Sequentially compact space

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Short description: Topological space where every sequence has a convergent subsequence

In mathematics, a topological space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in X.

Every metric space is naturally a topological space, and for metric spaces, the notions of compactness and sequential compactness are equivalent (countable choice suffices; see below). However, there exist sequentially compact topological spaces that are not compact, and compact topological spaces that are not sequentially compact.

Examples and properties

The space of all real numbers with the standard topology is not sequentially compact; the sequence (sn) given by sn=n for all natural numbers n is a sequence that has no convergent subsequence.

On a first countable space, a sequence xn has a convergent subsequence if and only if

n{xmmn}

is nomempty. Indeed, a limit of a convergent subsequence is necessarily in the above intersection (this direction holds for any topological space). Conversely, if x is in the above intersection, then let xU2U1 be a countable neighborhood base at x. Then, inductively, choose integers ni>0 such that ni is a least integer with the property (1) ni>ni1 and (2) xniUi, which is possible since is a well-ordered set. Then xnjx.

A point in the above intersection is called a cluster point. Thus, for first countable spaces, the definition of a sequentially compact space is the same as saying that each sequence in the space has a cluster point.

If a space is a metric space, then it is sequentially compact if and only if it is compact (cf. Heine–Borel theorem § Generalization).[1] Here is how to see this, using only the countable Choice. We have to show "sequentially compact" implies "compact". First, we note X is totally bounded, meaning for each ϵ>0, there is a finite cover of X consisting of open balls of radius ϵ. Indeed, if it fails for some ϵ, by countable Choice, choose a sequence xn such that

xn∉B(x1,ϵ)B(xn1,ϵ).

This sequence xn has no convergent subsequence, a contradiction. It follows that X has a countable base. Hence, it is enough to show X is countably compact; i.e., each descending sequence E1E2 of nonempty closed subsets has nonempty intersection. But this is clear since

n{xmmn}nEn

for some sequence xn with xnEn.

The first uncountable ordinal with the order topology is an example of a sequentially compact topological space that is not compact. The topological product of 20=𝔠 copies of the closed unit interval is an example of a compact space that is not sequentially compact.[2]

A topological space X is said to be limit point compact if every infinite subset of X has a limit point in X, and countably compact if every countable open cover has a finite subcover. In a metric space, the notions of sequential compactness, limit point compactness, countable compactness and compactness are all equivalent (if one assumes the axiom of choice).

In a sequential (Hausdorff) space sequential compactness is equivalent to countable compactness.[3]

There is also a notion of a one-point sequential compactification—the idea is that the non convergent sequences should all converge to the extra point.[4]

See also

Notes

  1. Willard, 17G, p. 125.
  2. Steen and Seebach, Example 105, pp. 125—126.
  3. Engelking, General Topology, Theorem 3.10.31
    K.P. Hart, Jun-iti Nagata, J.E. Vaughan (editors), Encyclopedia of General Topology, Chapter d3 (by P. Simon)
  4. Brown, Ronald, "Sequentially proper maps and a sequential compactification", J. London Math Soc. (2) 7 (1973) 515-522.

References

  • Munkres, James (1999). Topology (2nd ed.). Prentice Hall. ISBN 0-13-181629-2. 
  • Steen, Lynn A. and Seebach, J. Arthur Jr.; Counterexamples in Topology, Holt, Rinehart and Winston (1970). ISBN 0-03-079485-4.
  • Willard, Stephen (2004). General Topology. Dover Publications. ISBN 0-486-43479-6.