'compactness' of a category with respect to a family of functors I was wondering if the following idea is a notion that has been studied. I thought of it while thinking of functorial invariants (such as (co)homology functors) and wondering what kind of category has a 'compactness condition' for a collection of functors that are able to distinguish non-isomorphic elements in a category. I tried searching for some likely terminology with no luck.

Definition: Let $\mathcal{C}$ be a category, and $\mathfrak{F} = \{ F_i \}_{i \in I}$ be a family of functors $F_i: \mathcal{C} \to \mathcal{D}_i$ for all $i \in I$ for some index set $I$ (where $I$ may be infinite). Then we say that $\mathfrak{F}$ distinguishes objects of $\mathcal{C}$ if for all $a, b \in \mathcal{C}$, with $a$ not isomorphic to $b$, there exists $F_j \in \mathcal{F}$ such that $F_j(a) \neq F_j(b)$.

Definition: $\mathcal{C}$ is functorially compact if for all families of functors $\mathfrak{F}$ that distinguish objects of $\mathcal{C}$ there exists a finite subset $\mathfrak{G} \subseteq \mathfrak{F}$ that distinguishes objects of $\mathcal{C}$.

Question: What sort of category can be 'functorially compact'?

Thoughts: The identity functor always distinguishes objects of $\mathcal{C}$, so we know that every category has at least one family of functors that distinguishes objects in it. I believe categories with a finite number of objects would always be functorially compact as well. I imagine that this is a rather strict property that wouldn't allow sophisticated categories, but I'd like to be proven otherwise. But what about categories with an infinite number of objects (small or large)?

Note: These definitions may not be perfect - maybe some restriction must be put on what kind of functor is allowed to make things interesting, but I don't know enough about category theory to know where to begin with that.

Edit: I made corrections as per Qiaochu's comments below.
I think that a category is "functorially compact" iff it is finite.

It is easy to see that finite $\Rightarrow$ functorially compact: for any family of functors $\mathfrak{F}$, we construct a finite sub-family by picking one distinguishing functor from $\mathfrak{F}$ for each pair of objects in $\mathcal{C}$.

To show that functorially compact $\Rightarrow$ finite, consider the family $\mathfrak{F} = \{F_A\}_{A \in Obj(\mathcal{C})}$ indexed by the objects of $\mathcal{C}$, where each $F_A : \mathcal{C} \rightarrow \mathbf{2}$.  We take $\mathbf{2}$ to be the category consisting of two objects $X$ and $Y$, and two (non-identity) morphisms $f : X \rightarrow Y$ and $g : Y \rightarrow X$.  We define $F_A$ to be the functor that sends $A$ to $X$ and every other object to $Y$ (this map uniquely determines where the functor maps the morphisms of $\mathcal{C}$).  $\mathfrak{F}$ distinguishes objects of $\mathcal{C}$, but there can be no finite subfamily that does the same, unless $\mathfrak{F}$ was finite to begin with (if there exist objects $A$ and $B$ such that neither $F_A$ nor $F_B$ belong to this subfamily, then $A$ and $B$ are not distinguished).
I think that a category is "functorially compact" iff it is finite.

It is easy to see that finite $\Rightarrow$ functorially compact: for any family of functors $\mathfrak{F}$, we construct a finite sub-family by picking one distinguishing functor from $\mathfrak{F}$ for each pair of objects in $\mathcal{C}$.

To show that functorially compact $\Rightarrow$ finite, consider the family $\mathfrak{F} = \{F_A\}_{A \in Obj(\mathcal{C})}$ indexed by the objects of $\mathcal{C}$, where each $F_A : \mathcal{C} \rightarrow \mathbf{2}$.  We take $\mathbf{2}$ to be the category consisting of two objects $X$ and $Y$, and two (non-identity) morphisms $f : X \rightarrow Y$ and $g : Y \rightarrow X$.  We define $F_A$ to be the functor that sends $A$ to $X$ and every other object to $Y$ (this map uniquely determines where the functor maps the morphisms of $\mathcal{C}$).  $\mathfrak{F}$ distinguishes objects of $\mathcal{C}$, but there can be no finite subfamily that does the same, unless $\mathfrak{F}$ was finite to begin with (if there exist objects $A$ and $B$ such that neither $F_A$ nor $F_B$ belong to this subfamily, then $A$ and $B$ are not distinguished).