The category $\mathsf{Stone}$ has cofiltered limits

Question

Is there an easy way to prove that the category $\mathsf{Stone}$ of Stone spaces and continuous functions has cofiltered limits? Maybe using the fact that the forgetful functor $\mathsf{Stone} \to \mathsf{Set}$ has an adjoint?

Answer 1

The category $\mathsf{Stone}$ is equivalent to the opposite category $\mathsf{Bool}^{op}$ of the category of Boolean algebras. Since $\mathsf{Bool}$ is the category of models of an algebraic theory, it has all limits and colimits, so its opposite category also has all limits and colimits.

Alternatively, $\mathsf{Stone}$ is a reflective subcategory of $\mathsf{Top}$, the reflector taking a space $X$ to the closure of the image of the natural map $X\to \{0,1\}^S$ where $S$ is the set of all continuous maps $X\to\{0,1\}$. So limits in $\mathsf{Stone}$ are the same as limits in $\mathsf{Top}$ and colimits in $\mathsf{Stone}$ can be computed by taking the colimit in $\mathsf{Top}$ and then applying the reflector.

Answer 2

A third approach, more precisely connected to cofiltered limits, is that the category of Stone spaces is actually equivalent to $\mathrm{Pro}(\mathrm{FinSet})$, the category of pro-objects (that is, cofiltered diagrams) in the category of finite sets. The most straightforward way to see this is via Stone duality: Boolean algebras are models of a finitary theory, so they form the category of ind-objects (filtered diagrams) in their finite objects. But the finite Boolean algebras are just the powersets of finite sets, so dualizing again, the result follows. Now, any pro-completion certainly has cofiltered limits, just by turning a cofiltered diagram in cofiltered diagrams into a single cofiltered diagram indexed by the product category. So from this angle, Stone spaces have cofiltered limits for very general reasons; the other limits come because $\mathrm{FinSet}$ happens to be finitely complete, which also implies that colimits exist by the general theory of locally co-presentable categories.