> 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\$$.