Michael wrote:

> Can we enrich a preorder?

Yes, but not today.

> To me its seems the category itself is also a preorder or can be made into a preorder since we are trying to answer the quantified version of the same question. Why do we have to define it as category instead of a preorder?

An enriched category has an _element of \$$\mathcal{V}\$$_ for each pair of objects \$$x,y\$$. For a preorder we have a _truth value_ for each pair \$$x,y\$$: either \$$x \le y\$$ is true or \$$x \le y\$$ is false. (See [Lecture 30](https://forum.azimuthproject.org/discussion/2124/lecture-30-chapter-1-preorders-as-enriched-categories) for a detailed explanation of the second sentence here.)

If you have a way to turn elements of \$$\mathcal{V}\$$ into truth values, and this way obeys some nice properties, then you can turn your enriched category into a preorder. Sometimes this is an interesting thing to do! But not always.

Remember, \$$\mathcal{V}\$$ could be any symmetric monoidal preorder whatsoever. Think of all the dozens of examples you've memorized, now that you've resolved to acquire a good "zoo" of examples of every concept. For example, take the real numbers with addition and the usual notion of \$$\le\$$. How are you going to turn real numbers into truth values? The only way that has nice properties that are required to get a preorder is the completely boring way: send every real number to "true".