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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
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2 Answers2

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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
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    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
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    @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
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    @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
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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
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