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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeMay 16th 2013
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   Author: Mike Shulman
   Format: MarkdownItexThe 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.

   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.

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeMay 16th 2013
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   Author: Mike Shulman
   Format: MarkdownItexHmm, [wikipedia](http://en.wikipedia.org/wiki/Uniform_continuity#Relations_with_the_extension_problem) suggests that maybe Cauchy-continuity coincides with local uniform continuity when the completion is locally compact?

   Hmm, wikipedia suggests that maybe Cauchy-continuity coincides with local uniform continuity when the completion is locally compact?

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 16th 2013
   • (edited May 16th 2013)
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   Author: DavidRoberts
   Format: MarkdownItex>What are some examples of nonuniformly continuous but Cauchy-continuous maps? for the benefit of others (you've probably seen it, Mike), there's $f\colon \mathbb{Q} \to \mathbb{R}$, $f(x) = a^x$ for $a \gt 1$ (from [here](http://en.wikipedia.org/wiki/Cauchy-continuous_function#Examples_and_non-examples)).

   What are some examples of nonuniformly continuous but Cauchy-continuous maps?

   for the benefit of others (you’ve probably seen it, Mike), there’s $f\colon \mathbb{Q} \to \mathbb{R}$, $f(x) = a^x$ for $a \gt 1$ (from here).

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 16th 2013
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   Author: Todd_Trimble
   Format: MarkdownItex> for the benefit of others This is extremely well-known. Even $f: \mathbb{Q} \to \mathbb{Q}$, $f(x) = x^2$ is not uniformly continuous. (The rule of thumb is that for continuously differentiable functions $f$, uniform continuity of $f$ is closely connected to uniform boundedness of $f'$. Certainly uniform boundedness of $f'$ implies uniform continuity of $f$, in fact Lipschitz continuity of $f$. I'd have to think more on whether the converse is true, but the rule of thumb says don't expect $f(x) = x^2$ to be uniformly continuous, because there is no uniform bound on $f'(x) = 2 x$.) 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: $x y = \frac1{2}((x+y)^2 - x^2 - y^2)$. So I think it would suffice just to observe that squaring is uniformly continuous on every _bounded_ subset of $\mathbb{Q}$: 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_.)

   for the benefit of others

   This is extremely well-known. Even $f: \mathbb{Q} \to \mathbb{Q}$, $f(x) = x^2$ is not uniformly continuous.

   (The rule of thumb is that for continuously differentiable functions $f$, uniform continuity of $f$ is closely connected to uniform boundedness of $f'$. Certainly uniform boundedness of $f'$ implies uniform continuity of $f$, in fact Lipschitz continuity of $f$. I’d have to think more on whether the converse is true, but the rule of thumb says don’t expect $f(x) = x^2$ to be uniformly continuous, because there is no uniform bound on $f'(x) = 2 x$.)

   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: $x y = \frac1{2}((x+y)^2 - x^2 - y^2)$. So I think it would suffice just to observe that squaring is uniformly continuous on every bounded subset of $\mathbb{Q}$: 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.)

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeMay 16th 2013
   • (edited May 16th 2013)
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   Author: Mike Shulman
   Format: MarkdownItexYeah, 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?

   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?

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 17th 2013
   • (edited May 17th 2013)
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   Author: Todd_Trimble
   Format: MarkdownItex> 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 $S$ of a uniform space $X$ to be _bounded_ if its image $f(S)$ is bounded for any uniformly continuous function $f: X \to \mathbb{R}$. If all works out, this would give a canonical notion of [[bornology]] for a uniform space. Lemma: if $g: X \to Y$ is uniformly continuous, then $g$ takes bounded sets of $X$ to bounded sets of $Y$. The proof is routine. So then one might hope to prove that if $X$ is a uniform space, $Y$ is a complete uniform space, and if $f: X \to Y$ is a continuous function whose restriction to each bounded subset of $X$ is uniformly continuous, then $f$ extends uniquely to a continuous function $\bar{X} \to Y$ on the completion, again with the property of being uniformly continuous when restricted to bounded sets of $\bar{X}$. (I'd like to return to this later, but thought I'd jot down these thoughts now for what they're worth.)

   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 $S$ of a uniform space $X$ to be bounded if its image $f(S)$ is bounded for any uniformly continuous function $f: X \to \mathbb{R}$. If all works out, this would give a canonical notion of bornology for a uniform space. Lemma: if $g: X \to Y$ is uniformly continuous, then $g$ takes bounded sets of $X$ to bounded sets of $Y$. The proof is routine.

   So then one might hope to prove that if $X$ is a uniform space, $Y$ is a complete uniform space, and if $f: X \to Y$ is a continuous function whose restriction to each bounded subset of $X$ is uniformly continuous, then $f$ extends uniquely to a continuous function $\bar{X} \to Y$ on the completion, again with the property of being uniformly continuous when restricted to bounded sets of $\bar{X}$.

   (I’d like to return to this later, but thought I’d jot down these thoughts now for what they’re worth.)

7. • CommentRowNumber7.
   • CommentAuthorTobyBartels
   • CommentTimeMay 19th 2013
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   Author: TobyBartels
   Format: MarkdownItexBounded 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.

   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.

8. • CommentRowNumber8.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 20th 2013
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   Author: Todd_Trimble
   Format: MarkdownItex(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.

   (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.

9. • CommentRowNumber9.
   • CommentAuthorMike Shulman
   • CommentTimeMay 20th 2013
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   Author: Mike Shulman
   Format: MarkdownItexDo the totally bounded subsets of a uniform space form a bornology?

   Do the totally bounded subsets of a uniform space form a bornology?

10. • CommentRowNumber10.
    • CommentAuthorTobyBartels
    • CommentTimeMay 20th 2013
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    Author: TobyBartels
    Format: MarkdownItexYes, 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 $\mathbb{R}$ 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!

    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 $\mathbb{R}$ 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!

11. • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 20th 2013
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    Author: Todd_Trimble
    Format: MarkdownItexThanks 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.

    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.

12. • CommentRowNumber12.
    • CommentAuthorTobyBartels
    • CommentTimeMay 20th 2013
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    Author: TobyBartels
    Format: MarkdownItex>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.

    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.

13. • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeMay 21st 2013
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    Author: Mike Shulman
    Format: MarkdownItexIt 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, $f$ extends to the completion if every point of the completion lies in the closure of a set on which $f$ 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.

    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, $f$ extends to the completion if every point of the completion lies in the closure of a set on which $f$ 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.