# Completely metrizable space

(Redirected from Completely metrizable)

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]

## Properties

• 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 nonempty metrizable spaces is completely metrizable in the product topology if and only if each factor is completely metrizable.[9] Hence, a product of nonempty metrizable topological spaces is completely metrizable if and only if at most countably many factors have more than one point 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.

## Notes

1. ^ Willard, Definition 24.2
2. ^ Steen and Seebach, I §5: Complete Metric Spaces
3. ^ Kelley, Problem 6.K, p. 207
4. ^ Kelley, Problem 6.L, p. 208
5. ^ Willard, Chapter 24
6. ^ Willard, Exercise 25A
7. ^ Willard, Theorem 24.13
8. ^ Willard, Chapter 24
9. ^ Willard, Chapter 24
10. ^ Because a product of nonempty metrizable topological spaces is metrizable if and only if at most countably many factors have more than one point (Willard, Chapter 22).
11. ^ Willard, Chapter 24