Which simplicial sets are filtered colimits of standard simplices? The question is all in the title : every simplicial set is a colimit of the standard simplices $\Delta^n$, but I'm wondering which ones are filtered or directed colimits of these, if there's a nice characterization of these simplicial sets.
The motivation is the following : we can see by hand that $|\Delta^p\times \Delta^q|\simeq |\Delta^p|\times|\Delta^q$, and since geometric relization commutes with colimits as a left adjoint, if $X,Y$ are filtered colimits of standard simplices, since filtered colimits commute with products in $\mathbf{Set, Top}$, we get $|X\times Y|\simeq |X|\times |Y|$, but here the $\times$ would be the usual product topology, not the compactly generated product topology.
If the subcategory of filtered colimits of standard simplices is big enough, this would give an easy proof of this last result for a large chunk of simplicial sets and so would be interesting. 
(As I'm writing this I'm realizing that I'm not sure that filtered colimits commute with products in $\mathbf{Top}$; is it the case ? )
 A: Unfortunately, not a lot of them. The closure of $\Delta\subset Fun(\Delta^{op},\mathbf{Set})$ under filtered colimits is called the $\mathrm{Ind}$-completion of $\Delta$, and it turns out we can quite easily compute it to be the category of linearly ordered posets; the comparison functor simply being the nerve.
Indeed, let $\mathbf{LPos}$ be that category, then the category of filtered colimit-preserving functors $\mathrm{Ind}(\Delta)\to \mathbf{LPos}$ is equivalent to the category of functors $\Delta\to \mathbf{LPos}$. Thus the inclusion induces a filtered-colimit preserving functor $\mathrm{Ind}(\Delta)\to \mathbf{LPos}$.
Moreover, each $[n]\in\Delta$ is clearly compact in $\mathbf{LPos}$, in the sense that $\hom([n],-)$ commutes with filtered colimits, from which it follows that $\mathrm{Ind}(\Delta)\to \mathbf{LPos}$ is fully faithful.
Finally, any $P\in\mathbf{LPos}$ is the filtered colimit of its finite subposets, and therefore our functor is essentially surjective. It follows that it's an equivalence, and the inverse functor can easily be seen to be the nerve functor.
