Timeline for Pairwise non-isomorphic interval-isomorphic lattices
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2019 at 17:57 | comment | added | Goldstern | Gabor Czedli has several papers about such lattices, which he calls "fractal": math.u-szeged.hu/~czedli/listak/publist.html | |
| Sep 5, 2019 at 17:56 | comment | added | Goldstern | The generic (or "typical") $\{0,1\}$-lattice (the Fraisse limit of all finite $\{0,1\}$-lattices) is another example. | |
| Sep 5, 2019 at 17:35 | answer | added | Keith Kearnes | timeline score: 1 | |
| Sep 4, 2019 at 8:28 | comment | added | Emil Jeřábek | @SamHopkins Usually not. Since $L$ embeds as an interval into $L\times L$ (say, by $x\mapsto(x,a)$ for a fixed element $a$), $L\times L$ has the property only if and only if (1) $L$ has the property and (2) $L\simeq L\times L$. | |
| Sep 4, 2019 at 8:23 | comment | added | Emil Jeřábek | @Bullet51 Up to isomorphism, $[0,1]\cap\mathbb Q$ is the only linearly ordered example (being the unique countable dense linear order with endpoints). | |
| Sep 3, 2019 at 7:55 | comment | added | Dominic van der Zypen | @WillBrian Good point - first I only could think of $\mathbb{Q}\cap [0,1]$ and then somebody mentioned the countable atomless Boolean algebra and I couldn't come up with anything else... But I assumed there must be infinitely many pairwise non-isomorphic such lattices, and "often", if you have countably many, you can find $2^{\aleph_0}$ many. (But vague intuition often leads astray.) | |
| Sep 3, 2019 at 7:51 | comment | added | Dominic van der Zypen | @Bullet51 I assume you meant $\cap$ instead of $\cup$? | |
| Sep 3, 2019 at 6:28 | comment | added | LeechLattice | Would the linearly ordered set $\mathbb{Q}[\sqrt 2] \cup [0,1]$ serve as a 3rd example? | |
| Sep 2, 2019 at 23:29 | comment | added | Sam Hopkins | Is this property preserved under direct products? | |
| Sep 2, 2019 at 21:41 | comment | added | Gerhard Paseman | That's one more than I could imagine. Maybe lexicographic product can make more? Gerhard "Is Low On Imagination Today" Paseman, 2019.09.02. | |
| Sep 2, 2019 at 16:16 | comment | added | Will Brian | Are there $3$? Maybe I'm just being an idiot here, but the only ones I can think of are the countable atomless Boolean algebra, and the linearly ordered set $\mathbb Q \cap [0,1]$. | |
| Sep 2, 2019 at 15:25 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |