If $A$ is poset with all directed suprema, it is common to consider the Scott topology on $A$, whose open subsets are the $U \subset A$ such that $U$ is upward closed and if $\bigcup_I a_i \in U $ for some directed supremum then $\exists i, a_i \in U$.

It is a classical fact that the specialization order induced by this topology on $A$ is exactly the order relation on $A$.

There are some examples (See P.Johnstone's 'Scott is not always sober') where the Scott topology is not sober. But one can always consider $\mathcal{O}(A)$ the frame of open subsets for the Scott topology, and look at:

$$ \overline{A} = pt(\mathcal{O}(A)) $$

the poset of 'points' (i.e. frame homomorphisms $\mathcal{O}(A) \rightarrow \{0,1\}$) of this frame. The fact that the Scott topology is a topology on $A$ immediately implies that there is a map $A \rightarrow \overline{A}$, and the observation above implies that this is an order embeding, moreover $\overline{A}$ is again directed complete and this inclusion preserves directed suprema. The question of soberness of the Scott topology boils down to whether this map is an isomorphism.

I want to know if this construction $A \rightarrow \overline{A}$ can be seen as a "completion", i.e. whether it is idempotent. The following are (I think) equivalent ways of formulating this question:

- is the map $\overline{A} \rightarrow \overline{\overline{A}}$ a bijection.
When $B$ is the poset of points of a frame, do we always have that $B \rightarrow \overline{B}$ is an equivalence.

Does the restriction to $A$ induce a bijection between the Scott topology of $\overline{A}$ and the Scott topology of $A$.

I found some papers that seem to suggest that the question has been studied, but I couldn't find any precise claim anywhere.

complemented? $\endgroup$ – Emil Jeřábek Nov 14 '18 at 11:02