Talk:Completely metrizable space

"Complete topological space"

Is the term "complete topological space" used in some notable references? Otherwise I think it should be deleted from the article. Anyway, I think "complete topological space" should not used on Wikipedia, as it is misleading (explained below).

I think the term "complete topological space" is outright confusing: there exist complete topological vector spaces which are not metrizable (consider e. g. the space $D(\mathbb {R} )$ of all smooth functions on $\mathbb {R}$ with compact supports, used as a test function space in distribution theory --- this is an LF-space which is complete but not metrizable (as explained in Distribution space#Test function space)), so they are not "complete topological spaces" in the sense of this article (although they are complete and they are topological spaces!). Actually I like the notion "topologically complete" only a little more, because it would be as natural to use it for complete uniformizability (like Kelley does), but it is used by several notable authors.

So I would rather delete the phrase "complete topological space" from this article (and any redirects and links to it), or if some notable authors have used "complete topological space" in this sense, then it should be mentioned in this article but it should be pointed out that this usage is misleading. --Jaan Vajakas (talk) 01:37, 5 March 2013 (UTC)

Have you looked in the history to figure out who added that term? Certainly it's not one I would use, but usage varies. What is a complete topological vector space? Offhand, I would expect that phrase to mean that the vector space is normed, and the metric derived from the norm is Cauchy complete. If that's not what it means, then what? --Trovatore (talk) 02:23, 5 March 2013 (UTC)

A complete topological vector space is one whose uniformity (induced by its topology and addition operation) is complete, i. e. where every Cauchy net converges (in the uniformity of topological vector space, a net $(x_{a})$ is Cauchy iff for any zero-neighborhood $O$ there exists $\alpha$ s. t. $\forall \beta ,\gamma \geq \alpha \quad x_{\beta }-x_{\gamma }\in O$). (Complete uniformity is a generalization of complete metric: a metric is complete iff the uniformity induced by it is complete.) Jaan Vajakas (talk) 11:05, 5 March 2013 (UTC)

I posted a message on User_talk:Sullivan.t.j but haven't got any answer so far. So now I removed the synonym complete topological space from this article and made Complete topological space into a disambiguation page instead of redirect to Completely metrizable space. Jaan Vajakas (talk) 16:12, 9 April 2013 (UTC)

Willard (Exercise 24F): A topological space X is topologically complete iff X is a G_δ in its Stone–Čech compactification βX. It need not be metrizable. YohanN7 (talk) 11:47, 17 December 2013 (UTC)

Yes, that is yet another meaning of topologically complete (besides completely metrizable and completely uniformizable). That property (but often with the assumption that the space is Hausdorff) is also called Čech-completeness (e. g. [1] or Engelking's General Topology (1989)). Jaan Vajakas (talk) 09:54, 19 December 2013 (UTC)

(Willard seems not to assume Hausdorffness but only complete regularity in his definition of topologically complete.) Jaan Vajakas (talk) 10:33, 19 December 2013 (UTC)

Completion of completely metrizable space?

I am not happy with the sentences "Every metrizable space has a completion that is a completely metrizable space. The completion of Q is R." in the article. I guess metrizable spaces and metric spaces have been confused here. How would the completion of a metrizable topological space be defined? There are many choices for a complete metric space that is a completion of some metric on the given metrizable topological space. Maybe there indeed exists a natural choice of choosing a single one out of them, but I couldn't find such in Willard's book. E. g. the completion of (0, 1) with respect to one metric compatible with its topology is [0, 1], with respect to another one (0, 1) itself is complete, and these completions [0, 1] and (0, 1) are not topologically the same (I should come up with a better counterexample, since one could of course define the completion of a metrizable space to be the space itself in the special case when it is already completely metrizable). Jaan Vajakas (talk) 10:24, 19 December 2013 (UTC)

I think you are entirely correct. The sentences aren't incorrect as it stands, but very misleading. How about

"Every metrizable space has a completion, depending on the choice of metric, that is a complete metric space. The completion of Q with its usual metric is R."

Also, somewhere in the article the difference between metrizable and metric should be mentioned. But I don't think we should abandon metrizable in favor of metric simply because the title of the page is about metrizable spaces. YohanN7 (talk) 11:03, 19 December 2013 (UTC)

Now I reworded and restructured the article to my liking. I also explained the difference between metric spaces and metrizable spaces. I haven't heard of the term “completion of metrizable space” and Willard does not seem to use it either, so I avoided it and only used the well-known concept “completion of metric space”. Jaan Vajakas (talk) 19:05, 19 December 2013 (UTC)

Now that I look at it, my explanation of the difference between (complete) metric space and (completely) metrizable space is probably not at the best place. Perhaps it should be in some other article (maybe Metrizable space). Or on a talk page. This is standard stuff that occurs in many places in pure mathematics. But it won't hurt either and I don't have time to make it nicer. Hope it helps. Jaan Vajakas (talk) 19:20, 19 December 2013 (UTC)