Timeline for Spaces that can't be embedded in the plane
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 6, 2022 at 9:29 | answer | added | Agelos | timeline score: 2 | |
| Jun 9, 2015 at 20:09 | vote | accept | Will Brian | ||
| Jun 9, 2015 at 5:25 | answer | added | Martin Tancer | timeline score: 3 | |
| Jun 9, 2015 at 1:13 | comment | added | Will Brian | @MartinTancer: No worries! I started this whole discussion by missing something much more obvious, so I'll be the last to judge you. Your idea about modifying $K_5$ by throwing in some topologist's sine curves seems like a good one. Unless I'm mistaken, you should be able to use that idea to come up with an infinite collection of complete metric spaces, none of which embeds in another, none of which embeds in $\mathbb{R}^2$, and such that if $X$ is any space embedding in two of them, then $X$ embeds in $\mathbb{R}^2$. This would answer the question negatively. Please do write up an answer! | |
| Jun 8, 2015 at 22:03 | comment | added | Martin Tancer | I am sorry I did not think about the complexes from the list properly. I agree that all of them are ruled out by your candidate list. I also did not assume that the list would provide a complete solution immediately, but it could rule out some cases. By the way, what about taking $K_5$ (or subdivided $K_5$ for more examples), replacing each vertex with a segment and each edge with a double-sided topologist's sine curve. This one seems neither embeddable nor ruled out by the list. (I hope I do not overlook a trivial problem this time.) | |
| Jun 8, 2015 at 19:23 | comment | added | Will Brian | @MartinTancer: The "ultra" that you're referring to was a typo, and I'm sorry for the confusion. The spaces you're referring to are not ultrametrizable (e.g., because every ultrametrizable space has a basis of clopen sets). I don't read German, but (based on the pictures) the link you shared looks interesting, although I'm not sure it answers my question (because "embeds" and "is a subcomplex of" are very different). By the way, I'll point out that three of the four spaces on my list embed into three triangles that share an edge (all but the sphere). | |
| Jun 8, 2015 at 19:15 | history | edited | Will Brian | CC BY-SA 3.0 |
fixed a typo
|
| Jun 8, 2015 at 18:28 | comment | added | Martin Tancer | 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. | |
| Jun 8, 2015 at 16:38 | review | Close votes | |||
| Jun 8, 2015 at 19:39 | |||||
| Jun 8, 2015 at 16:10 | history | edited | Will Brian | CC BY-SA 3.0 |
added 4 characters in body
|
| Jun 8, 2015 at 16:03 | comment | added | Will Brian | @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 . . . $. | |
| Jun 8, 2015 at 16:00 | comment | added | Joel David Hamkins | 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. | |
| Jun 8, 2015 at 15:58 | history | edited | Will Brian | CC BY-SA 3.0 |
improved wording
|
| Jun 8, 2015 at 15:51 | history | edited | Will Brian | CC BY-SA 3.0 |
I have removed the order theory tag, which I think is no longer appropriate.
|
| Jun 8, 2015 at 15:34 | answer | added | Nash-Williams | timeline score: 1 | |
| Jun 8, 2015 at 15:27 | comment | added | Will Brian | @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. | |
| Jun 8, 2015 at 15:21 | comment | added | Tom Goodwillie | The closed surfaces contradict (1). | |
| Jun 8, 2015 at 15:11 | history | asked | Will Brian | CC BY-SA 3.0 |