ind-completion and ordinals

Question

Given a category $C$, we have its ind-completion $Ind(C)$ whose objects are filtered diagrams in $C$.

Assuming the axiom of choice, is any object in $Ind(C)$ isomorphic, in $Ind(C)$, to an ordinal indexed filter?

Answer 1

No. For instance, let $C$ be the category of finite sets, and let $X$ be an uncountable set. The collection of all finite subsets of $X$ (and their inclusion maps) form a filtered diagram in $C$ and hence an object of $Ind(C)$. However, this object cannot be represented by an ordinal-indexed diagram. Indeed, suppose $(Y_\beta)_{\beta<\alpha}$ is a diagram in $C$ indexed by an ordinal $\alpha$, with maps $f_{\beta\gamma}:Y_\beta\to Y_\gamma$ when $\beta<\gamma$. Then the colimit $Y$ of this diagram in $Set$ must be countable: $Y$ is the union of the images of each $Y_\beta$ in $Y$, so it is the union of a chain of finite subsets. But isomorphic objects of $Ind(C)$ have isomorphic colimits in $Set$ (the colimit is a functor $Ind(C)\to Set$, in fact an equivalence of categories), and the colimit of our diagram of finite subsets of $X$ is just $X$ which is uncountable. Thus $(Y_\beta)$ cannot be isomorphic to the diagram of finite subsets of $X$.