Timeline for Non-principal prime ideals in infinite distributive lattices
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 10, 2019 at 8:36 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
minor typo
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| Oct 9, 2015 at 18:05 | comment | added | Joseph Van Name | However, since $X$ is inifnite, the set $C$ must also be infinite. Recall that as a consequence of Ramsey's theorem, every poset must have an infinite ascending chain, an infinite descending chain, or an infinite antichain. Therefore, the set $C$ must have an infinite antichain $(c_{n})_{n\in\mathbb{N}}$. Take note that since each $c_{n}$ is join-prime, we have $c_{n}\not\leq c_{1}\vee...\vee c_{n-1}$. Therefore, we have $c_{1}\vee...\vee c_{n-1}\leq c_{1}\vee...\vee c_{n}$. We therefore conclude that $(c_{1}\vee...\vee c_{n})_{n\in\omega}$ is an infinite ascending chain after all. | |
| Oct 9, 2015 at 18:05 | comment | added | Joseph Van Name | Clearly not every infinite lattice has a non-principal ideal or a non-principal filter (for example, take the Dedekind-Macneille completion of any infinite antichain). Let me prove that every infinite distributive lattice has a non-principal ideal or a non-principal filter. Suppose that $X$ is an infinite distributive lattice with no non-principal ideals and no non-principal filters. Let $C$ be the set of all join-prime elements in $X$. Then take note that since $X$ has no infinite descending chains, every element in $X$ is the least upper bound of finitely any join-irreducible elements. | |
| Oct 9, 2015 at 3:26 | comment | added | Joseph Van Name | Dominic van der Zypen. I must ask what argument do you use to conclude that every infinite distributive lattice has an non-principal ideal or a non-principal filter? | |
| Oct 9, 2015 at 3:24 | comment | added | Joseph Van Name | მამუკა ჯიბლაძე. The proof still works if $\bigvee J$ does not necessarily exist. If $\bigvee J$ does not exist, then one simply defines $G$ to be the set of all upper bounds of $J$ and the proof goes through with no problem. | |
| Oct 7, 2015 at 14:18 | comment | added | მამუკა ჯიბლაძე | Does not your argument depend on completeness? What if $\bigvee J$ does not exist at all? | |
| Oct 7, 2015 at 13:02 | history | answered | Dominic van der Zypen | CC BY-SA 3.0 |