Let $P$ be a partially ordered set, and $Idl(P)$ be the partially ordered set of ideals over $P$.
Let $\downarrow : P \to Idl(P)$ be the function defined by $\downarrow x = \{ y \in P : y \le x \}$.
Question: is this function Scott-continuous?
My thoughts: Let $D$ be a directed subset of $P$ and assume that it has a supremum $d$. One must show that $\downarrow d$ is the supremum of $\{ \downarrow x : x \in D \}$. It is easy to show that $\downarrow d$ is an upper bound of this set. But I cannot figure out how to prove that it is the least one.
$\downarrow$ is almost never Scott continuous.
Suppose $D$ is a directed set with supremum $d$. In $\mathrm{Idl}(P)$, the supremum of $\{\downarrow x\mid x\in D\}$ is the union $\bigcup_{x\in D}\downarrow x$. If $\downarrow$ is Scott continuous, this is equal to $\downarrow d$. In particular, $d\in \bigcup_{x\in D}\downarrow x$, so there is some $x\in D$ such that $d\leq x$. Since $d$ is the supremum of $D$, $d=x\in D$.
Thus we see that $\downarrow$ is Scott continuous if and only if every directed set with a supremum in $P$ contains its supremum. Equivalently, every element of $P$ is compact. Equivalently, every upward-closed set in $P$ is Scott open.