<p>That is, considering the underlying poset of a domain, when does the order-dual poset also comprise a domain?<br>
Below's a little, not strictly necessary, elaboration of that question.</p>
<p>Usual straightforward def of domain (as per, e.g., def 3.1.8, pg 57, &quot;Mathematical Theory of Domains&quot;, Stoltenberg-Hansen, et al, CUP 1994, ISBN 0-521-38344-7) is that<br>
(i) the poset is a dcpo (lubs of all directed subsets exist), and is algebraic
(every element is the lub of the set of compact elements weaker than itself).<br>
(ii) every consistent set (a set with an upper bound) has a lub, i.e., if an upper bound exists then a least upper bound exists.</p>
<p>So, if a poset satisfies (i) and (ii) above, when does its order-dual also satisfy them?<br>
Also, since there are various slight variations of the definition of a domain, choose any other you prefer. Whatever definition, the same question remains: when is the order-dual poset also a domain?<br>
(This isn't really research, so I started posting it to cs.stackexchange, but only cstheory has a domain-theory tag)</p>

CC BY-SA 4.0
2021-06-03T08:29:08.210Z
No, usually not. Typical domains don’t even have a largest element (a domain with a largest element is in fact an algebraic complete lattice), hence there is no way for their dual to be a domain. Furthermore, the dual of an algebraic complete lattice is not in general algebraic either.
Emil Jeřábek
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2021-06-03T08:40:51.367Z
Oh, I missed the question is when, not whether. Well, just follow the definition: a poset and its dual are both domains iff it is an algebraic and dually algebraic complete lattice. Either way, this is not a research-level question.
Emil Jeřábek
CC BY-SA 4.0
2021-06-03T10:21:13.770Z
@EmilJeřábek Thanks (yeah, I know it's not research-level, which is why I wrote that explanatory parenthetical sentence at the very end). Nevertheless, re the question itself, what I was thinking is that maybe a domain whose underlying poset is orthocomplemented might remain a domain with respect to the order-dual poset. Clearly, $\perp\leftrightarrow\top$ isn't a problem, and I think there's a waybelow<-->wayabove (so to speak) correspondence that preserves the algebraic property, but haven't been able to prove it or to locate a proof. Any ideas along those (or any other) lines? Thanks again.
John Forkosh
CC BY-SA 4.0
2021-06-03T10:29:10.317Z
This *is* a research level question. A while ago people organized workshops in which they studied bi-continuous lattices, the OP is asking about an algebraic version of that.
Andrej Bauer
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2021-06-03T10:32:17.283Z
@AndrejBauer Clearly, you are reading into the question something that I do not see there. Care to explain what? How is the question not answered by “algebraic and dually algebraic complete lattice”, as I wrote above?
Emil Jeřábek
CC BY-SA 4.0
2021-06-03T10:33:16.637Z
I would expect that the obvious condition can be massaged into something a lot less obvious. For example, if you take the compact and co-compact elements, will they form a lattice?
Andrej Bauer
CC BY-SA 4.0
2021-06-03T10:37:27.090Z
@JohnForkosh Well, by definition, an orthocomplement is in particular an isomorphism of the lattice to its dual. If a poset is isomorphic to its dual, then of course it is a domain if and only if its dual is.
Emil Jeřábek
CC BY-SA 4.0
2021-06-03T10:51:53.313Z
@AndrejBauer Take, for example, an atomic Boolean algebra (or even just a powerset algebra). Then compact elements are atoms, and cocompact elements are coatoms. These just form two antichains, nothing like a lattice. I’m willing to get surprised, but I rather doubt that in general, these structures satisfy any useful conditions that do not easily follow from the definition.
Emil Jeřábek
CC BY-SA 4.0
2021-06-03T11:05:47.120Z
