6
$\begingroup$

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.]

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), together with the fact that every completely ultrametrizable space can be represented as a tree.

Improve this question
$\endgroup$
8
  • $\begingroup$ The closed surfaces contradict (1). $\endgroup$
    – Tom Goodwillie
    Commented Jun 8, 2015 at 15:21
  • $\begingroup$ @Tom: You're right! The sum of $n$ tori (for example) does not embed in the sum of $n+1$ tori (or vice versa). I feel like a bit of an idiot for missing such a natural example. Thanks for the fast response. $\endgroup$
    – Will Brian
    Commented Jun 8, 2015 at 15:27
  • $\begingroup$ I guess one can make antichains of size continuum, even amongst the Polish spaces: fix an almost-disjoint family of continuum many subsets $A\subset\mathbb{N}$, let $\Gamma_A$ be the sum of surfaces of genus $k$ for $k\in A$. So $A\neq B$ in the family implies $\Gamma_A$ and $\Gamma_B$ do not embed into each other. $\endgroup$
    – Joel David Hamkins
    Commented Jun 8, 2015 at 16:00
  • $\begingroup$ @Joel: Correct. And using that idea, one can also find infinite decreasing chains. Just use $\Gamma_{A_n}$ where $A_0 \supsetneq A_1 \supsetneq A_2 . . . $. $\endgroup$
    – Will Brian
    Commented Jun 8, 2015 at 16:03
  • $\begingroup$ There is a list of seven forbidden subcomplexes (up to a subdivision) for embedding 2-dimensional complexes into $\mathbb{R}^2$ by Halin and Jung [R. Halin and H. A. Jung. Charakterisierung der Komplexe der Ebene und der 2-Sphäre. Arch. Math., 15:466–469, 1964]. I am not familiar with ultrametrizable spaces, but this list could perhaps provide an answer to your question. It includes, for example, three triangles sharing an edge. If this is ultrametrizable, then your candidate list is perhaps incomplete. $\endgroup$
    – Martin Tancer
    Commented Jun 8, 2015 at 18:28

3 Answers 3

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
2
$\begingroup$

You get a positive answer if you restrict to compact, locally connected metric spaces.

See https://www.semanticscholar.org/paper/On-planarity-of-compact%2C-locally-connected%2C-metric-Richter-Rooney/70090a524f2509408e5170f814ed1b33654bb585 and references therein.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Interesting -- thanks very much for sharing this. $\endgroup$
    – Will Brian
    Commented Apr 6, 2022 at 12:24
1
$\begingroup$

Edit: This is not an answer to the question, but a pointer to the solution of a different problem (due to a misunderstanding on my part)

I'm not an expert on this, so possibly misunderstood something, but to my mind your result on ultrametric spaces appears to badly contradict Theorem 4.2 of a paper by Louveau and Rosendal, which says that the quasiordering of embeddability for Polish ultrametric spaces is a universal analytic quasi-order (for Borel reducibility), which implies that it is very far from being wqo.

The paper in question is available here: http://homepages.math.uic.edu/~rosendal/PapersWebsite/CompleteAnalytic.pdf

Improve this answer
$\endgroup$
3
  • $\begingroup$ Their result seems to be for isometric embeddings, whereas I'm just asking about embeddings. $\endgroup$
    – Will Brian
    Commented Jun 8, 2015 at 15:39
  • $\begingroup$ For the case of Polish ultrametric spaces, you can see a detailed proof of (a strengthening of) my assertion in this paper: logique.jussieu.fr/~carroy/papers/qoforfunctions.pdf. $\endgroup$
    – Will Brian
    Commented Jun 8, 2015 at 15:53
  • $\begingroup$ Oh, OK, I understand. When I read "embedding of metric spaces" I thought the embedding was isometric. $\endgroup$
    – Nash-Williams
    Commented Jun 8, 2015 at 16:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .