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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.)

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    $\begingroup$ It would be nice to include the two metriziability conditions in the text of the question. $\endgroup$ – Berci Jul 17 '19 at 12:06
  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet Jul 17 '19 at 14:57
  • $\begingroup$ @Berci thanks for the suggestion, I edited accordingly. $\endgroup$ – ChanaG Jul 17 '19 at 16:30
  • $\begingroup$ @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. $\endgroup$ – ChanaG Jul 17 '19 at 16:34
  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet Jul 18 '19 at 15:07
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Bing-Nagata-Smirnov's theorem (BNS) is a necessary and sufficient condition for being a metrisable space, while Urysohn's theorem is merely sufficient. Many other general criteria for metrisability have been proved on the basis of BNS, look at the Encyclopedia of General Topology (chapters on metrisation theorems). A classic application is the fact that a paracompact locally metrisable space is metrisable (which is often used in manifolds). So BNS is a convenient general tool.

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.

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  • $\begingroup$ Thanks, but I'm looking for the opposite case - not a concrete example for using Urysohn's thm, but a concrete example where Urysohn's thm fails and Nagata-Smirnov's thm is applicable. $\endgroup$ – ChanaG Jul 17 '19 at 16:37
  • $\begingroup$ @ChanaG BNS is never applied to concrete spaces. It’s a tool to prove other theorems. $\endgroup$ – Henno Brandsma Jul 18 '19 at 5:13
  • $\begingroup$ Why is that? Is it because proving metrizability of concrete spaces is uninteresting outside the framework of a general topology course (which is, however, my framework)? Or is it because there are no concrete spaces such as I look for, for which BNS is useful? $\endgroup$ – ChanaG Jul 18 '19 at 6:02
  • $\begingroup$ @ChanaG When constructing a concrete space, it's often much easier to just define a metric on it too. It's rare to construct a metrisable space from a topology definition alone. $\endgroup$ – Henno Brandsma Jul 18 '19 at 7:22
  • $\begingroup$ I understand. Well, I'm looking for such a rare example... and beginning to think it doesn't exist. $\endgroup$ – ChanaG Jul 18 '19 at 7:51

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