Re: Limit of finite sets

[Michael Mislove <mwm@math.tulane.edu>]

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
>
>