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.

Template:Collapse top (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. Template:Collapse bottom

Examples

Properties

See also

Notes

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References

  1. Steen & Seebach, p. 19
  2. Template:Cite web
  3. Steen & Seebach, p. 20
  4. Steen & Seebach, Example 105, p, 125
  5. Willard, problem 17G, p. 125
  6. Template:Citation, Theorem 1.20
  7. Willard, problem 17F, p. 125
  8. Willard, problem 17F, p. 125
  9. 9.0 9.1 Template:Cite web
  10. Steen & Seebach, Figure 7, p. 25
  11. Template:Cite web
  12. Willard, problem 17F, p. 125
  13. Template:Cite web
  14. Engelking, example 3.10.19