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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.

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  • $\begingroup$ Isn't your construction just the sobrification of a space, i.e., one part of the adjunction between spaces and locales? (So sobrification is adjoint to $\mathcal{O}$.) $\endgroup$ – Andrej Bauer Nov 13 '18 at 16:54
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    $\begingroup$ @AndrejBauer : Yes and no: $\overline{A}$ with the topology induced by the Scott topology on $A$ is indeed the soberification of $A$ with its Scott topology. What I'm wondering is if this topology on $\overline{A}$ is the same as the Scott topology of $\overline{A}$ or something else. The adjunction I'm looking at is the functor "Points" but from the category of frames to the category of dcpo and its left adjoints. $\endgroup$ – Simon Henry Nov 13 '18 at 17:00
  • $\begingroup$ Ah yes, thank you for the explanation. $\endgroup$ – Andrej Bauer Nov 13 '18 at 17:55
  • $\begingroup$ Lawson describes this as particular kind of completion in "The round ideal completion via sobrification" $\endgroup$ – მამუკა ჯიბლაძე Nov 13 '18 at 22:06
  • $\begingroup$ What do you mean by complemented? $\endgroup$ – Emil Jeřábek Nov 14 '18 at 11:02
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I believe the paper by Johnstone linked to in the question contains the answer, and it is negative.

In that paper, Johnstone constructs a Scott topology that is not sober as a byproduct of answering in the negative the following question (marked by (a) in the paper):

if $(X,\leqslant)$ has directed joins, is there a sober topology on $X$ inducing $\leqslant$?

(A counterexample $(X,\leqslant)$ to (a) evidently has non-sober Scott topology since the latter induces $\leqslant$.)

But further, he answers in the negative also the following question (marked (b)):

if $(X,\leqslant)$ is induced by some sober topology, is the Scott topology on $X$ sober?

And, what is crucial for the present question, his counterexample to (b) is obtained as the sobering of a non-sober Scott topology.

Thus, he has got $(X,\leqslant)$ with directed joins such that taking in succession Scott, then sobering, then specialization gives another poset with directed joins, and for this another poset, again taking Scott, sobering, specialization gives still something different.

In more detail -

Notation, terminology: I will call a poset with all directed joins/suprema a dcpo. For a dcpo $P$ I will denote by $S(P)$ the topological space with the underlying set $P$ and with the Scott topology of $P$, i. e. topology with closed sets = those downsets of $P$ which are sub-dcpos. For a topological space $X$ I will denote by $P(X)$ the specialization preorder of $X$, and by $s(X)$ the sobering of $X$, i. e. the space of irreducible closed subsets of $X$ with its sober topology. Note that $P(s(X))$ is given by the inclusion order of irreducible closed subsets, and $X$ maps to $s(X)$ by sending a point to its closure.

Thus, in the notation of the question, $\overline A$ is $P(s(S(A)))$, and the question asks whether $P(s(S(A)))\to P(s(S(P(s(S(A))))))$ is a bijection for every dcpo $A$.

Johnstone constructs a dcpo $X$ such that $X$ is the only irreducible closed subset of $S(X)$ that is not closure of any point. Hence the specialization order of its sobering, $X^+=P(s(S(X)))$, is obtained by adding a top element $\infty$ to $X$, with $X^+$ the closure of $\{\infty\}$ in $s(S(X))$. At the same time, $S(X^+)$ is not sober since in the Scott topology of $X^+$, the subset $X$ is closed (being a sub-dcpo downset of $X^+$) and irreducible (since a topology inducing a dcpo order is contained in the Scott topology of that order).

Look now at the sobering $s(S(X^+))$ of $S(X^+)$. Because of the above, it contains one extra point $\infty'$, such that $X\cup\{\infty'\}$ is the closure of $\infty'$ in $s(S(X^+))$. Then in $P(s(S(X^+))$, the point $\infty'$ is an upper bound of $X$ and $\infty'<\infty$ (since as irreducible closed subsets of $S(X^+)$, $\infty'$ is $X$ and $\infty$ is $X^+$ ($=$ closure of $\{\infty\}$)).

So the map $P(s(S(X)))\to P(s(S(P(s(S(X))))))$, i. e. $X^+\to P(s(S(X^+)))$, is not a bijection: $\infty'$ is not in its image.

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