Replacing a projective system with one indexed by an ordinal Consider some complete concrete category where the underlying set of an inverse limit is the inverse limit of underlying sets. Very often, one has an inverse limit $\varprojlim_{n\in\mathbb{N}}(X_n,p_{n+1,n})$ of a projective system indexed by the natural numbers. In categories where the objects are highly structured, one can often start with an element $a_1\in X_1$ and inductively construct a sequence $a_n\in X_n$ coherent with the bonding maps $p_{n+1,n}$ so that $(a_n)_{n\in\mathbb{N}}$ is an element of the inverse limit. Of course, the ability to do this uses the fact that $\mathbb{N}$ is well-ordered.
I'm interested to know how far this kind of approach can be applied to limits indexed by other infinite directed sets. 
Question: If one has an inverse limit $\varprojlim_{j\in J}(X_j,p_{j,k})$ indexed by some infinite directed set $J$, when is it possible to replace the inverse system $(X_j,p_{j,k})$ with another inverse system indexed by an ordinal $\kappa$ so that the new limit $\varprojlim_{\alpha\in\kappa}(X_\alpha,p_{\alpha,\beta})$ is isomorphic to the original limit? 
I suppose it would be enough to find an order-preserving cofinal map $\kappa\to J$ from some ordinal $\kappa$. However, this may be a lot to ask for. The existence of non-pseudoradial topological spaces makes me doubtful that such a replacement is always possible.
 A: You cannot do this if you want each of the objects and maps in your new system to be objects and maps of the old system.  For instance, if start with any inverse system of finite sets in which all the maps are surjections, this would imply that the ordinal-indexed system is essentially countable (i.e., if you discard maps which are isomorphisms, there are only countably many terms, so that there is a countable subsequence with the same limit), since the cardinality of $X_\alpha$ can increase only countably many times.  That would in particular imply the cardinality of the inverse limit is at most $2^{\aleph_0}$, since it is a countable inverse limit of finite sets.  But every infinite set can be written as an inverse limit of finite sets and surjections (see https://mathoverflow.net/questions/172707/maximum-cardinality-of-a-filtered-limit-of-finite-sets).
If you don't mind modifying the objects, then there are some constructions you can make.  For instance, let us suppose $J$ has finite joins (if it doesn't, you can formally adjoin them, and extend the system by mapping the formal join of $F\subseteq J$ to the limit of the diagram of elements that are below some element of $F$; alternatively, instead of using joins we can just arbitrarily pick some upper bounds in the construction below, with a bit of extra work).  Fix an enumeration $(j_\alpha)_{\alpha<\kappa}$ of $J$ indexed by some ordinal $\kappa$.  For each $\alpha<\kappa$, let $X_\alpha$ be the limit of the diagram formed by the objects $X_s$ where $s$ ranges over the set of finite joins of elements $j_\beta$ for $\beta<\alpha$ (when $\alpha$ is finite, this just means $X_{\alpha}=X_s$ where $s$ is the join of $j_0,\dots,j_{\alpha-1}$).  Then the $X_\alpha$ naturally form an inverse system whose inverse limit is the same as the inverse limit of the original system.
Note in particular that if we pick $\kappa$ to be a cardinal, then each $X_\alpha$ is the inverse limit of an inverse system of smaller cardinality.  So by induction on $\kappa$, this iteratively constructs our inverse limit out of ordinal-indexed inverse limits (a countable inverse limit is an ordinal-indexed limit built from the original diagram, a size $\aleph_1$ inverse limit is an ordinal-indexed limit where each term is a countable inverse limit built from the original diagram, etc).
This construction and variants on it are useful for some purposes; for instance, it can be used to prove that if a functor preserves ordinal-indexed limits (and those limits always exist) then it preserves all inverse limits.
A: A potentially useful fact: in N. Howes, A note on transfinite sequences on page 221 we find a lemma:

Let $(P,<)$ be a partial order and let $\prec$ be any well-order on $P$. Then there is a subset $S$ of $P$, cofinal w.r.t. $<$, such that $\prec$ and $<$ are compatible on $S$: $$\forall a,b \in S: a < b \implies a \prec b$$

I think this would give you the cofinal subset indexed by an ordinal (all well-orders are order isomorphic to ordinals).  
