[Michael Hong wrote](https://forum.azimuthproject.org/discussion/comment/18542/#Comment_18542):

> I think not knowing that we have to define another relation \\(\le\\) on \\( \mathrm{Ob}(\mathcal{X})\\) is what confused me the most when I first attempted to understand the past few lectures.

But we _don't_ have to define another relation on \\(\mathrm{Ob}(\mathcal{X})\\). Not in general, anyway. The theorem you're referencing is taking any given **Bool**-enriched category and constructing a preorder out of the information it contains. This is a theorem _about_ **Bool**-enriched categories, not a requirement of their definition. It wouldn't work this way for, say, **Cost**-categories. That's why we say that **Bool**-categories "are" preorders, and **Cost**-categories "are" Lawvere metric spaces -- each has a "straightforward" way to convert between the two concepts.

As John says in this lecture:

> We've managed to define preorders in a new, more abstract, more confusing way!

At present, \\(\mathcal{V}\\)-enriched categories are just sets \\(S\\) with an associated map \\(\mathcal{X} : S \times S \to \mathcal{V}\\) into a monoidal preorder. We require \\(\mathcal{X}\\) to satisfy some additional properties, so it can't be just any such map, but it _is_ a map. And for whatever reason, we tend to prioritize the map instead of the set in our notation, denoting \\(S\\) by \\(\mathrm{Ob}(\mathcal{X})\\).

When \\(\mathcal{V}\\) is **Bool**, \\(\mathcal{X}\\) has the form \\(S \times S \to \mathbf{Bool}\\), which [as Christopher notes](https://forum.azimuthproject.org/discussion/comment/18545/#Comment_18545) is the form of the characteristic function for a binary relation on \\(S\\). This gives a convenient stepping stone in the interconvertability between **Bool**-enriched categories and preorders (which are just particularly nice binary relations).

> I think not knowing that we have to define another relation \\(\le\\) on \\( \mathrm{Ob}(\mathcal{X})\\) is what confused me the most when I first attempted to understand the past few lectures.

But we _don't_ have to define another relation on \\(\mathrm{Ob}(\mathcal{X})\\). Not in general, anyway. The theorem you're referencing is taking any given **Bool**-enriched category and constructing a preorder out of the information it contains. This is a theorem _about_ **Bool**-enriched categories, not a requirement of their definition. It wouldn't work this way for, say, **Cost**-categories. That's why we say that **Bool**-categories "are" preorders, and **Cost**-categories "are" Lawvere metric spaces -- each has a "straightforward" way to convert between the two concepts.

As John says in this lecture:

> We've managed to define preorders in a new, more abstract, more confusing way!

At present, \\(\mathcal{V}\\)-enriched categories are just sets \\(S\\) with an associated map \\(\mathcal{X} : S \times S \to \mathcal{V}\\) into a monoidal preorder. We require \\(\mathcal{X}\\) to satisfy some additional properties, so it can't be just any such map, but it _is_ a map. And for whatever reason, we tend to prioritize the map instead of the set in our notation, denoting \\(S\\) by \\(\mathrm{Ob}(\mathcal{X})\\).

When \\(\mathcal{V}\\) is **Bool**, \\(\mathcal{X}\\) has the form \\(S \times S \to \mathbf{Bool}\\), which [as Christopher notes](https://forum.azimuthproject.org/discussion/comment/18545/#Comment_18545) is the form of the characteristic function for a binary relation on \\(S\\). This gives a convenient stepping stone in the interconvertability between **Bool**-enriched categories and preorders (which are just particularly nice binary relations).