> So maybe for cost it would be sum or maximum of the distance from F(x) to G(x) is the cost from F to G?

I think you right! In the case I gave, we have:

\[
\begin{align}
\forall x. f(x) \leq_{\mathcal{X}} g(x) & \iff \forall x\in \mathrm{Ob}(\mathcal{X}). \mathcal{X}(f(x), g(x)) \\\\
& \iff \bigwedge_{x\in \mathrm{Ob}(\mathcal{X})} \mathcal{Z}(f(x), g(x))
\end{align}
\]

Hence, it looks like \\(\mathcal{X}^\mathcal{Y}(f,g) = \bigwedge_{x\in \mathrm{Ob}(\mathcal{X})} \mathcal{X}(f(x), g(x))\\).

This is exactly the max distance between \\(f\\) and \\(g\\) in \\(\mathbf{Cost}\\).

So we just need the category to have arbitrary meets if we want to define \\(\mathcal{X}^\mathcal{Y}\\). For big-boy enriched categories, using monoidal categories rather than monoidal preorders for \\(\mathcal{V}\\), then I suspect if they are *complete* there's an analogue to the construction above.

We can always find a terminal \\(\mathcal{V}\\)-enriched category for any monoidal preorder \\(\mathcal{V}\\). Let \\(\mathrm{Ob}(\top) = \mathrm{Ob}(\mathcal{V})\\) an \\(\top(x,y) = I\\), where \\(I\\) is the identity element of \\(\mathcal{V}\\). We can make a functor \\(T\_{\mathcal{X}} : \mathcal{X} \to \mathbf{Top}\\) from any \\(\mathcal{V}\\)-category \\(\mathcal{X}\\) to \\(\top\\) by using \\(x \mapsto I\\).