I was wondering if there is a canonical topology induced by a partial order on a set and how that relates to the total ordering topology (if it can be extended to a total ordering).

I thought maybe the basis would be defined as in total orderings, but this wouldn't include elements that are incomparable to everything. I noticed another question mentions the space generated by upsets and downsets; would this be the natural topology induced? Or are others used?

((I haven't been able to find much by googling, and often the pages are about all topologies on a set being partially ordered))

Thanks in advance!

  • $\begingroup$ The principle that every partial order can be extended to a total order, called the order-extension principle, is a "choice principle" that holds in ZFC but not in ZF. It lies in strength strictly between the ultrafilter lemma and the ordering principle (which is itself strictly stronger than the axiom of choice for sets of finite sets). $\endgroup$ – dfeuer Jun 17 '13 at 16:18
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    $\begingroup$ Check out en.wikipedia.org/wiki/Alexandrov_topology but note that that topology is very different from the order topology for total orders. $\endgroup$ – dfeuer Jun 17 '13 at 16:21
  • $\begingroup$ Why is it very different? In a total ordering the upper sets are rays, so the order topology must be at least finer than the Alexandrov topology? Also, is this alexandrov topology the usual topology one would induce through a partial ordering? $\endgroup$ – user68193 Jun 17 '13 at 18:52
  • $\begingroup$ No. There are upper sets that are not open in the order topology, and there are open rays that are not open in the Alexandroff topology. Can you find simple examples? $\endgroup$ – dfeuer Jun 17 '13 at 19:00
  • $\begingroup$ I guess I forgot the inequality for upper sets isn't strict, so they can be closed rays. And I suppose similarly an open ray will not include it's left-boundary while an upper set will (at least if that element has an immediate successor?). $\endgroup$ – user68193 Jun 17 '13 at 20:00

There is no canonical topology defined on a partial order, but several topologies on them have been defined and studied; in this answer I mention several of them with links to further information.

  • $\begingroup$ Cool; thanks Brian! $\endgroup$ – user68193 Jun 20 '13 at 12:36
  • $\begingroup$ @Bleys: You’re welcome! $\endgroup$ – Brian M. Scott Jun 20 '13 at 19:51

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