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".