Christopher wrote:

> The \\( \mathbf{Bool} \times \mathbf{Bool} \\) interpretation seems to extend nicely to let one talk about \\(S\\) indexed preorders as categories enriched in the space of (monotonic?) functions from \\(S\\) to \\(\mathbf{Bool}\\).

Yes! You didn't say what \\(S\\) is, but if it's just a set we don't need "monotonic" functions from \\(S\\) to \\(\mathbf{Bool}\\); we can just look at _all_ functions from \\(S\\) to \\(\mathbf{Bool}\\). This is called \\(\mathbf{Bool}^S\\), because you can also think of it as a product of copies of \\(\mathbf{Bool}\\), one for each element of \\(S\\):

\[ \prod_{s \in S} \mathbf{Bool} . \]

\\(\mathbf{Bool}^S\\) becomes a symmetric monoidal poset in an obvious way, and an \\(\mathbf{Bool}^S\\)-category is the same as a set \\(X\\) with a bunch of different preorders, one for each element of \\(S\\).

The "obvious way": given \\(f, g : S \to \mathbf{Bool}\\), we say \\(f \le g\\) iff

\[ f(s) \le g(s) \textrm{ for all } s \in S .\]

This makes \\(\mathbf{Bool}^S\\) into a poset, and then we make it into a symmetric monoidal poset using \\( \wedge \\).

But in fact, the poset \\(\mathbf{Bool}^S\\) is isomorphic to \\(P(S)\\), the power set of \\(S\\), made into a poset using \\(\subseteq\\). It's easiest to understand this by looking at the case where \\(S\\) has 3 elements. Then \\(P(S)\\) is a cube:



and you can see it's isomorphic to \\(\mathbf{Bool}^S \cong \mathbf{Bool} \times \mathbf{Bool} \times \mathbf{Bool} \\). (Sorry, this picture calls \\(S\\) "\\(X\\)".)

It's also interesting to think about what happens when \\(S\\) is a _preorder_, and we look at _monotone_ maps \\(f : S \to \mathbf{Bool}\\).