# How to define compatible topology for first-order structures?

Background Because a bounded distributive lattice can be represented by the clopen sets of a Priestley space, I tried to learn some basics about Priestley spaces. After reading (on Wikipedia)

A Priestley space is an ordered topological space with special properties.

I googled for "ordered topological space", with confusing results. Even so I know that the Priestley separation axiom already ensures that the topology is compatible with the partial order, I still would like to know what it actually means for "the topology to be compatible with the partial order". I became confused when I read a reference ("On partially ordered sets possessing a unique order-compatible topology" by E. S. Wolk) which required that the topology must be coarser than the topology with the Dedekind-closed subsets as closed sets. It also required the topology to be finer than the interval topology, which is probably equivalent to the condition I would have guessed.

Question What is wrong with saying that a partially ordered set is a universal Horn structure, and that a compatible topology just needs to satisfy the "natural" conditions for a topology compatible with a universal Horn structure? Here I would just require that all the functions $f^n$ (from the signature of the Horn structure) are continuous functions from the argument space with the product topology into the object space, and that for all predicates $P^n$ (from the signature of the Horn structure) the sets $\{(x_1,\ldots,x_n):P^n(x_1,\ldots,x_n)\}$ are closed sets for the product topology. (If equality is part of the language, also the binary predicate $=$ must be included among these predicates. This means that the topology must be Hausdorff, as I learned now.)

I'm restricting myself to universal Horn structures here, because this is sufficient for my problem, and I would be surprised if one could come up with a useful definition of compatible topology for general first-order structures (without any additional restrictions like being universal Horn structures).

Many notions of compatibility between a partially ordered set and a topology on its underlying set are analogous to separation axioms and other well known concepts from general topology.

One can generalize the separation axioms and other notions such as $T_{2}$,complete regularity, and zero-dimensionality to axioms on ordered spaces that basically say that the order is compatible with the topology in some sense. For instance, following this trend, the notion of Stone-duality becomes the notion of Priestley duality. In this answer, I will outline some of the separation axioms and related axioms for ordered topological spaces. One familiar with general topology should be able to see the analogy between these results regarding ordered topological spaces and general topology.

For example, a topological space $X$ with a partial order $\leq$ is said to be order-Hausdorff if whenever $x\not\leq y$, there is an upper set $U$ and an lower set $V$ with $U\cap V=\emptyset$ and where $x\in U^{\circ},y\in V^{\circ}$. One can show that a partially ordered topology is Hausdorff if and only if $\leq$ is closed in $X^{2}$.

In general topology, we usually want more than simply the Hausdorff separation axiom (remember that most interesting topological spaces are completely regular), and the case is not much different for ordered sets. For ordered spaces, we sometimes want our spaces to be more than just Hausdorff.

An ordered topological space $X$ is said to be completely order-regular if

i. whenever $x\not\leq y$, then there is a continuous order preserving map $f:X\rightarrow[0,1]$ with $f(y)<f(x)$, and

ii. if $x\in X$ and $C\subseteq X$ is a closed set with $x\not\in C$, then there is some continuous order preserving map $f:X\rightarrow[0,1]$ and a continuous order reversing map $g:X\rightarrow[0,1]$ with $f(x)=g(x)=1$ and $C\subseteq f^{-1}[\{0\}]\cup g^{-1}[\{0\}]$.

An ordered topological space $X$ is said to satisfy the Priestley separation axiom if whenever $x\not\leq y$, then there is a clopen upper set $C$ with $x\in C,y\not\in C$. The ordered topological spaces that satisfy the Priestley separation axiom are said to be totally order disconnected. A Priestley space is a compact totally order disconnected space.

An order-compactification of an ordered topological space $X$ is a compact Hausdorff-ordered space $Y$ such that $X$ is a dense ordered-subspace of $Y$. An order-compactification $Y$ of an ordered topological space $X$ is said to be a Priestley order compactification if $Y$ is a Priestley space.

$\mathbf{Theorem}$(Nachbin) An ordered topological space has an order-compactification if and only if it is completely order regular.

An ordered topological space $X$ is said to be order-zero-dimensional if satisfies the Priestley separation axiom and the set $\{U\setminus V|U,V\,\textrm{are clopen upper sets}\}$ forms a basis for the topology on $X$.

One can generalize the notion of the Stone-Cech compactification to ordered spaces. If $X$ is a completely order regular space, then there is an order compactification denoted by $n(X)$ called the Nachbin compactification where every continuous order preserving mapping $f:X\rightarrow Y$ such that $Y$ is a compact order-Hausdorff space can be extended to a continuous order preserving map $\overline{f}:n(X)\rightarrow Y$.

A completely order regular space $X$ is said to be strongly-order-zero-dimensional if whenever $f:X\rightarrow[0,1]$, then there is some clopen upper set $U$ with $f^{-1}[\{1\}]\subseteq U$ and $f^{-1}[\{0\}]\subseteq U^{c}$.

$\mathbf{Theorem}$ Let $X$ be a completely-order-regular space. Then the Nachbin order-compactification $n(X)$ is a Priestley space iff $X$ is strongly order-zero-dimensional.

$\large\textbf{The Specialization Ordering}$ It is often the case that an ordered topological space does not satisfy any strong separation axioms, but one should never say that these ordered sets have an incompatible topology. It therefore seems like the notion of compatibility between a topology and a partial order depends on context. Every topological space $X$ has an inherent preordering $\leq$ called the specialization ordering where $x\leq y$ if and only if $\overline{x}\subseteq\overline{y}$. Said slightly differently, $x\leq y$ if and only if $x\in\overline{y}$. This preordering is a partial ordering if and only if the space $X$ is a $T_{0}$-space. However, a space $X$ is a $T_{1}$-space if and only if $x\leq y\Rightarrow x=y$ for $x,y\in X$. In other words, a space is a $T_{1}$-space if and only if the specialization ordering is trivial. Often, when one studies ordered topological spaces, the topology coincides with the specialization ordering and thus satisfies no interesting separation axioms. Nevertheless, these ordered topological spaces are important. My answer here also gives information on the specialization ordering in the more general context of closure systems and closure operators while discussing the importance and intuition behind spaces that do not satisfy strong separation axioms.

It is possible to obtain ordered topological spaces that satisfy strong order-separation axioms from topological spaces that are not even $T_{1}$. Furthermore, one obtains dualities between categories of ordered topological spaces satisfying strong separation axioms and ordered topological spaces that do not satisfy strong separation axioms. One should therefore give an equal level of importance to ordered topological spaces that do not satisfy strong separation axioms to ordered topological spaces that do satisfy strong separation axioms.

In my own personal research, I have developed the following duality between ordered topological spaces satisfying strong separation axioms and those that do not satisfy strong separation axioms.

I define a lower limit space to be a partially ordered space $X$ such that each set $\downarrow x=\{y\in X|y\leq x\}$ is closed and where the sets $\{\downarrow x\cap U|x\in X,U\,\textrm{is an open upper set}\}$ form a basis for the topology on $X$. It is easy to show that every lower limit space is order-zero-dimensional. If one has a lower limit space, then the collection of open upper sets forms a topology on the set. Furthermore, this restriction to this new topology gives us a duality between the category of lower limit spaces and certain topologies called topological LUB-systems. The importance of topological LUB-systems lies in the fact that LUB-systems are essentially partially ordered sets (the partial order on a LUB-system is the specialization ordering) along with a notion of which least upper bounds are important and which least upper bounds are unimportant.

There is also a duality between order-Hausdorff compact spaces and stably compact spaces. Here when we use compactness or local compactness, we shall not assume that the Hausdorff separation axiom. Recall that a topological space $X$ is locally compact if for each $x\in X$ and open set $U$ with $x\in U$ there is an open $V$ and a compact $K$ with $x\in V\subseteq K\subseteq U$. A subset of a topological space is said to be saturated if it is the intersection of open sets. A space is said to be coherent if the intersection of two compact saturated sets is compact. A space is said to be stably compact if it is compact, locally compact, coherent, and sober.

If $(X,\leq,\mathcal{T})$ is a compact order-Hausdorff ordered topological space, then define a new topology $\mathcal{T}^{s}$ where $\mathcal{T}^{s}$ consists of all open upper sets. Then $(X,\mathcal{T}^{s})$ is a stably compact space and $\leq$ is the specialization ordering on $\mathcal{T}^{s}$. If $(X,\mathcal{T})$ is a topology, then let $\mathcal{T}^{d}$ denote the topology generated by a subbase consisting of the complements of the compact saturated sets. If $(X,\mathcal{T})$ is a topology, then the patch topology $\mathcal{T}^{p}$ is the topology on $X$ generated by the union $\mathcal{T}^{d}\cup\mathcal{T}$. If $(X,\mathcal{T})$ is a stably compact space, then $(X,\mathcal{T}^{p},\leq)$ is a compact order-Hausdorff space where $\leq$ is the specialization ordering in the space $(X,\mathcal{T})$. These two constructions are inverses and they yield an equivalence between the category of stably compact spaces (with perfect maps) and compact order-Hausdorff spaces.

