Timeline for Are lattices quotients of their Dedekind-MacNeille completion?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Oct 4, 2015 at 16:45 | vote | accept | Dominic van der Zypen | ||
| Oct 3, 2015 at 21:32 | answer | added | Joseph Van Name | timeline score: 4 | |
| Oct 2, 2015 at 0:29 | answer | added | David E Speyer | timeline score: 4 | |
| Oct 1, 2015 at 19:06 | comment | added | მამუკა ჯიბლაძე | A free Boolean algebra on infinitely many generators is not a quotient of any complete Boolean algebra. (If it would be, it would be a retract of it, and would be complete itself.) | |
| Oct 1, 2015 at 15:56 | comment | added | Gerhard Paseman | For that matter, there is no surjective order-preserving map from the completion of the naturals N to N as a lattice. It should be clear that unbounded and countably (maybe uncountably?) cofinal lattices cannot have such maps from their completions. Gerhard "Where Does The Top Go?" Paseman, 2015.10.01 | |
| Oct 1, 2015 at 15:03 | comment | added | David E Speyer | @Dominic Do you want the composition $P \to DM(P) \to P$ to be the identity? | |
| Oct 1, 2015 at 15:03 | comment | added | David E Speyer | @darijgrinberg For the record, no, there is no order preserving surjection $\phi: \mathbb{R} \cup \{ \pm \infty \} \to \mathbb{Q}$. The image of such a map would land in $[\phi(- \infty), \phi(\infty) ]$, which is not all of $\mathbb{Q}$. There are also no order preserving surjections $\mathbb{R} \to \mathbb{Q}$, although it takes a bit more work. Might be a good problem for an analysis exam... | |
| Oct 1, 2015 at 13:31 | comment | added | David E Speyer | The DM completion of a finite lattice is the lattice itself. (The DM completion of a poset $P$ is the smallest complete lattice containing $P$ as a subposet. For finite posets, "lattice" and "complete lattice" are equivalent, so the DM completion of a lattice is itself.) | |
| Oct 1, 2015 at 12:40 | comment | added | darij grinberg | Is $\mathbb{Q} $ a quotient of $\mathbb{R}\cup\left\{\pm\infty\right\} $ ? I don't know the answer, but it doesn't look like it works this way. (That said finite posets might be different.) | |
| Oct 1, 2015 at 11:24 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |