Lecture 30  Preorders as Enriched Categories
Okay, last time I dumped the definition of "enriched category" on you. Now let's figure out what it does! To get going we need to pick a monoidal preorder \(\mathcal{V}\). Then:
Definition. A \(\mathcal{V}\)enriched category, say \(\mathcal{X}\), consists of:

a set of objects, \(\mathrm{Ob}(\mathcal{X})\), and

an element \(\mathcal{X}(x,y) \in \mathcal{V}\) for any pair of objects \(x,y\).
such that:
a) \( I\leq\mathcal{X}(x,x) \) for any object \(x\), and
b) \( \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \) for any objects \(x,y,z\).
We often say "\(\mathcal{V}\)category" instead of \(\mathcal{V}\)enriched category, just to save time. Life is short.
To understand this definition, we need examples! The simplest interesting example is \(\mathcal{V} = \mathbf{Bool}\). This is the set of Booleans, or "truth values":
[ \mathbf{Bool} = \{ \texttt{true}, \texttt{false} \} . ]
It's a poset where \(x \le y\) means "\(x\) implies \(y\)". Namely, \( \texttt{false} \le \texttt{true} \) and everything is less than or equal to itself, and that's all. We make it monoidal by taking \(x \otimes y\) to mean "\(x\) and \(y\)", so the unit \(I\) is \(\texttt{true}\). Let's see what this gives.
Theorem. A \(\mathbf{Bool}\)category is a preorder.
Proof. Let's start with a \(\mathbf{Bool}\)category and see what it actually amounts to. A \(\mathbf{Bool}\)enriched category, say \(\mathcal{X}\), consists of:

a set of objects, \(\mathrm{Ob}(\mathcal{X})\) and

an element \(\mathcal{X}(x,y) \in \mathbf{Bool} \) for any two objects \(x,y\)
So, we get a truth value for each pair \(x,y\). Let's make up a relation \(\le\) on \( \mathrm{Ob}(\mathcal{X})\) such that \(x \le y\) is true when \( \mathcal{X}(x,y) = \texttt{true}\), and false when \( \mathcal{X}(x,y) = \texttt{false}\).
Next:
a) \( I\leq\mathcal{X}(x,x) \) for any object \(x\).
In our example this means
[ \texttt{true} \textrm{ implies that } x \le x . ]
In other words, \(x \le x\) is true. So this is just the reflexive law for \(\le\)!
b) \( \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \) for any objects \(x,y,z\).
In our example this means
[ x \le y \textrm{ and } y \le z \textrm{ implies } x \le z .]
So this is just the transitive law for \(\le\)!
It's easy to turn this argument around and get the converse: any preorder gives a \(\mathbf{Bool}\)category \(\mathcal{X}\) if we define \( \mathcal{X}(x,y) = \texttt{true}\) when \(x \le y\) and \( \mathcal{X}(x,y) = \texttt{false}\) otherwise. \( \qquad \blacksquare \)
Yeehah! We've managed to define preorders in a new, more abstract, more confusing way!
But of course that's not our goal. That was just a warmup. A much cooler example occurs when we take \(\mathcal{V} = \mathbf{Cost}\). This is the set of costs, or nonnegative real numbers including infinity:
$$ \mathbf{Cost} = [0,\infty] . $$
It's a poset where the order is defined in the opposite of the usual way: we use \(\ge\) instead of \(\le\). It becomes a monoidal using addition, with \(0\) as the unit. In case you're nervous about infinity, this means that \(x + \infty = \infty = \infty + x\) and \(x \le \infty\) for all \(x\).
Puzzle 88. Show that these choices actually makes \( \mathbf{Cost}\) into a symmetric monoidal poset.
Puzzle 89. Figure out exactly what a \(\mathbf{Cost}\)category is, starting with the definition above. Again, what matters is not your final answer so much as your process of deducing it!
Hint: given a \(\mathbf{Cost}\)category \(\mathcal{X}\), write \(\mathcal{X}(x,y)\) as \(d(x,y)\) and call it the distance between the objects \(x,y \in \mathrm{Ob}(\mathcal{X})\). The axioms of an enriched category then say interesting things about "distance".
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