> **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:

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

**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:

$$
\begin{align}
\mathcal{X}(\hat{a},\hat{b}) + \mathcal{X}(\hat{b},\hat{a}) & \leq \mathcal{X}(\hat{a},\hat{a}) \\\\
\implies \mathcal{X}(\hat{a},\hat{b}) & \leq \mathcal{X}(\hat{a},\hat{a})
\end{align}
$$

However, then we have

$$
\begin{align}
\mathcal{X}(\hat{a},\hat{b}) + \mathcal{X}(\hat{a},\hat{a}) & \leq \mathcal{X}(\hat{a},\hat{b}) \\\\
\implies 2 \mathcal{X}(\hat{a},\hat{b}) & \leq \mathcal{X}(\hat{a},\hat{b})
\end{align}
$$

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:

$$
\mathcal{X}(a,\hat{a}) + \mathcal{X}(\hat{a},\hat{b}) \leq \mathcal{X}(a,\hat{b})
$$

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

$$
\mathcal{X}(a,\hat{b}) + \mathcal{X}(\hat{b},b) \leq \mathcal{X}(a,b)
$$

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

[**Edit**: Updated following Christopher's suggestion. Thanks Chris!]