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

Properties

See also

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. Kremsater, Terry Philip (1972), Sequential space methods (Thesis), University of British Columbia, doi:10.14288/1.0080490, Theorem 1.20
  8. Willard, problem 17F, p. 125
  9. Willard, problem 17F, p. 125
  10. Engelking 1989, Theorem 3.10.3(ii).
  11. 11.0 11.1 "Countably compact paracompact space is compact".
  12. Engelking 1989, Theorem 5.1.20.
  13. Engelking 1989, Theorem 5.3.2.
  14. Steen & Seebach, Figure 7, p. 25
  15. "Prove that a countably compact, first countable T2 space is regular".
  16. Willard, problem 17F, p. 125
  17. "Is the Product of a Compact Space and a Countably Compact Space Countably Compact?".
  18. Engelking, example 3.10.19

References