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


Earlier today, I asked the question:



> 
> *Is this a well-quasi-order on the completely metrizable spaces?*
> 
> 
> 


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, for any completely metrizable space $X$, $X \preceq \mathbb{R}^2$ if and only if $Y \not\preceq X$ for every $Y \in F$?
> 
> 
> 


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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*The following was a comment to the original question. It is not relevant to the modified question, but I am keeping it to explain the post of Nash-Williams below:*


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](http://web.mat.bham.ac.uk/D.Kuehn/bqofinal.pdf)), together with the fact that every completely ultrametrizable space can be represented as a tree.



Spaces that can't be embedded in the plane
It seems that the answer to the question is negative. However, it should be noted that I did not work out all details for the following counterexample.


Consider a graph $G$ that is not planar. Replace every vertex of $G$ by a segment and every edge of $G$ by a double-sided topologist's sine curve, obtaining a space $X\_G$. Then


1. $X\_G$ does not seem embeddable into $\mathbb R^2$.
2. If we only consider $G$'s which are subdivisions of $K\_5$, then they do not embed one into another. And it also seems that they cannot be forbidden by a finite list of non-embeddable spaces.


Both claims would require some bit of work to be proved properly.


