It is easy to see that the disjoint union $\bigsqcup_i X_i$ of a collection of metric spaces is metrizable, simply by rescaling or chopping off the individual metrics to have diameter at most one, and then making points in the various components distance two apart. This produces a metric on the disjoint union giving the disjoint union topology.

A subtle point arises, however, if we are given not a collection of metric spaces, but instead merely a collection of metrizable topological spaces $X_i$, where a particular metric has not yet been chosen. In this case, in order to carry out the construction above, we would need first to choose a particular metric realizing the metrizability of each space, and this would seem to require the axiom of choice.

Question. Without assuming the axiom of choice, when is the disjoint union of a set of metrizable spaces metrizable?

One can see that some choice is definitely required by the following argument. Suppose that we are given a collection of countable sets $S_i$. Let's regard these sets as discrete topological spaces and let $S_i^+$ be the corresponding one-point compactification, which is metrizable, since each $S_i$ is countable. If we had a metric on the disjoint union $\bigcup S_i^{+}$, however, we could consider the largest ball at the new point in each $S_i^+$ to omit some elements of $S_i$, and this would provide us with a canonical choice of non-empty finite sets $T_i\subset S_i$. But the existence of such a kind of choice surely requires some form of choice. Thus, in general, we cannot prove without any AC that the disjoin union of metrizable spaces is metrizable.

So what I seek is a nice sufficient condition for making this conclusion. Here is one answer: given a family of metrizable spaces $X_i$ suppose there exists a finite family of distinct metrizable spaces $M_j$ such that

(1) each $M_j$ has a metric that makes every topological automorphism of $M_j$ an isometry; and

(2) but for finitely many $i$, each $X_i$ is homeomorphic to some $M_j$.

(Given an $X_i$ and a homeomorphic $M_j$, avoid choice by working with all the homeomorphisms from $X_i$ to $M_j$.)

This condition seems to me far too restrictive. For example one could have infinite sets of such $M_j$s as long as they came pre-equipped with a selection of appropriate metrics.

So what I seek are general conditions on such $M_j$'s (or even for say just rigid metrizable spaces) that permit the selection of a canonical metric based solely on knowledge of the topology.

If the general question is still too vague, I don't even know the answer to this: does a compact, rigid space homeomorphic to some subset of the unit disk in ${\Bbb R}^2$ admit a distinguished metric (perhaps optimal for some objective function)?

Contrarywise, I'd be interested in models where a countable disjoint union of rigid metrizable spaces might have no compatible metric.

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    $\begingroup$ Doesn't the subset of $\mathbb{R}^2$ inherit the Euclidean metric from $\mathbb{R}^2$? $\endgroup$ – Andrej Bauer Sep 21 '12 at 18:11
  • $\begingroup$ @Andrej I think we are supposed to forget how the space is embedded in $\mathbb{R}^2$ $\endgroup$ – Trevor Wilson Sep 21 '12 at 20:17
  • $\begingroup$ Trevor Wilson has it right! $\endgroup$ – David Feldman Sep 21 '12 at 20:36
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    $\begingroup$ @David: I have deleted my answer, but I don't understand why you said that it was "simply incorrect." As far as I can tell, I did not make any false statements (after the edit). $\endgroup$ – Qiaochu Yuan Sep 22 '12 at 6:13
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    $\begingroup$ Thank you Qiaochu Yuan! And I do take your point...you made no false statements. My concern was that an "answer" here purports to satisfy the demands of the question, and that you did not succeed in doing (yet?). So I feared your leaving it up would turn into a distraction. $\endgroup$ – David Feldman Sep 22 '12 at 17:03

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