[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).