Timeline for Is my open set a ball?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2022 at 13:56 | comment | added | Timothy Chow | @SamHopkins Oh, I see..."contained in $\cal C$" is supposed to mean that it's an element of $\cal C$ and not just that it's a subset of the union of the elements in $\cal C$. | |
| Apr 16, 2022 at 13:26 | comment | added | Timothy Chow | @SamHopkins Ah yes, that makes sense! In my defense, the definition of $\cal C$ as written in the question does not say this, or use the word "complex." | |
| Apr 16, 2022 at 13:10 | comment | added | Sam Hopkins | @TimothyChow: maybe what's tripping you up is that the polyhedral complex $\mathcal{C}$ consists of faces (polytopes) of all dimensions. So the $\mathrm{Int}(Q)$ are relatively open sets (in $Q$), but their union forms a connected, open set. | |
| Apr 16, 2022 at 12:33 | comment | added | Timothy Chow | There's something about the definitions that I don't understand. Isn't $A$ disconnected, with one connected component for each $Q$? | |
| Apr 16, 2022 at 12:01 | comment | added | Sam Hopkins | You are of course correct. I think the 2nd condition should at least be true for Q a vertex, and there may be ways to amend it in higher dimensions… | |
| Apr 16, 2022 at 4:18 | comment | added | Nicholas Proudfoot | I believe that the examples that I really care about do have the property that you describe, but I don't think this property follows from the hypotheses that I wrote in the question. | |
| Apr 16, 2022 at 4:18 | comment | added | Nicholas Proudfoot | @SamHopkins I agree with your first statement; this is how I know that $A$ is open. I don't think that the second statement is true, though. For example, suppose that $P$ is a triangle, and $\mathcal{C}$ is a decomposition of $P$ into three triangles meeting at a single vertex in the middle (so the 1-skeleton is $K_4$). Then it is possible to choose $U$ in such a say that $A$ consists of all three (open) triangles along with two of the three interior edges, but not the third. | |
| Apr 16, 2022 at 2:47 | comment | added | Sam Hopkins | Let $\mathcal{D}$ be the set of $Q$ you are taking union over to define $A$ (i.e., the $Q\in \mathcal{C}$ with $Q\cap U\neq \varnothing$). It seems that two important properties of $\mathcal{D}$ are: if $Q \in \mathcal{D}$ then $Q' \in \mathcal{D}$ for all $Q' \supset Q$; and if $Q$ is a face not on the boundary of $P$ with $Q' \in \mathcal{D}$ for all $Q' \supset Q$, then $Q \in \mathcal{D}$. Is it possible that these properties together already force your $A$ to be an open ball? | |
| Apr 14, 2022 at 20:01 | comment | added | Sam Hopkins | Sorry this is a vague comment but I think discrete Morse theory (en.wikipedia.org/wiki/Discrete_Morse_theory) is a powerful tool for thinking about this kind of thing | |
| Apr 14, 2022 at 18:03 | history | asked | Nicholas Proudfoot | CC BY-SA 4.0 |