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

Tom, The classical context I know is that in the category of compact Hausdorff spaces and continuous maps, limits exist, and are non-empty if each component space is non-empty. This uses the Axiom of Choice (and the Tychonoff Theorem) to show that the product of the component spaces $S_i$ is non-empty and compact. The generalization just embeds Set in Top by regarding sets as discrete spaces, so all sets are locally compact, and the compact sets are exactly the finite ones. The limit can be defined as the family $ L = \{ (s_i) \mid p_ji(s_j) = s_i, for i \leq j \in I\}. $ This limit is then the filtered intersection of the sets $T_F = \prod_{i \not\in F} S_i \times \{(s_i)_{i \in F} \mid p_{ji}(s_j) = s_i, i \leq j \in F\},$ where $F\subseteq I$ is finite. The Axiom of Choice implies $F, \prod_{i \not\in F} S_i$ is non-empty, and it's easy to show that $\{(s_i)_{i \in F} \mid p_{ji}(s_j) = s_i, i \leq j \in F\}$ is non-empty since $F$ is finite, so $T_F$ is non-empty. Since the bonding maps $p_{ji} \colon S_j \to S_i$ are continuous, the set $T_F$ is closed, so it's a non-empty closed subset of the compact Hausdorff space $\prod_i S_i$. Since the family is filtered, its intersection is non-empty. This generalization also suggests why the sets have to be finite - limits of non-empty locally compact spaces and continuous maps can be empty, because the limit object may be empty. Indeed, the same construction as given above defines the limit object L, but L may be empty because the filtered intersection of a family of closed, non-empty subsets of a locally compact space may be empty. Best regards, Mike On Jul 29, 2004, at 7:37 PM, 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