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I've been going back over some results from Munkres's Topology, and I'm curious about some things. (I originally posted this on M.SE, but I think it is probably a better fit here.)

I know that Choice principles have some connection to the separation axioms (in ZF, at least)--for example:

Locally compact Hausdorff spaces are Baire if and only if the Axiom of Dependent Multiple Choices holds. [Due to Fossy and Morillon, I believe.]

Complete pseudometric spaces are Baire if and only if the Axiom of Dependent Choices holds.

Still, it seems likely that compactness (or the weaker condition of completeness) and "Baireness" may be playing a substantial part, here.

I know that the Urysohn Metrization Theorem--which states that regular, Hausdorff, second-countable spaces are metrizable (or equivalently, that a space is separable and metrizable if and only if it is regular, Hausdorff, and second countable)--holds in ZF, though the Urysohn Lemma--which states that a space is normal Hausdorff if and only if any two disjoint closed sets can be separated by a continuous function--does not. That's a nice result.

First, the usual proof of the Urysohn Lemma (cf. Munkres) uses Dependent Choice, and it has been suggested to me that Dependent Multiple Choices may do the trick. Does anyone know a proof for this?

Second, I wonder what is known about how much Choice is needed to prove the following metrization theorems, or whether any of them are known to be equivalent in ZF:

Nagata-Smirnov Metrization Theorem: A topological space $X$ is metrizable if and only if it is $T_3$ and has a basis that is countably locally finite [= $\sigma$-locally finite base].

Smirnov Metrization Theorem: A topological space is metrizable if and only if it is paracompact Hausdorff and locally metrizable.

Bing Metrization Theorem: A topological space $X$ is metrizable if and only if it is $T_3$ and has a basis that is countably locally discrete [= $\sigma$-discrete base].

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By the way. I think that the question about Urysohn came up here before and it was pointed out that the answer is currently unknown, but suspected something like Dependent Multiple Choice (which is weaker than DC itself). This is just a vague memory. Searching the site seems like a better idea. –  Asaf Karagila Jul 19 '13 at 13:17
    
I'm not finding anything else, so far, but I'll keep checking for a bit. Thanks, Asaf! –  Cameron Buie Jul 19 '13 at 13:22
    
Here's a start: mathoverflow.net/questions/99376/… (there are links in the comments which you can follow as well). –  Asaf Karagila Jul 19 '13 at 13:29
    
Oh, for Pete's sake! I searched "Urysohn Lemma Dependent Choice" (and variations) and got this question exactly. Apparently that "apostrophe s" is a big deal to the search engine.... –  Cameron Buie Jul 19 '13 at 13:34
    
Well, I used Google with site:mathoverflow.net DMC Urysohn and there was only one result. This one. –  Asaf Karagila Jul 19 '13 at 13:35

2 Answers 2

up vote 3 down vote accepted

They show that the Nagata–Smirnov and Bing metrization theorems are equivalent to Stone's Theorem (paracompactness of metric spaces) though the backward implications are provable in ZF.

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(+1) So, if I understand you correctly, Nagata-Smirnov and Bing are both equivalent (in ZF) to the statement "all metric spaces are paracompact"? What do you mean by "the backward implications"? –  Cameron Buie Jul 19 '13 at 13:15
    
Yes. The backward half of the equivalence, in the order you wrote them. –  François G. Dorais Jul 19 '13 at 13:17
    
Ah! Of course. Many thanks! –  Cameron Buie Jul 19 '13 at 13:19

To prove Urysohn's Lemma using Dependent Multiple Choice (DMC), proceed as in the usual proof, using Dependent Choice (DC), with the following modification. At each stage of the usual inductive construction, you need to find some open sets (usually $2^n$ of them at stage $n$) subject to requirements that look like this: You already have open sets $U$ and $V$ with the closure $\overline U$ of $U$ included in $V$, and you want an open $W$ with $\overline U\subseteq W\subseteq \overline W\subseteq V$. Normality of thespace gives you the existence of such a $W$, but you use DC to choose appropriate $W$'s for all the stages together. Using DMC instead of DC, you get, instead of a particular $W$ for given $U$ and $V$, a finite set of such $W$'s. But then you can replace those finitely many $W$'s with their intersection (or their union) to get a single appropriate $W$. That's the essential idea of the proof; one still needs to be careful about some technical details, but, if I remember correctly, they're rather straightforward.

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Thanks, Andreas! I'll sit down and try to hammer out the details. –  Cameron Buie Jul 19 '13 at 21:16

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