Does the category of pro-sets have a generator?

Consider for simplicity the category of pointed sets $Set_*$. Let's call an object $G$ of a pointed category $C$ a generator if for every nonzero object $X \in C$ there is a nonzero map $G \to X$. In the category of pointed sets, for example, the two-point set is a generator (and a cogenerator as well).

Does the category $Pro-Set_*$ of cofiltered diagrams of pointed sets have a generator as well? If $X$ is a pro-pointed set with nontrivial limit then there is a nontrivial map $G \to X$, where $G$ is a generator of pointed sets, considered as a one-object diagram. However, there are many pro-sets with trivial limit. For example, the pro-set $X\colon \mathbb N \to Set$ given by $X(i) = \mathbb N$ and $X(i+1 \to i)(n) = n+1$ is a pro-set with empty limit, but which is nontrivial -- add a point in every degree to get an example in pointed sets.

Thus, my question is: Does the category of pro-pointed sets have a generator?

And, if the answer was yes, does more generally $Pro-C$ have a generator whenever $C$ has one?

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No, I think. Look at your $X$ a little differently: $X(i)$ is the set of all $n>i$ and the maps $X(j)\to X(i)$ are inclusions. You can make something similar for other directed sets than $mathbb N$, in particular for other ordinals. It seems to me that, just as $X$ with basepoint added has no nontrivial map from a constant diagram, a similar thing based on an uncountable ordinal with have no nontrivial map from anything with countable index set. And so on. –  Tom Goodwillie Jul 28 '10 at 16:32