@EmilJeřábek (and (at)AndrejBauer) Thanks so much, guys, for the very useful and insightful comments (and please keep them coming if you've got anything additional you'd care to say). So, Emil, are you saying that orthocomplemented is indeed sufficient (but maybe not necessary)? And, if so, does your high-level argument constitute a bona-fide proof? Or do I need a constructive proof, explicitly demonstrating the requisite properties (which I tried and failed to do on a few sheets of scrap paper)? I'd also tried and failed googling. Are there maybe any "close" references I missed? Thanks.
John Forkosh
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2021-06-03T11:16:11.417Z
@Emil: Aren't the compact elements the finite subsets and the cocompact ones the cofinite subsets, and they form a Boolean algebra? In any case, your doubting that there is a useful answer is a poor reason for shutting down the question, especially since someone who actually does domain theory stated that a while ago the topic was of some research interest. I can probably dig up the relevant Dagstuhl workshop number.
Andrej Bauer
CC BY-SA 4.0
2021-06-03T11:21:02.117Z
[Bi-Continuous Domains and Some Old Problems in Domain Theory](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.331.4197) by Klaus Keimel has some basic observations about bi-continuous posets. This might be a good starting point, and it shouldn't be so difficult to port it to the bi-algebraic case.
Andrej Bauer
CC BY-SA 4.0
2021-06-03T11:35:23.607Z
@JohnForkosh What I wrote is a proof. You don’t even need the “complemented” part of orthocomplemented, just the existence of a bijection $\neg$ such that $x\le y\iff\neg y\le\neg x$.
Emil Jeřábek
CC BY-SA 4.0
2021-06-03T12:00:46.350Z
@EmilJeřábek: are you having a bad day?
Andrej Bauer
CC BY-SA 4.0
2021-06-03T12:23:37.520Z
@EmilJeřábek Thanks, again. So I take it that $\neg:P\to P$ such that $\forall x,y\in P: x\sqsubseteq y \Longleftrightarrow \neg y\sqsubseteq \neg x$ is the necessary and sufficient condition? Terrific!!! And thanks for the upvote (which I assume is yours or maybe Andrej's). So, if you'd like, post that as an answer and I'll accept it (not that you need the points). By the way, your remark "compact elements are atoms" isn't applicable to >>all<< atomic lattices, is it? I'd assume atoms are always compact, but that there can, in the general case, be compact elements that aren't atoms.
John Forkosh
CC BY-SA 4.0
2021-06-03T12:37:29.053Z
@AndrejBauer Thanks for your remarks and the Klaus Keimel reference, which I've downloaded but not thoroughly read yet. But I did notice his last appendix uses the formal ball model, which I'd studied from Edalat's use of that idea in https://www.researchgate.net/publication/2356551_Domains_for_Computation_in_Mathematics_Physics_and_Exact_Real_Arithmetic for his similar proof for Hilbert spaces. And I also noticed he has a Keye Martin reference involving domain theory that I'd also studied a bit. So, while I can't be sure about Emil, I've been having a very good day thanks to the both of you:)
John Forkosh
CC BY-SA 4.0
2021-06-03T12:42:16.090Z
The condition is sufficient, but *not* necessary. Also, the comment about compact elements and atoms (which was supposed to apply only in the specific example of powerset algebras) was completely wrong, as pointed out by Andrej.
Emil Jeřábek
CC BY-SA 4.0
2021-06-04T06:42:31.983Z
Martin Escardo tells me you should look up "bicontinuous" in the Compendium of Continuous Lattices and Domains.
Andrej Bauer
CC BY-SA 4.0
2021-06-05T04:25:45.287Z
@AndrejBauer Thanks again, Andrej. Yeah, I picked up on that from your previous comment referencing Keimel's "Domain IX" talk, which I also found extremely interesting. But I hadn't looked up bicontinuous in the compendium -- actually, I have a hardcover copy of its successor, Continuous Lattices and Domains, where page 501 def VII-2.5 has an elementary (i.e., just right for me) definition of bicontinuous lattices. And that indeed looks relevant; like you said above, "people organized workshops in which they studied bi-continuous lattices, the OP is asking about an algebraic version of that".
John Forkosh
<order-theory><domain-theory>
What conditions are necessary (and sufficient) for the order-dual of a Scott-Ershov domain to also be a domain?
John Forkosh
# What conditions are necessary (and sufficient) for the order-dual of a Scott-Ershov domain to also be a domain?

