I was not understanding what was being 'enriched' and what was doing the 'enriching'. Based on comments I think I am not the only one. Here is my informal definition of \\( \mathcal{X} \\) a \\( \mathcal{V}\\)-enriched category or \\( \mathcal{V}\\)-category. [Which may not turn out to be a category at all :P .]

You start with a container \\( \mathcal{X} \\) which contains, in part, of a set of objects \\( \text{Ob}(\mathcal{X}) \\). You want to say something about the structure of objects in this container. You decide that structure has something to do with the relationship between
pairs of objects \\( \mathcal{X}(x, y) | x, y \in Ob(\mathcal{X}) \\) .

Stop for a moment and notice that we have not mentioned the enriching \\( \mathcal{V} \\) yet.

We introduce the enriching \\( \mathcal{V} := (V, \le, I, \otimes ) \\). Now we map each of the pairs \\( \mathcal{X}(x, y) \rightarrow V \\) and in doing we say that \\( \mathcal{X} \\) is enriched in \\( \mathcal{V} \\).
[Why don't we say 'enriched by'?] Well almost enriched, the mapping is constrained by two properties.

a) We want the enrichment to reinforce the identity of each object. For the identity pair \\( \mathcal{X}(x, x) | x \in Ob( \mathcal{X} ) \\) we have \\( I \le \mathcal{X}(x, x) \\) where the \\(I\\) and \\( \le \\) come from \\( \mathcal{V} \\).

b) We also want to have some notion of composition. When we have three objects \\( x, y, z \in \text{Ob}( \mathcal{X} ) \\), with pairs \\( \mathcal{X}(x, y) \text{ and } \mathcal{X}(y, z) \\). Using the order relation and monoidal operation \\( \le , \otimes \\) from \\( \mathcal{V} \\) we require that they compose in the following way \\( \mathcal{X}(x, y) \otimes \mathcal{X}(y, z) \le \mathcal{X}(x, z) \\).

**Puzzle 87.** A preorder is isomorphic with **Bool**-category. What makes this odd it that the number of pairs in the **Bool**-category appears to be greater than the number of arrows in the corresponding preorder.
This discrepancy is explained by the fact that 'false' pairs are not expressed in the preorder.