Seven Sketches in Compositionality - Exercise 2.73.1

\section*{Problem}

Show that a skeletal dagger Cost-category is an extended metric space.

\section*{Solution}

We already know that a Cost-category is a Lawvere metric space. We only need to prove that the distance function $d$ of such metric space is symmetric and that $d(x,y) = 0$ implies $x = y$.

The first condition is proved by appealing to the dagger condition. Since $Id: \mathcal{X} \to \mathcal{X}^{op}$ is a Cost-functor, $\mathcal{X} (x,y) \geq \mathcal{X}^{op}(x,y) = \mathcal{X} (y,x)$. Dually, $\mathcal{X} (y,x) \geq \mathcal{X} (x,y)$. Thus, since the ordering on the extended real is a total order, we can deduce that $\mathcal{X} (y,x) = \mathcal{X} (x,y)$. This is true for all objects $x$ and $y$ in $\text{Ob}(\mathcal{X})$.

The second condition is proved by appealing to skeletality. Suppose $d(x,y) = 0$. Symmetry tells us that $d(y,x) =0$. Then, $0 \geq d(x,y)$ and $0 \geq d(y,x)$. Then, $x = y$ by the skeletality condition.

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