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: [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
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(1) [math]\displaystyle{ \Rightarrow }[/math] (2): Suppose (1) holds and A is an infinite subset of X without [math]\displaystyle{ \omega }[/math]-accumulation point. By taking a subset of A if necessary, we can assume that A is countable. Every [math]\displaystyle{ x\in X }[/math] has an open neighbourhood [math]\displaystyle{ O_x }[/math] such that [math]\displaystyle{ O_x\cap A }[/math] is finite (possibly empty), since x is not an ω-accumulation point. For every finite subset F of A define [math]\displaystyle{ O_F = \cup\{O_x: O_x\cap A=F\} }[/math]. Every [math]\displaystyle{ O_x }[/math] is a subset of one of the [math]\displaystyle{ O_F }[/math], so the [math]\displaystyle{ O_F }[/math] cover X. Since there are countably many of them, the [math]\displaystyle{ O_F }[/math] form a countable open cover of X. But every [math]\displaystyle{ O_F }[/math] intersect A in a finite subset (namely F), so finitely many of them cannot cover A, let alone X. This contradiction proves (2). (2) [math]\displaystyle{ \Rightarrow }[/math] (3): Suppose (2) holds, and let [math]\displaystyle{ (x_n)_n }[/math] 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 [math]\displaystyle{ A=\{x_n: n\in\mathbb N\} }[/math] 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) [math]\displaystyle{ \Rightarrow }[/math] (1): Suppose (3) holds and [math]\displaystyle{ \{O_n: n\in\mathbb N\} }[/math] is a countable open cover without a finite subcover. Then for each [math]\displaystyle{ n }[/math] we can choose a point [math]\displaystyle{ x_n\in X }[/math] that is not in [math]\displaystyle{ \cup_{i=1}^n O_i }[/math]. The sequence [math]\displaystyle{ (x_n)_n }[/math] has an accumulation point x and that x is in some [math]\displaystyle{ O_k }[/math]. But then [math]\displaystyle{ O_k }[/math] is a neighborhood of x that does not contain any of the [math]\displaystyle{ x_n }[/math] with [math]\displaystyle{ n\gt k }[/math], so x is not an accumulation point of the sequence after all. This contradiction proves (1). (4) [math]\displaystyle{ \Leftrightarrow }[/math] (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.[3]
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.[4] The converse does not hold. For example, the product of continuum-many closed intervals [math]\displaystyle{ [0,1] }[/math] with the product topology is compact and hence countably compact; but it is not sequentially compact.[5]
- For first-countable spaces, countable compactness and sequential compactness are equivalent.[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.[9]
- Every countably compact paracompact space is compact.[9]
- 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
- ↑ Steen & Seebach, p. 19
- ↑ "General topology - Does sequential compactness imply countable compactness?". https://math.stackexchange.com/a/718043/52912.
- ↑ Steen & Seebach 1995, example 42, p. 68.
- ↑ Steen & Seebach, p. 20
- ↑ Steen & Seebach, Example 105, p, 125
- ↑ Willard, problem 17G, p. 125
- ↑ Willard, problem 17F, p. 125
- ↑ Willard, problem 17F, p. 125
- ↑ 9.0 9.1 "General topology - Countably compact paracompact space is compact". https://math.stackexchange.com/q/171182.
- ↑ Steen & Seebach, Figure 7, p. 25
- ↑ "General topology - Prove that a countably compact, first countable T2 space is regular". https://math.stackexchange.com/q/2379365.
- ↑ Willard, problem 17F, p. 125
- ↑ "General topology - is the Product of a Compact Space and a Countably Compact Space Countably Compact?". https://math.stackexchange.com/q/3486708.
- ↑ Engelking, example 3.10.19, p. 205
References
- Engelking, Ryszard (1989). General Topology. Heldermann Verlag, Berlin. ISBN 3-88538-006-4.
- James Munkres (1999). Topology (2nd ed.). Prentice Hall. ISBN 0-13-181629-2.
- Steen, Lynn Arthur; Seebach, J. Arthur Jr. (1995) [1978]. Counterexamples in Topology (Dover reprint of 1978 ed.). Berlin, New York: Springer-Verlag. ISBN 978-0-486-68735-3.
- Willard, Stephen (2004), General Topology (Dover reprint of 1970 ed.), Addison-Wesley
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