# Nagata-Smirnov vs. Urysohn metrization theorems - an example?

I'm looking for an example to demonstrate that fact that the Nagata-Smirnov thm is "more useful" than Urysohn's. That is, I'm looking for a space that you can prove is metrizable using Nagata-Smirnov but not using Urysohn.

I've found (in Steen & Seebach's "Counterexamples") some examples of spaces which have $$\sigma$$-locally finite bases but are not second-countable. Problem is, they all have an explicit metric... In fact, most of them are only described as metric spaces, with no metric-independent description of the topology. So these examples are not very motivating...

I know there are theoretical implications of NG thm that provides motivation, but right now I'm looking specifically for a down-to-earth, tangible example, preferably one that the average undergrad can cope with.

Edit: here is the formulation of the thms:

Urysohn: A $$T_{3}$$, second-countable space (i.e. $$X$$ has a countable base for its topology) is metrizable.

Nagata-Smirnov: A topological space is metrizable iff it's $$T_{3}$$ and has a base for its topology that is a countable union of locally finite families. (where a family $$A_i, i \in I$$ of subsets of $$X$$ is locally finite in $$X$$ iff every $$x \in X$$ has a neighbourhood $$O_x$$ such that $$\{i \in I: O_x \cap A_i \neq \emptyset\}$$ is finite.)

• It would be nice to include the two metriziability conditions in the text of the question. – Berci Jul 17 '19 at 12:06
• In an old issue circa 1970, of Amer. Math. Monthly, there is a proof, using both directions of Nagata-Smirnov, that a non-compact metrizable space has an incomplete metric. I can post the proof from memory if you want. – DanielWainfleet Jul 17 '19 at 14:57
• @Berci thanks for the suggestion, I edited accordingly. – ChanaG Jul 17 '19 at 16:30
• @DanielWainfleet I think I've come across something like that while searching for an example. While interesting by itself, it falls under "theoretical implications" which is not what I'm looking for. Thanks anyway. – ChanaG Jul 17 '19 at 16:34
• According to the Answer posted, Urysohn's theorem is only sufficient. The proof of the result that I cited uses the necessity (of the Bing-Nagata-Smirnov condition) at one step, and the sufficiency is used on another space at a later step..... I am unsure what you are looking for. – DanielWainfleet Jul 18 '19 at 15:07

It has been the basis of a quite extensive literature on metrisability, and the more specialised Urysohn's theorem rather less so. One of the few concrete examples of an application of Urysohn is that the unit ball in $$X^\ast$$ in the weak-star topology is metrisable when $$X$$ is a separable Banach space. Not a concrete example, but useful nonetheless. In that case the metric on that ball is easy enough to construct without Urysohn, but at least we know a metric exists, so even without knowing it we can apply sequential compactness of that ball (based on Alaoglu) instead of "mere" compactness.