Английская Википедия: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.
Шаблон:Collapse top (1) <math>\Rightarrow</math> (2): Suppose (1) holds and A is an infinite subset of X without <math>\omega</math>-accumulation point. By taking a subset of A if necessary, we can assume that A is countable. Every <math>x\in X</math> has an open neighbourhood <math>O_x</math> such that <math>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>O_F = \cup\{O_x: O_x\cap A=F\}</math>. Every <math>O_x</math> is a subset of one of the <math>O_F</math>, so the <math>O_F</math> cover X. Since there are countably many of them, the <math>O_F</math> form a countable open cover of X. But every <math>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>\Rightarrow</math> (3): Suppose (2) holds, and let <math>(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>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>\Rightarrow</math> (1): Suppose (3) holds and <math>\{O_n: n\in\mathbb N\}</math> is a countable open cover without a finite subcover. Then for each <math>n</math> we can choose a point <math>x_n\in X</math> that is not in <math>\cup_{i=1}^n O_i</math>. The sequence <math>(x_n)_n</math> has an accumulation point x and that x is in some <math>O_k</math>. But then <math>O_k</math> is a neighborhood of x that does not contain any of the <math>x_n</math> with <math>n>k</math>, so x is not an accumulation point of the sequence after all. This contradiction proves (1).
(4) <math>\Leftrightarrow</math> (1): Conditions (1) and (4) are easily seen to be equivalent by taking complements. Шаблон:Collapse bottom
Examples
- The first uncountable ordinal (with the order topology) is an example of a countably compact space that is not compact.Шаблон:Sfn
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 <math>[0,1]</math> 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]
- 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.[6]
- The continuous image of a countably compact space is countably compact.[7]
- Every countably compact space is pseudocompact.
- In a countably compact space, every locally finite family of nonempty subsets is finite.[8]
- Every countably compact paracompact space is compact.[8]
- Every countably compact Hausdorff first-countable space is regular.[9][10]
- Every normal countably compact space is collectionwise normal.
- The product of a compact space and a countably compact space is countably compact.[11][12]
- The product of two countably compact spaces need not be countably compact.[13]
See also
Notes
References
- ↑ Steen & Seebach, p. 19
- ↑ Шаблон:Cite web
- ↑ 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
- ↑ 8,0 8,1 Шаблон:Cite web
- ↑ Steen & Seebach, Figure 7, p. 25
- ↑ Шаблон:Cite web
- ↑ Willard, problem 17F, p. 125
- ↑ Шаблон:Cite web
- ↑ Engelking, example 3.10.19, p. 205
