Is it true that in a partial order the intersection of two upper cones is either disjoint or again an upper cone?

Question

Suppose $(P, \leq)$ is a partially ordered set.

For $x \in P$, define $U_x := \{ y \in P \ | \ y \geq x\}$.

Is it true that for any $x,y \in P$, either $U_x \cap U_y = \emptyset$ or $U_x \cap U_y = U_z$ for some $z \in P$ ?

Answer 1

No. Consider the set $P=\{\{a\},\{b\},\{a,b,c\},\{a,b,d\}\}$ partially ordered by inclusion. $U_{\{a\}}\cap U_{\{b\}}=\{\{a,b,c\},\{a,b,d\}\}$, which is neither empty nor of the form $U_z$ for some $z\in P$.

Answer 2

What does it meant that $U_x\cap U_y$ is empty? It means that no element is larger than both of them.

What does it meant that $U_x\cap U_y=U_z$? It means that any element which is larger than both is also larger than $z$ (or $z$ itself).

So in order to find a counterexample, we need to engineer a partial order in which there are two elements which are themselves not comparable, but have a common upper bound, yet they do not have a least upper bound. What would that mean?¹ It means that if $z\geq x,y$ then there is some $z'$ such that $x,y\leq z'<z$. So there is at least a decreasing sequence of upper bounds.

So we can start with something that looks like a decreasing sequence, and put two (or more) elements below that sequence. I'll leave it for you to come up with such an example.

---

1. It could also mean that there are two incomparable "minimal upper bounds" to $x$ and $y$. So it would look like

    *   *
    | X |
    *   *