It appears that comments from new users are collapsed by default, and cannot be replied to without a “Like”. These seem like bad features.
Your proof that there’s no uniformly continuous on bounded sets function f:X×X→[0,1] admitting all uniformly continuous on bounded sets functions X→[0,1] as fibers looks correct now. It also looks like it can be easily adapted to show that there is no uniformly continuous f:X×X→[0,1] admitting all uniformly continuous functions X→[0,1] as fibers. Come to think of it, your proof works for arbitrary metric spaces X, not just complete separable metric spaces, though those are nicer.
I see what you mean now about uniform continuity giving you an algorithm, but I still don’t think that’s specific to uniform continuity in an important way. After all, if you have an algorithm for computing images of points in the countable dense set, and a computable “local modulus of continuity” in the sense of a computable function h:X×[0,∞)→[0,∞) with h(x,0)=0 and d(x,y)<r⟹|f(y)−f(x)|<h(x,r), then f is computable, and this does not require f to be uniformly continuous. Although I suppose you could object that this is a bit circular, in that I’m assuming the “local modulus of continuity” is computable only in the standard sense, which does not require uniform continuity.
I’m not sure why you would allow singularities at some points (presumably a uniformly discrete set, or something like that) while still insisting on uniform continuity elsewhere. It still seems to me that the arguments for uniform continuity rather than continuity all point to wanting uniform continuity entirely, rather than some sense of local uniform continuity in most places.
Thanks for pointing out the error in my definition of fn; I’ve fixed it.
In your argument that locally compact Polish spaces can be given metrics with respect to which they are proper, it isn’t true that d′ is necessarily a proper metric. For instance, consider a countably infinite set with d(x,y)=1 for x≠y. This is a locally compact Polish space, but f(x)=1 for every x, so d′=d, and the space is not proper.
Your last proposition looks correct (though with a typo: last ⊆ in the proof should be ⊇). However, if X is not locally compact, then the compact-open topology isn’t necessarily the right topology to consider on [0,1]X. We want a topology making [0,1]X into an exponential object, and it isn’t clear that such a topology even exists, or that it is the compact-open topology if it does exist (though it must be a refinement of the compact-open topology if it does exist). Maybe asking about non-locally compact Polish spaces X with a Polish exponential space [0,1]X is a kind of weird question, though, and if we’re even considering non-locally compact Polish spaces, we should turn to the version of the question where we just want a continuous function X×X→[0,1] admitting all continuous functions X→[0,1] as fibers.
I will have to think more about the issue of continuity vs uniform continuity. I suppose my last remaining argument would be the fact that Bishop—Bridges’ classic book on constructive analysis uses uniform continuity on bounded sets rather than continuity, which suggests that it is probably better for constructive analysis at least. But maybe they did not analyze the issue carefully enough, or maybe the relevant issues here are for some reason different.
To fix the argument that every locally compact Polish space admits a proper metric, let f be as before and let F(x,y)=∞ if d(x,y)≥f(x) and F(x,y)=f(x)/[f(x)−d(x,y)] if d(x,y)<f(x). Next, let g(y)=minn[n+F(xn,y)], where (xn) is a countable dense sequence. Then g is continuous and everywhere finite. Moreover, if S=g−1([0,N]), then S⊆⋃n≤NB(xn,(1−1/N)f(xn)) and thus S is compact. It follows that the metric d′(x,y)=d(x,y)+|g(y)−g(x)| is proper.
Hm, perhaps I should figure out what the significance of uniform continuity on bounded sets is in constructive analysis before dismissing it, even though I don’t see the appeal myself, since constructive analysis is not a field I know much about, but could potentially be relevant here.
f is the reciprocal of what it was before, but yes, this looks good. I am happy with this proof.
It appears that comments from new users are collapsed by default, and cannot be replied to without a “Like”. These seem like bad features.
Your proof that there’s no uniformly continuous on bounded sets function f:X×X→[0,1] admitting all uniformly continuous on bounded sets functions X→[0,1] as fibers looks correct now. It also looks like it can be easily adapted to show that there is no uniformly continuous f:X×X→[0,1] admitting all uniformly continuous functions X→[0,1] as fibers. Come to think of it, your proof works for arbitrary metric spaces X, not just complete separable metric spaces, though those are nicer.
I see what you mean now about uniform continuity giving you an algorithm, but I still don’t think that’s specific to uniform continuity in an important way. After all, if you have an algorithm for computing images of points in the countable dense set, and a computable “local modulus of continuity” in the sense of a computable function h:X×[0,∞)→[0,∞) with h(x,0)=0 and d(x,y)<r⟹|f(y)−f(x)|<h(x,r), then f is computable, and this does not require f to be uniformly continuous. Although I suppose you could object that this is a bit circular, in that I’m assuming the “local modulus of continuity” is computable only in the standard sense, which does not require uniform continuity.
I’m not sure why you would allow singularities at some points (presumably a uniformly discrete set, or something like that) while still insisting on uniform continuity elsewhere. It still seems to me that the arguments for uniform continuity rather than continuity all point to wanting uniform continuity entirely, rather than some sense of local uniform continuity in most places.
Thanks for pointing out the error in my definition of fn; I’ve fixed it.
In your argument that locally compact Polish spaces can be given metrics with respect to which they are proper, it isn’t true that d′ is necessarily a proper metric. For instance, consider a countably infinite set with d(x,y)=1 for x≠y. This is a locally compact Polish space, but f(x)=1 for every x, so d′=d, and the space is not proper.
Your last proposition looks correct (though with a typo: last ⊆ in the proof should be ⊇). However, if X is not locally compact, then the compact-open topology isn’t necessarily the right topology to consider on [0,1]X. We want a topology making [0,1]X into an exponential object, and it isn’t clear that such a topology even exists, or that it is the compact-open topology if it does exist (though it must be a refinement of the compact-open topology if it does exist). Maybe asking about non-locally compact Polish spaces X with a Polish exponential space [0,1]X is a kind of weird question, though, and if we’re even considering non-locally compact Polish spaces, we should turn to the version of the question where we just want a continuous function X×X→[0,1] admitting all continuous functions X→[0,1] as fibers.
I will have to think more about the issue of continuity vs uniform continuity. I suppose my last remaining argument would be the fact that Bishop—Bridges’ classic book on constructive analysis uses uniform continuity on bounded sets rather than continuity, which suggests that it is probably better for constructive analysis at least. But maybe they did not analyze the issue carefully enough, or maybe the relevant issues here are for some reason different.
To fix the argument that every locally compact Polish space admits a proper metric, let f be as before and let F(x,y)=∞ if d(x,y)≥f(x) and F(x,y)=f(x)/[f(x)−d(x,y)] if d(x,y)<f(x). Next, let g(y)=minn[n+F(xn,y)], where (xn) is a countable dense sequence. Then g is continuous and everywhere finite. Moreover, if S=g−1([0,N]), then S⊆⋃n≤NB(xn,(1−1/N)f(xn)) and thus S is compact. It follows that the metric d′(x,y)=d(x,y)+|g(y)−g(x)| is proper.
Anyway I have fixed the typo in my previous post.
Hm, perhaps I should figure out what the significance of uniform continuity on bounded sets is in constructive analysis before dismissing it, even though I don’t see the appeal myself, since constructive analysis is not a field I know much about, but could potentially be relevant here.
f is the reciprocal of what it was before, but yes, this looks good. I am happy with this proof.