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The page Cauchy space has a definition of ’Cauchy-continuous map’ between Cauchy spaces, and remarks that while every uniformly continuous map is Cauchy-continuous, the converse is not true. What are some examples of nonuniformly continuous but Cauchy-continuous maps? For instance, does it suffice to be locally uniformly continuous? Can the Cauchy-continuous maps between (say) metric spaces be characterized any more concretely?
I am wondering (in connection with the higher inductive-inductive Cauchy reals) to what extent the extension of the algebraic structure on the rational numbers to their Cauchy completion is a formal consequence of the universal property of completion. While addition is uniformly continuous, multiplication is not, so this requires knowing how completion behaves on some nonuniformly continuous maps.
Hmm, wikipedia suggests that maybe Cauchy-continuity coincides with local uniform continuity when the completion is locally compact?
What are some examples of nonuniformly continuous but Cauchy-continuous maps?
for the benefit of others (you’ve probably seen it, Mike), there’s , for (from here).
for the benefit of others
This is extremely well-known. Even , is not uniformly continuous.
(The rule of thumb is that for continuously differentiable functions , uniform continuity of is closely connected to uniform boundedness of . Certainly uniform boundedness of implies uniform continuity of , in fact Lipschitz continuity of . I’d have to think more on whether the converse is true, but the rule of thumb says don’t expect to be uniformly continuous, because there is no uniform bound on .)
Speaking of this example, though, with regard to the second paragraph of #1: notice that multiplication can be defined in terms of squaring and linear maps: . So I think it would suffice just to observe that squaring is uniformly continuous on every bounded subset of : that would be enough to show that squaring extends continuously to its Cauchy completion. (I’m not sure it’s worth the bother getting into issues of local compactness.)
Yeah, I guess when I asked for examples I wanted general classes of examples (like ’uniformly continuous on bounded subsets’), not random particular ones.
Anyway, that’s exactly what Andrej did to construct multiplication on the HIIT Cauchy reals. It works, but I want a general theory. Possibly ’uniformly continuous on every bounded subset’ is what I meant instead of ’locally uniformly continuous’ – that’s also what Wikipedia says. I guess that this condition should work for every metric space. Can the concept be extended to uniform spaces somehow?
Wikipedia only mentions local compactness in connection with the converse, that if a function extends to the completion then it is uniformly continuous on bounded subsets. I guess the idea is that the completion of a bounded subset will then be compact, so that any continuous function is uniformly so. So for a counterexample to the converse, we just need a nonuniformly continuous function on a bounded and complete but non-locally-compact space. Maybe the closed unit ball in a Hilbert space?
Can the concept be extended to uniform spaces somehow?
Interesting question. My guess is “probably so”, along lines of the idea that every uniform space embeds uniformly into a product of (pseudo)metric spaces.
Again, this is just a guess, but one might try defining a subset of a uniform space to be bounded if its image is bounded for any uniformly continuous function . If all works out, this would give a canonical notion of bornology for a uniform space. Lemma: if is uniformly continuous, then takes bounded sets of to bounded sets of . The proof is routine.
So then one might hope to prove that if is a uniform space, is a complete uniform space, and if is a continuous function whose restriction to each bounded subset of is uniformly continuous, then extends uniquely to a continuous function on the completion, again with the property of being uniformly continuous when restricted to bounded sets of .
(I’d like to return to this later, but thought I’d jot down these thoughts now for what they’re worth.)
Bounded subspaces are a red herring. In the real line, a subspace is totally bounded iff it's bounded, and one traditionally speaks of bounded subspaces instead of totally bounded ones in this context. (Similarly, one traditionally speaks of closed subspaces instead of complete ones; thus, undergraduates learn that compact means closed and bounded instead of complete and totally bounded.) Since every metric space is uniformly equivalent to a bounded metric space (in which every subspace is therefore bounded), uniform continuity on bounded subspaces is not even a uniform property and so can't possibly mean Cauchy continuity.
(I’ve been away from the internet for a few days.) I’m pretty sure that my last comment did define a bornology on a uniform space. It’s also true that this bornology need not coincide with the collection of “bounded spaces” (as ordinarily understood) in a metric space. So I think we might be talking about different things, and I’m not yet convinced that this bornology is a “red herring”. But perhaps I should put this to the test and see what if anything in #6 pans out.
Do the totally bounded subsets of a uniform space form a bornology?
Yes, the totally bounded subsets of a uniform space form a bornology.
My statement that boundedness is a red herring is not a direct response to Todd's bornology but a general principle. When generalising from to arbitrary uniform spaces or beyond, total boundedness is usually the way to go. In particular, I'm not claiming that Todd's bornology is a red herring. It can't possibly extend boundedness of metric spaces (since it's a manifestly uniform property), as Todd suggested; but since boundedness of metric spaces is a red herring, that's a good thing!
Thanks for clarifying, Toby.
Total boundedness is closely connected with compactness (for a complete metric space, the compact subsets are the closed totally bounded subsets), so I thought it might be worthwhile to have a look at a less restrictive concept which could apply to non-(locally compact) examples like Hilbert spaces. But as I say, I need to think more about how good a concept it is.
for a complete metric space, the compact subsets are the closed totally bounded subsets
More generally, for any metric space (or uniform space, or more generally), the compact subsets are the complete totally bounded subsets. In a complete Hausdorff space, complete = closed; but more generally, it's completeness that matters.
It seems to me that what we really need, in order for a continuous function to extend to the completion, is for it to be uniformly continuous on the bounded sets in some bornology with the property that every point of the completion lies in the closure of a bounded set. Or to put it another way, extends to the completion if every point of the completion lies in the closure of a set on which is uniformly continuous. This latter property is what isn’t obvious to me for the totally bounded sets or those in Todd’s bornology — in fact at the moment it seems unlikely to me to be true.
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