# In which posets is the set of compact elements downwards closed?

In a poset $(D, \sqsubseteq)$, a compact element is an element $d \in D$ such that for every directed set $A$ which happens to have a supremum $\bigsqcup A \in D$ with $d \sqsubseteq \bigsqcup A$, it is $d \sqsubseteq a$ for some $a \in A$ (e.g., Definition I-4.1 of [Gierz et al 2003]).

My (elementary) question. Suppose $d$ is compact and $c \sqsubseteq d$. What are typical extra impositions on the poset, that ensure that $c$ must also be compact?

• (hmm, now that I think of it, it might have been better to ask this in math.se...) – Basil Feb 12 '15 at 13:55
• I added some more tags, using only "domain-theory" may have limited the number of people viewing the question. – Damiano Mazza Feb 13 '15 at 13:04

The above condition is the defining property of Berry's dI-domains, which are distributive (that's what the "d" stands for) algebraic domains satisfying condition I. This is a widely known and well studied class of domains, which served as the basis for what is known as "stable" semantics (an attempt to capture the sequential behavior of the $\lambda$-calculus). Girard's coherence spaces [2] give a particularly simple and useful example of dI-domains. A coherence space may be seen as the set of cliques of a reflexive graph, ordered by inclusion. The compact elements are the finite cliques, which obviously satisfy your requirement.
I don't know if generalizations of the I condition have been studied. A counterexample to your property is given by the ordinal $\omega+2$ ($\omega+1$ is compact but $\omega$ is not), so certainly generalizing "finite" to "well-founded+no infinite antichains" does not work.