Timeline for Is there an elementary proof that distal maps are invertible?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jun 16, 2021 at 10:39 | vote | accept | Nate River | ||
| Jun 16, 2021 at 9:43 | answer | added | Ville Salo | timeline score: 8 | |
| May 26, 2021 at 2:04 | vote | accept | Nate River | ||
| S Jun 16, 2021 at 10:39 | |||||
| May 25, 2021 at 16:26 | comment | added | Benjamin Steinberg | @YCor, I think the definition of distal for topological semigroups is slightly more complicated. You can find it in arxiv.org/pdf/1708.00996.pdf and the conclusion is you have a group. They show the whole enveloping semigroup is a group as well | |
| May 25, 2021 at 16:10 | comment | added | YCor | @BenjaminSteinberg sorry but I see what the conclusion is but not what the assumption purports to be. I'd assume part of the setting is:"let $S$ be a submonoid of $\mathrm{Homeo}(X)$, $X$ compact metric space, such that $D(x,y)=\inf_{s\in S}d(sx,sy)>0$ for all $x\neq y$. The conclusion is that something (what? certainly not $S$) is a group. | |
| May 25, 2021 at 15:20 | comment | added | Benjamin Steinberg | @YCor, well I believe the general result is for topological semigroups and the conclusion is being a topological group. So I suppose the continuity of the inverse is not free without further topological assumptions. | |
| May 25, 2021 at 15:00 | comment | added | Piotr Hajlasz | What is $X$ is say a circle or a Euclidean sphere? Is it easier then? It would be much more surprising if the result was equally difficult for a simple space. | |
| May 25, 2021 at 14:26 | comment | added | YCor | @BenjaminSteinberg in which sense? | |
| May 25, 2021 at 11:56 | comment | added | Benjamin Steinberg | There is a more general result that distality of a semigroup flow implies being a group. | |
| May 25, 2021 at 11:32 | answer | added | Ville Salo | timeline score: 7 | |
| May 25, 2021 at 10:54 | vote | accept | Nate River | ||
| May 25, 2021 at 10:55 | |||||
| May 25, 2021 at 8:35 | history | became hot network question | |||
| May 25, 2021 at 8:18 | answer | added | Nicolast | timeline score: 8 | |
| May 25, 2021 at 6:20 | history | edited | YCor | CC BY-SA 4.0 |
added tag, added remark
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| May 25, 2021 at 2:50 | comment | added | Piotr Hajlasz | Wow! Very surprising. I would expect that this is a very elementary exercise. | |
| May 25, 2021 at 1:45 | comment | added | Nate River | @rpotrie Yes, surjectivity is the nontrivial part. | |
| May 25, 2021 at 1:45 | comment | added | Nate River | It is! You can find the proof in Bergelson’s survey article Ergodic Ramsey Theory - an Update. It’s on page 33. | |
| May 25, 2021 at 1:43 | comment | added | rpotrie | If $T$ is not inyective, there are points $x\neq y$ such that $T(x)=T(y)$ and thus $inf_n d(T^n(x),T^n(y))=0$. I guess you ask about surjectivity? | |
| May 25, 2021 at 1:38 | comment | added | Asvin | That's a very nice result. Do you have a reference for the Stone Cech compactification proof? | |
| May 25, 2021 at 0:25 | history | asked | Nate River | CC BY-SA 4.0 |