1
$\begingroup$

Let $(X,\leq)$ be a partially ordered set such that each nonempty subset of $X$ has a supremum in $X$.

I have two related questions regarding such structures as $(X,\leq)$:

Question 1: Is $(X,\leq)$ equivalent to being a directed-complete partial order that is also a join-semilattice?

Since any nonempty subset of $X$ has a supremum, this in particular holds for any nonempty finite subset of $X$, and so it is a join-semilattice.

Let $U$ be an upward directed set in $X$. Then it is a nonempty subset of $X$, and so it has a supremum. It follows that $(X,\leq)$ is a directed-complete partial order.

Is this reasoning correct?

As the converse also holds, a partially ordered set in which all nonempty sets have a supremum must be equivalent to a directed-complete partial order that is also a join-semilattice.

Question 2: How to name such a poset in which all nonempty subsets have a supremum?

The structure of $(X,\leq)$ is dual to that of posets in which all nonempty subsets have an infimum. The latter is equivalent to bounded completeness, and according to Wikipedia "there is no common name for the dual property" of bounded completeness, i.e., for the structure of $(X,\leq)$ considered here. So, what would be a good name for the structure of $(X,\leq)$? Is there a better name than a directed-complete partial order (a.k.a. dcpo) that is also a join-semilattice?

$\endgroup$
3
  • $\begingroup$ Complete lattice? $\endgroup$
    – Berci
    Commented Aug 8, 2021 at 9:11
  • $\begingroup$ @Berci I think that is strictly stronger. In a complete lattice also the empty set has a supremum, which is the least element of the lattice. But I do not require the existence of a least element. For example, take for $X$ the collection of all nonempty subsets of the set $\{1,2\}$ and set inclusion as ordering, then each nonempty subset of $X$ has a supremum, but $X$ has no least element. $\endgroup$
    – Bart
    Commented Aug 8, 2021 at 9:38
  • $\begingroup$ @Ali Do I understand correctly that a suplattice from a categorical prespective is like a complete lattice, except that the morphisms are different? In my (perhaps ignorant) way of thinking I only look at objects, not morphisms, so it's basically the same? Sorry, my understanding of category theory is very limited. $\endgroup$
    – Bart
    Commented Aug 8, 2021 at 10:04

1 Answer 1

2
$\begingroup$

If you are making some research on this topic, you can come up with the terminology you think makes better sense to you, preferably without crashing with existing one.
I think you could call these complete semi-lattices.

While some would argue that a complete semi-lattice is just the algebraic reduct of a complete lattice, that can be disputed on the ground that it just have to be complete relative to the relevant operation.

There is a paper by Mai Gehrke and Hilary Priestley where they study a hierarchy of completions of posets:
Canonical extensions and completions of posets and lattices, in Reports on Mathematical Logic, 43 (2008), pages 133-152.
There they introduce concepts such as directed join completion, and duals, and although I only read it some ten years ago, I think they have the concept of completion of a poset which is not a complete lattice, unless that poset was already a lattice. It might worth reading, at least to justify the use of the expression "complete" in this context.

$\endgroup$
2
  • $\begingroup$ +1 (or "complete join-semilattice") $\endgroup$ Commented Aug 9, 2021 at 11:42
  • $\begingroup$ @WilliamDeMeo Yes! But if we only care about one operation, then meet or join are interchangeable. But you're right since the OP asks about the situation in which the join is the relevant one. $\endgroup$
    – amrsa
    Commented Aug 9, 2021 at 12:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .