9

On page 172, James Munkres' textbook Topology(2ed), there is a theorem about compact subspaces of the real line:

Let $X$ be a simply-ordered set having the least upper bound property. In the order topology, each closed interval in $X$ is compact.

My question is whether there is a generalized theorem about a Poset(or a lattice, complete lattice, maybe). Is there some elegant way to define a topology on Poset?

Metta World Peace
  • 6,683

2 Answers2

18

Many topologies have been defined on partial orders and lattices of various types. One of the most important is the Scott topology. Let $\langle P,\preceq\rangle$ be a partial order. A set $A\subseteq P$ is an upper set if ${\uparrow\!\!x}\subseteq A$ whenever $x\in A$, where ${\uparrow\!\!x}=\{y\in P:x\preceq y\}$. A set $U\subseteq P$ is open in the Scott topology iff $U$ is an upper set with the following property:

if $D\subseteq P$ is a directed set in $P$, and $\bigvee D\in U$, then $D\cap U\ne\varnothing$. (In this case $U$ is said to be inaccessible by directed joins.)

The upper topology is the topology that has $\{P\,\setminus\!\downarrow\!\!x:x\in P\}$ as a subbase. The lower topology is generated by the subbase $\{P\setminus{\uparrow\!\!x}:x\in P\}$.

The Lawson topology is the join (coarsest common refinement) of the Scott and lower topologies.

A number of interval topologies have also been defined, the first ones by Frink and by Birkhoff; this paper deals with a number of such topologies.

These terms should at least give you a start for further search if you’re interested.

Brian M. Scott
  • 631,399
  • 5
    I can't resist noticing the irony that "...one of the most important is the Scott topology, Brian M. Scott! :-) – amWhy Jan 02 '13 at 17:03
  • 3
    @amWhy: Yes, that struck me, too. But I assure you that no one will ever confuse my mathematical achievements with Dana Scott’s! – Brian M. Scott Jan 02 '13 at 17:05
  • @BrianM.Scott Are you sure your definition of upper topology listed above is correct? It doesn't match the definition that appears in the page you link to, and it's not symmetrical with your definition of lower topology. It also seems like it might not be equivalent (I could only prove one direction of the equivalence). – cjerdonek Sep 06 '21 at 17:11
  • 1
    @cjerdonek: You’re absolutely right; I have no idea why I wrote that (or how it’s gone uncorrected for so long!). Fixed now; thanks! – Brian M. Scott Sep 07 '21 at 19:47
  • @BrianM.Scott Thanks, Brian! By the way, do you know if the topology you had listed as the previous definition has a name, and if not, why it might not? – cjerdonek Sep 07 '21 at 20:32
  • @cjerdonek: You’re welcome! It’s the Alexandrov topology, which consists of all of the upper sets. – Brian M. Scott Sep 08 '21 at 18:43
  • This is very interesting! What happens if you have both the upper and lower sets as the subbase? That would seem to be what the regular order topology has. Is this, I guess, guaranteed to be the Lawson topology? – Mike Battaglia Apr 08 '24 at 00:52
  • @MikeBattaglia: $\uparrow!!x,\cap\downarrow!!x={x}$, so you get the discrete topology. – Brian M. Scott Apr 08 '24 at 03:08
  • Oops! I meant, I guess, the "open" upper/lower sets - the sets ${y \in P: x \prec y}$ and ${y \in P: y \prec x}$. Basically just the usual order topology, with open sets generated by "open rays" at each point. – Mike Battaglia Apr 08 '24 at 17:11
  • @MikeBattaglia: Take $P$ to be $\Bbb R\times\Bbb R$ with the natural product partial order $\preceq$. For $p\in P$ let $\uparrow!!p={q\in P:p\prec q}$ and $\downarrow!!p={q\in P:q\prec p}$. Then $\uparrow!!\langle a,b\rangle\cap\downarrow!!\langle c,b\rangle=(a,c)\times{b}$ if $a,b,c\in\Bbb R$ with $a<c$, and open vertical segments can be obtained similarly, so your idea generates the discrete topology on $P$. The upper and lower topologies on $P$ are refined by the Euclidean topology, so the Lawson topology is as well and thus cannot be discrete. – Brian M. Scott Apr 09 '24 at 02:38
  • Yes, for that particular partial order you do get the discrete topology. However, if you use a related "strict" version of that partial order, I think you get the Euclidean topology. This is the one where we start with $(a, b) < (c, d)$ iff $a < c$ and $b < d$, and thus we have $(a, b) ≤ (c, d)$ iff $(a, b) < (c, d)$ or $(a, b) = (c, d)$. The difference is that we have $(0, 0) ≤ (0, 1)$ in the product order, but the two are incomparable in this order. So, upper and lower sets are "open quadrants" in the plane. – Mike Battaglia Apr 09 '24 at 07:34
  • @MikeBattaglia: Let $U$ be the union of the open first, second, and fourth quadrants minus the origin; then $U$ is open in the upper topology and hence in the Lawson topology but not in the Euclidean topology, so your interval topology can’t be the Lawson topology. My example already showed that what you suggested in your second comment doesn’t work in general to yield the Lawson topology, and it appears not to work for this modified partial order, either. – Brian M. Scott Apr 10 '24 at 05:08
  • Yes, I see. Well, the basic idea is just to use the usual order topology, but instead of "open rays" to instead use "open lower and upper sets." This seems like a good idea and with the strict product order appears to give the Euclidean topology on $\Bbb R^2$. Is it any of the topologies you listed? It seems like a very natural idea and so I would guess it's something that has a name. – Mike Battaglia Apr 10 '24 at 09:18
  • @MikeBattaglia: (I was hoping to get here first: those should have been closed quadrants in the definition of $U$.) I don’t think that it’s any of the ones that I listed in the original answer. I also note that its definition depends on the fact that you’re dealing with a product; it’s not clear that the idea would generalize to arbitrary partial orders. – Brian M. Scott Apr 10 '24 at 16:46
  • You can look at the topology generated by open lower and upper sets for any partial order - it's just the usual order topology with partial orders. Why would it depend on a product? It's true that the usual product order gives the discrete topology for $\Bbb R^2$ but there are probably many other partial orders which aren't products that give interesting results with this topology. – Mike Battaglia Apr 10 '24 at 20:06
  • @MikeBattaglia: I think that you may have lost sight of your original question, which was whether taking the upper and lower sets as subbase would produce the Lawson topology. I showed that it wouldn’t necessarily, so you changed the partial order to get slightly different upper and lower sets. I noted that this didn’t give you the Lawson topology for the new partial order and further noted that the change that you made depended on the product structure of $\Bbb R\times\Bbb R$, so that even if it had worked in this case, it wasn’t clear that the idea could be extended to other partial orders. – Brian M. Scott Apr 10 '24 at 21:37
  • I see the confusion. I'll rephrase it like this: suppose that you take the definition of the order topology and just use it for posets, with a subbase the open upper and open lower sets. This seems like a useful topology in its own right. Does it have a name? It gives sensible results for at least some partial orders, and is "backward-compatible" with the order topology if your poset is a linear order. Is that even true for any of the others listed above? – Mike Battaglia Apr 10 '24 at 23:11
  • @MikeBattaglia: How are you going to define open upper (lower, resp.) sets in general? With the usual definition of upper and lower sets you always just get the discrete topology. – Brian M. Scott Apr 11 '24 at 02:13
  • You just look at ${y \in P: x \prec y}$ and ${y \in P: y \prec x}$ for any $x$. How does this always give you the discrete topology? I thought I'd given an example of a partial order above where it didn't ($\Bbb R^2$ with the modified product order). – Mike Battaglia Apr 11 '24 at 02:26
  • @MikeBattaglia: The usual definition of upper and lower sets, which is what I was talking about there, includes the sets ${y\in P:x\preceq y}$ and ${y\n P:y\preceq x}$, and their intersection is ${x}$. How are you going to define open upper and lower sets so as to exclude these? Remember, the defining characteristic of an upper set is simply that it’s upward closed; it can be much more complicated than ${y\in P:x\prec y}$. How do you decide which ones are open? – Brian M. Scott Apr 11 '24 at 04:50
  • "How are you going to define open upper and lower sets so as to exclude these?" -You take as subbase for the topology the sets ${y \in P: x \prec y}$ and ${y \in P: y \prec x}$ for all $x$, in exactly the same way that you do with the order topology. The set ${y \in P: y \preceq x}$ is not one of these sets in general. – Mike Battaglia Apr 11 '24 at 08:24
  • @MikeBattaglia: In other words, you’re not actually defining open upper and lower sets in any real generality: you’re simply taking as a subbase for a topology the union of a very limited subset of the family of upper sets and a similarly limited subset of the family of lower sets. And in the case of the usual product partial order on $\Bbb R\times\Bbb R$ you get the discrete topology, as in my comment of 9 April at 2:38. It’s been years since I thought about such things, but at the moment I don’t see any nice general statement to be made about the topologies produced in this way. – Brian M. Scott Apr 13 '24 at 08:05
  • @BrianM.Scott maybe here's a better question. The thing that I'm talking about is "backwards-compatible" with the order topology in that if your poset is also a linearly ordered set, the result is the order topology. Do any of these have that property? – Mike Battaglia Apr 21 '24 at 03:29
4

You could use the following topology: a set $V$ is closed if and only if for all $x\in V$ and $y\geq x$, $y\in V$. (Check for yourself that this gives a well-defined topology!)

This is very different from the order topology, though, and what follows does not generalize to the order topology when your poset is an ordered set.

Unless your order relation is trivial, in which case our topology will be discrete (check this!), the topology will not be Hausdorff, so the most you can ask for is quasi-compactness (every open cover has a finite subcover).

Then a "half-open" interval $\{x: x\leq b\}$ will be quasi-compact since any open set containing $b$ contains the whole set. The "closed" interval $\{x: a\leq x$, $x\leq b\}$ (which is not generally closed in this topology) will be quasi-compact, since it is relatively closed in the half-open interval. You'll notice that no completeness axiom was necessary for this argument.

Brett Frankel
  • 10,085