All metric spaces are Lawvere metric spaces18 Apr 2020 ⇐ Notes archive
(This is an entry in my technical notebook. There will likely be typos, mistakes, or wider logical leaps — the intent here is to “let others look over my shoulder while I figure things out.”)
I’m definitely the rookie in my research group, so the notebook will be a bit math-heavy as I try to catch up.
nLab’s definiton is a bit impenetrable. At a glance, it seems like tacking on Lawvere’s name, to an already general concept, means added axioms.
It’s…surprisingly the opposite.
All metric spaces are Lawvere metric spaces—that is, we lift some of the constraints on plain ol’ metrics.
Recapping, a metric space is a set $X$ equipped with a distance function $d: X \times X \rightarrow [0, \infty)$ under the following coherences:
Assuming $x, y, z \in X$,
- $d(x, y) = 0 \iff x = y$ (zero-distance coincides with equality).
- $d(x, y) = d(y, x)$ (symmetry).
- $d(x, y) + d(y, z) \geq d(x, z)$ (the triangle inequality).
And Lawvere relaxed a few bits. A Lawvere metric space has a distance function
- that respects the triangle inequality,
- whose codomain includes $\infty$ (which is helpful when we want a “disconnectedness” between points),
- and $d(x, x) = 0$ (points are zero-distance from themselves).
We’re dropping the symmetry requirement and allowing for possibly zero distances between distinct points.
The former lets us represent, e.g. in a distance as cost situation, non-symmetric costs. Borrowing from Baez, imagine the commute from $x$ to $y$ being cheaper than from $y$ to $x$.
The easing of zero-distance being necessary and sufficient for equality to only one side of the implication adds the ability to reach points “for free” (continuing with the transportation theme).
I need to read up on more applications this freedom affords us. In the meantime, here’s some links I’ve come across: