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.

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.