# What else (besides the usual) can be said about a Scott Information System if the constructed domain is required to be Hausdorff?

As per subject, if a Scott domain with $T_2$ topology is to be constructed from a Scott information system, then what besides the usual definitional requirements, e.g., https://en.wikipedia.org/wiki/Scott_information_system , does the information system have to satisfy?

Information systems are more completely discussed in (Scott's original paper) https://www.researchgate.net/publication/220897586_Domains_for_Denotational_Semantics (note: the original paper is followed by a three-page bibliography missing from this pdf) containing a section "Topological Considerations" starting on page 586, where "Hausdorff" is briefly mentioned on page 588 with respect to (what Scott calls) "total elements", $Tot_A$, assuming no unique top. But the domain itself, what Scott calls $\mid A\mid$, is still $T_0$ in general. And I'm asking for additional definitional requirements so that $\mid A\mid$, when contsructed from the corresponding information system, is "automatically" $T_2$.

• As Andrew points out, the only way to get $T_2$ for the entire domain is to make it trivial. But also note that apart from Scott topology on a domain there is also the Lawson topology on a domain, which is Hausdorff. – Andrej Bauer Apr 12 '17 at 18:50
• Thanks, @Kaveh for your comment about Stone duality, though you seem to have deleted it. Scott's "brief mention" that I cited above indeed also says "compact totally disconnected Hausdorff space", and I'd imagine you're likely right that this suggests some connection with the Stone representation theorem (though google's not immediately coughing it up). – John Forkosh Apr 13 '17 at 7:52
• Thanks @AndrejBauer (and also thanks for comment below). And I think you may be right, as follows. If for every data token $a\in A$ there's a complement $a^\perp\in A$, maybe analogous to $a\subseteq\mathbb{N}\;\mbox{r.e.}\Longrightarrow \mathbb{N}\backslash a\;\mbox{r.e.}$, then would that induce a Lawson topology on the corresponding constructed domain??? (Haven't tried to answer that myself yet; only just now saw your comment and had the preceding thought.) – John Forkosh Apr 13 '17 at 8:04
• Note that the maximal set of an (algebraic) domain is totally disconnected, whereas Stone duality gives you zero-dimensional spaces. These are two different concepts, and indeed we get different kinds of spaces. There is no direct relation to Stone duality, but of course one can cook up a slightly less direct one (as always). – Andrej Bauer Apr 13 '17 at 8:27
• Thanks again, @AndrejBauer . I'm now thinking your and A.P.'s remarks below are likely setting me on the right track -- think about the space of total elements rather than the domain. That gives more opportunity/flexibility to (your words) "cook up" interesting stuff. – John Forkosh Apr 14 '17 at 8:24

Since the least element $\bot$ of any Scott domain is a compactification point $-$ the only open set containing it is the whole space $-$ the Scott topology is never Hausdorff, unless it is trivial.
An information system $(A,Con,\vdash)$ induces a trivial domain iff it is itself trivial: $Con = \{\emptyset\}$. Otherwise, there exists at least one point above $\bot$.
The total elements of the Scott domain can indeed be given their own topology sometimes, which may be $T_2$ or even metrizable.