If $X$ and $Y$ are completely metrizable spaces, let us write $X \preceq Y$ whenever $X$ embeds in $Y$.

Earlier today, I asked the question:

>*Is this a well-quasi-order?*

This was short-sighted, as Tom Goodwillie has pointed out in the comments that the closed surfaces give an easy counterexample.

Since I can't accept Tom's comment as an answer, I'd like to modify the question to make it more interesting (while still being very closely related to the original):

> Is there a finite list $F$ of completely metrizable spaces such that a given completely ultrametrizable space embeds in $\mathbb{R}^2$ if and only if it does not contain a topological copy of a member of $F$?

An affirmative answer would be something analogous to Wagner's Theorem, but with a more topological flavor.

[Considering this question was part of what led me to ask my other question: if embeddability were a wqo (which it isn't), then the answer to the present question would be yes.]

**Candidate list:** the topological graphs $K_5$ and $K_{3,3}$, the sphere $S^2$, and the subspace of $\mathbb{R}^3$ obtained by taking the X-Y plane and a sequence converging to the origin along the Z axis.

[Notice that every closed surface contains one of these.]

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**Comment to the original question:** Embeddability is *not* a well-quasi-order for metric spaces generally. An easy way to get a counterexample is to build one by transfinite recursion: you can find infinitely many subsets of $\mathbb R$ that violate either/both of the conditions listed above. The examples you build will be very far from $G_\delta$, so not completely metrizable.

Completely ultrametrizable spaces *are* well-quasi-ordered by embedability. This follows (with a little bit of work) from a version of the Nash-Williams Tree Theorem (see Theorem 11 [here][1]), together with the fact that every completely ultrametrizable space can be represented as a tree.


  [1]: http://web.mat.bham.ac.uk/D.Kuehn/bqofinal.pdf