Countably compact space

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:

Steen & Seebach, p. 19{{Cite web|url=https://math.stackexchange.com/a/718043/52912|title=General topology - Does sequential compactness imply countable compactness?}}

:(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.

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(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.

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Examples

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.Steen & Seebach, p. 20 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.Steen & Seebach, Example 105, p, 125
  • For first-countable spaces, countable compactness and sequential compactness are equivalent.Willard, problem 17G, p. 125 More generally, the same holds for sequential spaces.{{Citation |last1=Kremsater |first1=Terry Philip |title=Sequential space methods |type=Thesis |date=1972 |publisher=University of British Columbia |doi=10.14288/1.0080490}}, Theorem 1.20
  • 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.Willard, problem 17F, p. 125
  • The continuous image of a countably compact space is countably compact.Willard, problem 17F, p. 125
  • Every countably compact space is pseudocompact.
  • In a countably compact space, every locally finite family of nonempty subsets is finite.{{sfn|Engelking|1989|loc=Theorem 3.10.3(ii)}}
  • Every countably compact paracompact space is compact.{{sfn|Engelking|1989|loc=Theorem 5.1.20}}{{Cite web|url=https://math.stackexchange.com/q/171182 |title=Countably compact paracompact space is compact}} More generally, every countably compact metacompact space is compact.{{sfn|Engelking|1989|loc=Theorem 5.3.2}}
  • Every countably compact Hausdorff first-countable space is regular.Steen & Seebach, Figure 7, p. 25{{Cite web|url=https://math.stackexchange.com/q/2379365|title=Prove that a countably compact, first countable T2 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.Willard, problem 17F, p. 125{{Cite web|url=https://math.stackexchange.com/q/3486708|title=Is the Product of a Compact Space and a Countably Compact Space Countably Compact?}}
  • The product of two countably compact spaces need not be countably compact.Engelking, example 3.10.19

See also

Notes

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References

  • {{cite book|last=Engelking|first=Ryszard| author-link=Ryszard Engelking|title=General Topology|publisher=Heldermann Verlag, Berlin|year=1989| isbn=3-88538-006-4}}
  • {{cite book

| author = James Munkres

| author-link = James Munkres

| year = 1999

| title = Topology

| edition = 2nd

| publisher = Prentice Hall

| isbn = 0-13-181629-2

}}

  • {{cite book | last1=Steen | first1=Lynn Arthur | author1-link=Lynn Arthur Steen | last2=Seebach | first2=J. Arthur Jr. | author2-link=J. Arthur Seebach, Jr. | title=Counterexamples in Topology | origyear=1978 | publisher=Springer-Verlag | location=Berlin, New York | edition=Dover reprint of 1978 | isbn=978-0-486-68735-3 | year=1995}}
  • {{Citation | last=Willard | first=Stephen | title=General Topology | orig-year=1970 | publisher=Addison-Wesley | edition=Dover reprint of 1970 | year=2004}}

Category:Properties of topological spaces

Category:Compactness (mathematics)