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Stream: learning: questions

Topic: Equivalent metric spaces categorically

fosco (Jun 28 2025 at 16:24):

Is there a particular meaning, in terms of enriched category theory, for two metric spaces equipped with equivalent metrics (=metrics defining the same topology)? It seems a very coarse equivalence relation on metric spaces, which does not possess a neat characterization in pure terms of $([0,\infty],\ge)$-enriched categories.

Todd Trimble (Jun 28 2025 at 21:42):

I'm not aware of any, although I've not tried hard to think about it. Cauchy completion doesn't have an invariant sense in the topological category, for example. An example is the space of irrational numbers, which is homeomorphic to $\mathbb{N}^\mathbb{N}$ via [[continued fractions]]. Under the metrization as a subset of the reals, this isn't Cauchy complete, but under a different metrization on sequences $(a_0, a_1, a_2, \ldots) \in \mathbb{N}^\mathbb{N}$, say $d(a, b) = 2^{-n}$ where $n$ is the first index where $a_n \neq b_n$, the irrationals are Cauchy complete. Same topology, different completions.

There might be an increased chance of saying something meaningful if you switch from continuity to uniform continuity, which has greater global engagement with the actual metrics (Cauchy completion is a meaningful concept in the context of uniform spaces). This is a little vague, but if you rehash the definition of $f: X \to Y$ being uniformly continuous,

$\forall \varepsilon > 0 \exists \delta > 0\; d_X(x, y) < \delta \Rightarrow d_Y(f(x), f(y)) < \varepsilon$

then it's like every uniformly continuous function $f$ has a modulus function $\mu_f: (0, \infty) \to (0, \infty)$, something a little looser than a Lipschitz constant, call it a Lipschitz variable maybe, where $d_Y(fx, fy) \leq \mu_f(d_X(x, y)) \cdot d_X(x, y)$ [so the modulus wouldn't depend on $x, y$ themselves, but only the distance between them]. I'm not sure any of this is completely accurate; I'm just spitballing as they say.

Mike Shulman (Jun 28 2025 at 21:48):

IIRC uniform spaces can be defined as monads in a certain bicategory whose objects are sets and whose morphisms are filters of some kind. Metric spaces are monads in the bicategory of $[0,\infty]$-valued matrices, and the "underlying uniform space" functor of a metric space is induced by a functor between these bicategories, so you can think of it as a sort of "change of enrichment". Thus saying that two metric spaces are uniformly equivalent is analogous to saying, say, that two topologically enriched categories have equivalent underlying ordinary categories.

fosco (Jun 29 2025 at 06:36):

@Mike Shulman I was thinking the same, in particular I suspect that one can say something like, two equivalent metrics on the same set equip the set with two category structures that differ by an isomorphism of the base of enrichment

fosco (Jun 29 2025 at 06:37):

(multiplying by $\alpha > 0$ is a quantale isomorphism of $[0,\infty]$)

fosco (Jun 29 2025 at 06:58):

Well, those are the only quantale isomorphisms of nonnegative reals

Mike Shulman (Jun 29 2025 at 08:21):

Being topologically equivalent is much weaker than just differing by a constant scaling.

fosco (Jun 29 2025 at 08:37):

Ah, yeah, I was focusing on the simpler sufficient condition of having positive constants $A d_1 \le d_2 \le B d_1$...

John Baez (Jun 29 2025 at 09:30):

Yes, that's very different; for example the function $x \mapsto \sqrt{x}$ is a homeomorphism from $[0,1]$ to itself that violates this condition, and you can replace the square root function by $x^{1/n}$ for any $n > 1$ to get scarier examples. All these functions are still Hölder continuous, but there are even worse homeomorphisms between metric spaces.