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In Johnstone's Stone Spaces it is proved that the category of profinite partial orders is (equivalent to) the category of ordered Stone spaces (also called Priestley spaces) and that the obvious embedding from finite partial orders into Priestley spaces is the pro-completion of finite partial orders.

Main question: Does it follow that there exists a profinite completion functor from partial orders to Priestley spaces?

If yes, is there an explicit construction of the profinite completion of a partial order?

For example, does anyone know what does the profinite completion of the partial order $(\omega,=)$ look like?

Moreover, profinite quasi-orders are the the Priestley quasi-orders, i.e. quasi-ordered Stone spaces such that if $x\not\leq y$ then there is a clopen downset $D$ such that $y\in D$ and $x\not\in D$. I have the same questions about the profinite completion of a quasi-order. In this context, does one gain anything by considering quasi-order instead of partial orders?

For the record I asked this question on MathSE, here.

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  • $\begingroup$ I would guess that the profinite completion of a partial order is a sort of generalized Stone Cech compactification of the partially ordered set $P$. More specifically, if $P$ is a poset, then let $\mathcal{L}(P)$ be the collection of all downwards closed subsets of $P$. Then let $\iota:P\rightarrow 2^{\mathcal{L}(P)}$ be the mapping where $\iota(x)(L)=1$ iff $x\in L$. Then let $X=\overline{\iota[P]}$. Then $X$ becomes a Priestley space with its natural partial ordering and it seems like $X$ is the profinite completion of $P$. $\endgroup$ – Joseph Van Name Jul 17 '14 at 19:03
  • $\begingroup$ @JosephVanName Thanks for your guess. I need to think about it. I gave another more pedestrian try. I don't know yet if your construction is equivalent. $\endgroup$ – Yann Pequignot Jul 18 '14 at 6:49
  • $\begingroup$ @JosephVanName the description you give of the profinite completion of $P$ is simply the Priestley dual of the lattice of downsets of $P$. I wrote a sketch of a proof that this indeed the case. Your comments are welcome! $\endgroup$ – Yann Pequignot Jul 18 '14 at 14:03
  • $\begingroup$ Yann Pequignot. You should post your proof as an answer to this question. Users are encouraged to answer their own questions if they find an answer after they ask the question. $\endgroup$ – Joseph Van Name Jul 18 '14 at 18:41
  • $\begingroup$ Also, it seems as if the compactification that we are talking coincides with the Nachbin compactification. I gave an answer here mathoverflow.net/a/140625/22277 sketching basic facts about ordered topological spaces and I talked a little about the Nachbin compactification there. Also, Guram Bezhanishvili and Patrick Morandi both from New Mexico State University have written several relevant papers on partially ordered spaces and their ordered compactifications. The paper sierra.nmsu.edu/gbezhani/tos.pdf describes the Nachbin compactification and other ordered compactifications. $\endgroup$ – Joseph Van Name Jul 18 '14 at 18:44
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Let $\mathbf{Po}$ be the category of posets with monotone maps, $\mathbf{PrSp}$ the category of Priestley spaces. Then let $U:\mathbf{PrSp}\to \mathbf{Po}$ be the forgetful functor. The profinite completion functor $P:\mathbf{Po}\to\mathbf{PrSp}$ would be the left adjoint of $U$. Let $\mathbf{DLat}$ be the category of bounded distributive lattices. Now by Birkhoff duality $\mathbf{DLat}_\text{fin}$ is dual to $\mathbf{Po}_\text{fin}$ and Priestley duality is that $\mathbf{DLat}=\text{Ind-}\mathbf{DLat}_\text{fin}$ is dual to $\mathbf{PrSp}=\text{Pro-}\mathbf{Po}_\text{fin}$. Let $D:\mathbf{Po}\to \mathbf{DLat}$ be the contravariant functor sending a po $P$ to the lattice $\mathcal{D}(P)$ of its downsets, and a monotone map $f:P\to Q$ to the preimage map $f^{-1}:\mathcal{D}(Q)\to\mathcal{D}(P)$. Let $H:\mathbf{DLat}\to \mathbf{PrSp}$ be the contravariant functor in Priestley duality. Now I claim that $P:\mathbf{Po}\to \mathbf{PrSp}$ can be defined as $P=H\circ D$.

Sketch of the proof

Let $Q$ be a partial order and $\langle p_{i,j}:Q_i\to Q_j \rangle$ the cofiltered diagram of its finite po quotients. Let $P(Q)=\lim_i Q_i$ be the limit in the category of Priestley spaces, each $Q_i$ considered discrete of course, and $i_Q:Q\to P(Q)$ the natural monotone mapping. Using Birkhoff duality, we may consider the diagram $\langle h_j,i:D_j\to D_i\rangle$ in $\mathbf{DLat}_\text{fin}$ dual to $\langle p_i:Q\to Q_i \rangle$ and its colimit $D=\text{colim}_j D_j$ in $\mathbf{DLat}$. This distributive lattice is just the lattice $\mathcal{D}(Q)$ of downsets of $Q$. Indeed $D$ is the union of all downsets of $Q$ of the form $p^{-1}(E)$ for a monotone map $p:Q\to F$ with $F$ finite and $E$ downset of $F$. But for every downset $L$ of $Q$ the characteristic function $\chi_L:Q\to 2$ into $2=\{1<0\}$ is monotone and $\chi_L^{-1}(\{1\})=L$. It follows by Priestley duality that the Priestley space $P(Q)=\lim_i Q_i$ is dual to the lattice $\mathcal{D}(Q)$. From this point of view we get that $i_Q:Q\to P(Q)$, $q\mapsto \{D\in \mathcal{D}(Q)\mid q\in D\}$ is the unit of the adjunction.

Now let $f:Q\to X$ be a monotone map into a Priestley space $X$. Then $X$ is dual to the lattice $\mathcal{CD}(X)$ of clopen downsets of $X$ under Priestley duality. Since $f$ is monotone the preimage map goes $f^{-1}:\mathcal{D}(X)\to\mathcal{D}(Q)$ and restricts to a homomorphism $f^{-1}:\mathcal{CD}(X)\to\mathcal{D}(Q)$ whose Priesltey dual is a monotone continuous map $\bar{f}:P(Q)\to X$. Clearly we have $\bar{h}\circ i_Q=h$. Moreover $\bar{h}$ is the unique Priestley map from $P(Q)\to X$ with this property since $Q$ is dense in $P(Q)$.

Hence the functor $P:\mathbf{Po}\to\mathbf{PrSp}$, $Q\mapsto P(Q)$ and for a monotone $f:Q\to R$ in $\mathbf{Po}$ then $P(f):P(Q)\to P(R)$ is the Priestley dual map of $f^{-1}:\mathcal{D}(R)\to\mathcal{D}(Q)$, is left adjoint to the forgetful functor $U:\mathbf{PrSp}\to \mathbf{Po}$.

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  • $\begingroup$ A loosely related question: How do we know that $\mathbf{DLat}$ is the ind-completion of $\mathbf{DLat}_{\operatorname{fin}}$? Is this easy to see? $\endgroup$ – Math Student 020 Nov 19 '18 at 23:39

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