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].

`site:mathoverflow.net DMC Urysohn`

and there was only one result. This one. $\endgroup$1more comment