gn.general topology - What is a monoidal metric space?

For the general question about how categorical concepts look when applied to metric spaces, one place to look is Lawvere's paper 'Taking categories seriously', section 6 onwards.

Aside from that, here are a few examples.

Functor categories became function spaces with the uniform or sup metric. That is, if $A$ and $B$ are metric spaces construed as enriched categories, then the functor category $B^A$ is the set of distance-decreasing maps $A to B$ with the sup metric. (I use "decreasing" in the non-strict sense.)

The (cartesian) product $A times B$ of two metric spaces --- that is, their product in the category of metric spaces --- has the '$infty$-metric':
$$
d((a, b), (a', b')) = max{d(a, a'), d(b, b')}.
$$
The same goes for infinite products --- remembering that $infty$ is allowed as a distance. Once you know this, limits in general work in the obvious way.

The coproduct $A + B$ of two metric spaces $A$ and $B$ is their disjoint union, with $d(a, b) = d(b, a) = infty$ for all $a in A, b in B$. Again, it's crucial here to allow $infty$ as a distance. Otherwise, your category of metric spaces will lack lots of limits and colimits. The coequalizer of two maps $f, g: A to B$ is $B$ quotiented out by the usual equivalence relation $sim$ (as in the category of sets), and metrized by
$$
d([b], [b']) = inf{ d(y, y'): y sim b, y' sim b'}
$$
where $[b]$ denotes the equivalence class of $b$. General colimits work similarly.

I mentioned the cartesian product, but there's another kind of product. Generally, if $mathbf{V}$ is a monoidal category then any two $mathbf{V}$-enriched categories, $A$ and $B$, have a tensor product $A otimes B$. Its set of objects is the product of the sets of objects of $A$ and $B$. Its hom-objects are given by
$$
(A otimes B)((a, b), (a', b')) = A(a, a') otimes B(b, b').
$$
This gives us a tensor product of metric spaces. Given metric spaces $A$ and $B$, the point-set of $A otimes B$ is the product of the point-sets of $A$ and $B$. The distance is given by
$$
d((a, b), (a', b')) = d(a, a') + d(b, b').
$$
In other words, it's the '$1$-metric', also known as the taxicab metric, Manhattan metric, etc.

So, Andrew, when you ask 'What is a monoidal metric space?', you have to say which product you want to be monoidal with respect to. That is, are you asking about (weak) monoids in $(mathbf{Met}, times)$ or in $(mathbf{Met}, otimes)$?

From the tone of your question, I would guess: both. So here goes.

The answer for cartesian product $times$ doesn't seem so interesting. Assuming that your metric spaces satisfy the classical skeletality axiom ($d(a, b) = 0 Rightarrow a = b$), a monoidal category for $times$ is a metric space $A$ equipped with a monoid structure on its set of points such that
$$
d(a cdot b, a' cdot b') leq max{d(a, a'), d(b, b')}.
$$
I can't think of anything more to say about that.

The answer for tensor product $otimes$ seems more interesting. A monoid in $(mathbf{Met}, otimes)$ is a metric space $A$ equipped with a monoid structure on its set of points such that for all $a$, the maps $acdot -$ and $- cdot a$ are distance-decreasing. For example, if it's a group then this says that left or right translation is always an isometry. This often happens: consider the underlying additive group of a normed vector space, for instance.