Timeline for Does the collection of coverings on a set $X$ form a lattice when ordered by refinement?
Current License: CC BY-SA 3.0
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| Jul 6, 2016 at 6:26 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Jul 6, 2016 at 6:25 | comment | added | Fedor Petrov | Ops. You are right, of course. | |
| Jul 6, 2016 at 0:43 | comment | added | Joel David Hamkins | It follows that in general, it is an upper semi-lattice. And therefore, if there are only finitely many points, it is a lattice. Namely, there is a least cover, consisting of the singletons, and any finite upper semi-lattice with a least element is a lattice: the meet of any two elements is the join of their lower bounds. So any counterexample must be infinite. | |
| Jul 5, 2016 at 23:33 | comment | added | François G. Dorais | It seems to me that the unique supremum in this case is the cover consisting of all four sides. In fact, I think that if $A$ and $B$ are two covers, then the supremum is the proper cover that consists of all inclusion maximal elements of $A \cup B$. | |
| Jul 5, 2016 at 20:03 | history | answered | Fedor Petrov | CC BY-SA 3.0 |