I asked for an explanation of the following result:

the limit in Set of any diagram

... ---> S_3 ---> S_2 ---> S_1

of finite nonempty sets is nonempty

Thanks very much to all who replied. I'll summarize some of the points made to me in private replies: 1. This is called Koenig's Lemma, and is usually stated in the form "any finitely-branching infinite tree contains an infinite (positively oriented) path". 2. This also follows from a general result in topology by regarding each S_n as a discrete space. The general result is that any "suitably-shaped" limit of nonempty compact Hausdorff spaces is nonempty. For Bourbaki (General Topology), "suitably-shaped" means indexed by a directed poset. More generally still, it could be any componentwise cofiltered limit, i.e. any limit for which each connected-component of the indexing category I is cofiltered (or equivalently, every finite connected diagram in I admits a cone). The proof of the general topological result specializes to give a nice topological proof of Koenig. For each n, let V_n be the subset of the product \prod_n S_n consisting of those sequences whose first n terms are compatible; then \lim_n S_n is the intersection of the (V_n)s. But with the discrete topology on each S_n, Tychonoff says that \prod_n S_n is compact, and (V_n) is a nested sequence of nonempty closed subsets so has nonempty intersection. Tom