Completely metrizable space
In mathematics, a completely metrizable space[1] (topologically complete space[2] or metrically topologically complete space[3]) is a topological space (X, T) for which there exists at least one metric d on X such that (X, d) is a complete metric space and d induces the topology T. The term topologically complete is sometimes also used for other classes of topological spaces, like completely uniformizable spaces[4] or Čech-complete spaces.
Difference between complete metric space and completely metrizable space
The difference between completely metrizable space and complete metric space is in the words there exists at least one metric in the definition of completely metrizable space, which is not the same as there is given a metric (the latter would yield the definition of complete metric space). Once we make the choice of the metric on a completely metrizable space (out of all the complete metrics compatible with the topology), we get a complete metric space. In other words, the category of completely metrizable spaces is a subcategory of that of topological spaces, while the category of complete metric spaces is not (instead, it is a subcategory of the category of metric spaces).
Examples
- The space (0,1) ⊂ R, the open unit interval, is not a complete metric space with its usual metric inherited from R, but it is completely metrizable since it is homeomorphic to R.[5]
- The set Q of rational numbers is metrizable but not completely metrizable.[6]
Properties
- A topological space X is completely metrizable if and only if X is metrizable and a Gδ in its Stone–Čech compactification βX.[7]
- A subspace of a completely metrizable space X is completely metrizable if and only if it is Gδ in X.[8]
- A countable product of metrizable spaces is completely metrizable in the product topology if and only if each factor is completely metrizable.[9] Hence, a product of metrizable topological spaces, each of which contains at least two elements, is completely metrizable if and only if the number of factor spaces is countable and each factor is completely metrizable.[10]
- For every metrizable space there exists a completely metrizable space containing it as a dense subspace, since every metric space has a completion.[11] In general, there are many such complete metrizable spaces, since completions of a topological space with respect to different metrics compatible with its topology can give topologically different completions.
See also
Notes
- ^ Willard, Definition 24.2
- ^ Steen and Seebach, I §5: Complete Metric Spaces
- ^ Kelley, Problem 6.K, p. 207
- ^ Kelley, Problem 6.L, p. 208
- ^ Willard, Chapter 24
- ^ Willard, Exercise 25A
- ^ Willard, Theorem 24.13
- ^ Willard, Chapter 24
- ^ Willard, Chapter 24
- ^ Because a product of metrizable topological spaces, each of which contains at least two elements, is metrizable if and only if the number of factors is countable (Willard, Chapter 22).
- ^ Willard, Chapter 24
References
- Kelley, John L. (1975). General Topology. Springer. ISBN 0-387-90125-6.
- Steen, Lynn Arthur; Seebach, J. Arthur Jr. (1970). Counterexamples in Topology. Holt, Rinehart and Winston, Inc. ISBN 978-0-03-079485-8.
- Willard, Stephen (1970). General Topology. Addison-Wesley Publishing Company. ISBN 978-0-201-08707-9.
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