(Asked by: username_0 on 2021-06-03)

That is, considering the underlying poset of a domain, when does the order-dual poset also comprise a domain?  
Below's a little, not strictly necessary, elaboration of that question.

Usual straightforward def of domain (as per, e.g., def 3.1.8, pg 57, "Mathematical Theory of Domains", Stoltenberg-Hansen, et al, CUP 1994, ISBN 0-521-38344-7) is that  
(i) the poset is a dcpo (lubs of all directed subsets exist), and is algebraic
(every element is the lub of the set of compact elements weaker than itself).  
(ii) every consistent set (a set with an upper bound) has a lub, i.e., if an upper bound exists then a least upper bound exists.

So, if a poset satisfies (i) and (ii) above, when does its order-dual also satisfy them?  
Also, since there are various slight variations of the definition of a domain, choose any other you prefer. Whatever definition, the same question remains: when is the order-dual poset also a domain?  
(This isn't really research, so I started posting it to cs.stackexchange, but only cstheory has a domain-theory tag)

**Tags:** order-theory, domain-theory

> **Comments:**
> **username_1**: No, usually not. Typical domains don’t even have a largest element (a domain with a largest element is in fact an algebraic complete lattice), hence there is no way for their dual to be a domain. Furthermore, the dual of an algebraic complete lattice is not in general algebraic either.  
> **username_1**: Oh, I missed the question is when, not whether. Well, just follow the definition: a poset and its dual are both domains iff it is an algebraic and dually algebraic complete lattice. Either way, this is not a research-level question.  
> **username_0**: @EmilJeřábek Thanks (yeah, I know it's not research-level, which is why I wrote that explanatory parenthetical sentence at the very end). Nevertheless, re the question itself, what I was thinking is that maybe a domain whose underlying poset is orthocomplemented might remain a domain with respect to the order-dual poset. Clearly, $\perp\leftrightarrow\top$ isn't a problem, and I think there's a waybelow<-->wayabove (so to speak) correspondence that preserves the algebraic property, but haven't been able to prove it or to locate a proof. Any ideas along those (or any other) lines? Thanks again.  
> **username_2**: This *is* a research level question. A while ago people organized workshops in which they studied bi-continuous lattices, the OP is asking about an algebraic version of that.  
> **username_1**: @AndrejBauer Clearly, you are reading into the question something that I do not see there. Care to explain what? How is the question not answered by “algebraic and dually algebraic complete lattice”, as I wrote above?  
> **username_2**: I would expect that the obvious condition can be massaged into something a lot less obvious. For example, if you take the compact and co-compact elements, will they form a lattice?  
> **username_1**: @JohnForkosh Well, by definition, an orthocomplement is in particular an isomorphism of the lattice to its dual. If a poset is isomorphic to its dual, then of course it is a domain if and only if its dual is.  
> **username_1**: @AndrejBauer Take, for example, an atomic Boolean algebra (or even just a powerset algebra). Then compact elements are atoms, and cocompact elements are coatoms. These just form two antichains, nothing like a lattice. I’m willing to get surprised, but I rather doubt that in general, these structures satisfy any useful conditions that do not easily follow from the definition.  
> **username_0**: @EmilJeřábek (and (at)AndrejBauer) Thanks so much, guys, for the very useful and insightful comments (and please keep them coming if you've got anything additional you'd care to say). So, Emil, are you saying that orthocomplemented is indeed sufficient (but maybe not necessary)? And, if so, does your high-level argument constitute a bona-fide proof? Or do I need a constructive proof, explicitly demonstrating the requisite properties (which I tried and failed to do on a few sheets of scrap paper)? I'd also tried and failed googling. Are there maybe any "close" references I missed? Thanks.  
> **username_2**: @Emil: Aren't the compact elements the finite subsets and the cocompact ones the cofinite subsets, and they form a Boolean algebra? In any case, your doubting that there is a useful answer is a poor reason for shutting down the question, especially since someone who actually does domain theory stated that a while ago the topic was of some research interest. I can probably dig up the relevant Dagstuhl workshop number.  
> **username_2**: [Bi-Continuous Domains and Some Old Problems in Domain Theory](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.331.4197) by Klaus Keimel has some basic observations about bi-continuous posets. This might be a good starting point, and it shouldn't be so difficult to port it to the bi-algebraic case.  
> **username_1**: @JohnForkosh What I wrote is a proof. You don’t even need the “complemented” part of orthocomplemented, just the existence of a bijection $\neg$ such that $x\le y\iff\neg y\le\neg x$.  
> **username_2**: @EmilJeřábek: are you having a bad day?  
> **username_0**: @EmilJeřábek Thanks, again. So I take it that $\neg:P\to P$ such that $\forall x,y\in P: x\sqsubseteq y \Longleftrightarrow \neg y\sqsubseteq \neg x$ is the necessary and sufficient condition? Terrific!!! And thanks for the upvote (which I assume is yours or maybe Andrej's). So, if you'd like, post that as an answer and I'll accept it (not that you need the points). By the way, your remark "compact elements are atoms" isn't applicable to >>all<< atomic lattices, is it? I'd assume atoms are always compact, but that there can, in the general case, be compact elements that aren't atoms.  
> **username_0**: @AndrejBauer Thanks for your remarks and the Klaus Keimel reference, which I've downloaded but not thoroughly read yet. But I did notice his last appendix uses the formal ball model, which I'd studied from Edalat's use of that idea in https://www.researchgate.net/publication/2356551_Domains_for_Computation_in_Mathematics_Physics_and_Exact_Real_Arithmetic for his similar proof for Hilbert spaces. And I also noticed he has a Keye Martin reference involving domain theory that I'd also studied a bit. So, while I can't be sure about Emil, I've been having a very good day thanks to the both of you:)  
> **username_1**: The condition is sufficient, but *not* necessary. Also, the comment about compact elements and atoms (which was supposed to apply only in the specific example of powerset algebras) was completely wrong, as pointed out by Andrej.  
> **username_2**: Martin Escardo tells me you should look up "bicontinuous" in the Compendium of Continuous Lattices and Domains.  
> **username_0**: @AndrejBauer Thanks again, Andrej. Yeah, I picked up on that from your previous comment referencing Keimel's "Domain IX" talk, which I also found extremely interesting. But I hadn't looked up bicontinuous in the compendium -- actually, I have a hardcover copy of its successor, Continuous Lattices and Domains, where page 501 def VII-2.5 has an elementary (i.e., just right for me) definition of bicontinuous lattices. And that indeed looks relevant; like you said above, "people organized workshops in which they studied bi-continuous lattices, the OP is asking about an algebraic version of that".  

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## 0 answers

