[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:

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

**Proof**.

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} \begin{align} \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}) \end{align}

However, then we have

\tag{c} \begin{align} \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}) \end{align}

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!]