[The original version of the following proof can be found here](https://forum.azimuthproject.org/discussion/comment/18540/#Comment_18540)

> **Puzzle 90.** What's a \\(\mathbf{Cost}^{\text{op}}\\)-category, and what if anything are they good for?

The story doesn't look very good for them :(

**Theorem.** If \\(\mathcal{X}\\) is a \\(\mathbf{Cost}^{\text{op}}\\)-enriched category, then:

\[ \tag{a} \forall a,b. \mathcal{X}(a,b) = 0 \text{ or } \mathcal{X}(a,b) = \infty \]


If every element \\(\mathcal{X}(a,b) = 0\\), we are done.

Next assume to the contrary. We must show that for an arbitrary \\(a\\) and \\(b\\) that \\(\mathcal{X}(a,b) = \infty\\).

So observe there must be some \\(\hat{a}\\) and \\(\hat{b}\\) such that \\(\mathcal{X}(\hat{a},\hat{b}) > 0\\).

It must be \\(\mathcal{X}(\hat{a},\hat{b}) = \infty\\). To see this, we know from the laws of enriched categories (part (b)) that:

\[ \tag{b}
\mathcal{X}(\hat{a},\hat{b}) + \mathcal{X}(\hat{b},\hat{a}) & \leq \mathcal{X}(\hat{a},\hat{a}) \\\\
\implies \mathcal{X}(\hat{a},\hat{b}) & \leq \mathcal{X}(\hat{a},\hat{a})

However, then we have

\[ \tag{c}
\mathcal{X}(\hat{a},\hat{b}) + \mathcal{X}(\hat{a},\hat{a}) & \leq \mathcal{X}(\hat{a},\hat{b}) \\\\
\implies 2 \mathcal{X}(\hat{a},\hat{b}) & \leq \mathcal{X}(\hat{a},\hat{b})

This can only happen if \\(\mathcal{X}(\hat{a},\hat{b}) = \infty\\) or \\(\mathcal{X}(\hat{a},\hat{b}) = 0\\). But we know \\(\mathcal{X}(\hat{a},\hat{b}) > 0\\) so it must be \\(\mathcal{X}(\hat{a},\hat{b}) = \infty\\) .

Next observe from the enriched category theory law (b) that:

\[ \tag{d}
\mathcal{X}(a,\hat{a}) + \mathcal{X}(\hat{a},\hat{b}) \leq \mathcal{X}(a,\hat{b})

So it must be that \\(\mathcal{X}(a,\hat{b}) = \infty\\). But then

\[ \tag{e}
\mathcal{X}(a,\hat{b}) + \mathcal{X}(\hat{b},b) \leq \mathcal{X}(a,b)

Hence \\(\mathcal{X}(a,b) = \infty\\). \\(\qquad \square \\)

[**Edit**: Updated following [Christopher's suggestion](https://forum.azimuthproject.org/discussion/comment/18543/#Comment_18543). Thanks Chris!]