It is purely a property of compactness. Take the product of compact spaces. The condition of an element of the product being in the limit is the conjunction of infinitely many conditions each of which says that two coordinates agree and if it is a chain (or, in fact, an inverse filtered diagram) of non-empty compact sets, each of those finitely many conditions defines a closed non-empty subset with the finite intersection property. So the general context is that an inverse filtered limit of non-empty compact Hausdorff spaces is non-empty. Although I don't seem to have used Hausdorffness, it is required to show those sets are closed. And without it, you could get the example Tom described using the trivial topology, which is certainly compact. I guess you could use the finite complement topology to get a T1 counterexample. Michael On Fri, 30 Jul 2004, Tom Leinster wrote:
I've recently come across the following curious little result. I know how to prove it and have a use for it, but my question is: can anyone supply a wider context or explanation?
The result is that the limit in Set of any diagram
... ---> S_3 ---> S_2 ---> S_1
of finite nonempty sets is nonempty. Note that finiteness cannot be dropped: for instance, take each S_n to be the natural numbers and each map to be addition of 1.
Thanks, Tom