Several of the results and definitions in this answer (especially regarding spaces satisfying higher separation axioms) can be found in papers by Guram Bezhanishvili and Patrick Morandi. The reader should also consult [2] or [3] for information about ordered topological spaces. I am unsure of whether there is an interesting generalization of these concepts from ordered sets to a broader class of structures.

1. Nachbin, Leopoldo. Topology and Order. Princeton, NJ: Van Nostrand, 1965.

2. Gierz, Gerhard; Hofmann, Karl Heinrich; Keimel, Klaus; Lawson, Jimmie D.; Mislove, Michael W.; Scott, Dana S. compendium of continuous lattices. Springer-Verlag, Berlin-New York, 1980.

3. Gierz, G., Hofmann, K. H. , Keimel, K., Lawson, J. D. , Mislove, M. Scott, D. S.; Continuous lattices and domains. Cambridge University Press, Cambridge, 2003.

• Did you have some sleep? Independent of whether you will finish editing the answer or just remove that line, I have a question. I thought about the second paragraph now and agree, except for the sentence "This is equivalent to saying $\leq$ is closed in $X^2$." I don't say that it is false, just that I don't understand it yet. To simplify, assume we say that $=$ is closed in $X^2$. Since this is a special case of your statement, this should be equivalent to $X$ being Hausdorff. I remember that there was some connection between the product topology and Hausdorff spaces, but was it so simple? – Thomas Klimpel Aug 29 '13 at 16:26
• And the connection between the product topology and Hausdorff spaces without order is also simple since a space is Hausdorff if and only if the diagonal is closed in $X\times X$, so this result on ordered topological spaces is a generalization of the characterization on Hausdorff spaces. – Joseph Van Name Aug 30 '13 at 2:26

EDIT: Well, this was impressively wrong: not only did I somehow spend most of the answer thinking "ordered space" meant "linearly ordered space," I also missed the significance of the OP's restriction to universal Horn structures. Below is a heavily edited version of my original post. As it stands, this is not an answer, just a too-long comment. (My apologies to the OP.)

First, Wikipedia has the definition of ordered space: http://en.wikipedia.org/wiki/Partially_ordered_space.

As to first-order structures which are "compatible" with a given topology, let me say three things:

• When studying topological structures, we don't always care about any sort of compatibility. See See Ziegler's article "Topological model theory" (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?handle=euclid.pl/1235417281&view=body&content-type=pdf_1) in Model-Theoretic Logics. Obviously, this isn't what you're talking about, but I think it's worth mentioning - especially if you start looking around for topological model theory, just so one is aware that frequently no such compatibility is required.

• If the complexity of the structure is not restricted, then we don't always want relations to be closed; sometimes we want relations to be open. This is the approach taken by Robinson (http://matwbn.icm.edu.pl/ksiazki/fm/fm81/fm81115.pdf). On the other hand, we could simply demand that every predicate be closed; the price of that would be that some reasonable topologies on some natural first-order structures would be ommitted.

• I don't know of a structure $M$ with a natural topology $\tau$ where $M$ is universal Horn and $\tau$ does not make all predicates closed; however, this doesn't mean that such structures aren't interesting. As an example, given a universal Horn structure $\mathcal{M}$ with a topology $\tau$, it is reasonable to consider adjoining a predicate $U$ to $\mathcal{M}$ naming a specific open subset of $\mathcal{M}$. The resulting expansion $\mathcal{M}^+$ might still be universal Horn, but now it's not clear to me that it's natural to want $U^{\mathcal{M}^+}$ to now be clopen. I think even in the universal Horn cases, demanding all predicates be closed might kill off some neat objects.

• My understanding is limited but I believe the brilliant breakthrough of Priestley's approach resides in combining the upper and lower views into one topology and therefore understanding the inherent duality present in (bounded) lattices. The key to understanding her approach is consumed by understanding the celebrated Priestley duality (which I unfortunately do not yet understand very well). – François G. Dorais Aug 28 '13 at 2:06
• I focus on universal Horn structures, because this invalidates the most obvious counterexamples. You stated: "However, note that many intuitively compatible structures are ruled out by your approach", but none of the counterexamples you gave was a universal Horn structure. This is no coincidence, because focusing on universal Horn structures allows us to "essentially" ignore negated relations. (If you define $x<y$ via $(x \leq y) \land \lnot (x=y)$, you see how negation entered the picture for your counterexamples.) – Thomas Klimpel Aug 28 '13 at 8:11