Example of a bicontinuous poset which is not jointly bicontinuous?

Question

Recall that a poset $P$ is said to be continuous if, for every $p \in P$, the set $\{q \in P \mid q \ll p \}$ is directed with supremum $p$. Here $q \ll p$ is the "way below" relation (see the above nlab link for complete definitions). Following Scott, continuous posets are a widely used concept in programming language semantics.

Following Martin and Panangaden, a poset $P$ is said to be bicontinuous if $P$ and $P^{op}$ are both continuous. Most familiar continuous posets are not bicontinuous, but Martin and Panangaden give an interesting family of examples: if $M$ is a nice time-oriented Lorentzian manifold, then the points of $M$, ordered by the causality relation $J^+$ (where $p \leq q$ if there is a future-oriented causal curve from $p$ to $q$), form a bicontinuous poset. The way-below relation is the chronology relation $I^+$ (so that $p \ll q$ if there is a future-oriented timelike curve from $p$ to $q$).

In this example, the way-below relation coincides with the dual "way-above relation" (there is a future-oriented timelike curve from $p$ to $q$ iff there is a past-oriented timelike curve from $q$ to $p$). Ebrahimi calls a bicontinuous poset with this property jointly bicontinuous.

Question: What is an example of a bicontinuous poset which is not jointly bicontinuous?

That is, what is an example of a poset $P$ such that $P$ and $P^{op}$ are both continuous, but where it is not the case that $x \ll y \Leftrightarrow y \ll^{op} x$ (where $\ll^{op}$ is the way-below relation on $P^{op}$)?

Answer 1

It turns out there are some easy examples, as indicated by Keimel in Bicontinuous Domains and Some Old Problems in Domain Theory:

• The unit interval $[0,1]$ is bicontinuous, but not quite jointly bicontinuous, because we have $0 \ll 0$ but not $0 \ll^{op} 0$ (and dually for $1$). Similar examples having to do with boundaries abound.

• If $X$ is a set, then the powerset lattice $P(X)$ is bicontinuous but not joint bicontinuous for $X$ infinite: we have $A \ll B$ iff $A$ is finite and $A \subseteq B$, whereas $B \ll^{op} A$ iff $B$ is cofinite and $A \subseteq B$. This example is harder to interpret in terms of "boundaries".