ox is a subset of one of the OF, so the OF cover X. Since there are countably many of them, the OF form a countable open cover of X. But every OF 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 (xn)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={xn: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 {On:n} is a countable open cover without a finite subcover. Then for each n we can choose a point xnX that is not in i=1nOi. The sequence (xn)n has an accumulation point x and that x is in some Ok. But then Ok is a neighborhood of x that does not contain any of the xn 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.Script error: No such module "Footnotes".Script error: No such module "Check for unknown parameters".

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.Script error: No such module "Footnotes".Script error: No such module "Check for unknown parameters".[9]
  • Every countably compact paracompact space is compact.Script error: No such module "Footnotes".Script error: No such module "Check for unknown parameters".[9] More generally, every countably compact metacompact space is compact.Script error: No such module "Footnotes".Script error: No such module "Check for unknown parameters".
  • 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

Notes

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  1. Steen & Seebach, p. 19
  2. Script error: No such module "citation/CS1".
  3. Steen & Seebach, p. 20
  4. Steen & Seebach, Example 105, p, 125
  5. Willard, problem 17G, p. 125
  6. Script error: No such module "citation/CS1"., Theorem 1.20
  7. Willard, problem 17F, p. 125
  8. Willard, problem 17F, p. 125
  9. a b Script error: No such module "citation/CS1".
  10. Steen & Seebach, Figure 7, p. 25
  11. Script error: No such module "citation/CS1".
  12. Willard, problem 17F, p. 125
  13. Script error: No such module "citation/CS1".
  14. Engelking, example 3.10.19

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References

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