Definition talk:Complete Metric Space

I'm staring at the refactor template and wondering why this definition page is supposed to be refactored differently from all other definition pages. Why not just make definition 1, definition 2, etc.? Nested Sphere Theorem is getting close to becoming a large part of the equivalence proof, which apparently has never actually been posted. --Dfeuer (talk) 07:13, 28 January 2013 (UTC)


 * If you can't work out what to do, pass on it and work on something you *do* know how to do. --prime mover (talk) 07:28, 28 January 2013 (UTC)


 * What? I'm not saying "I don't know how to do what this says", I'm saying "Why does this say to do something different from how the entire rest of the site operates?" --Dfeuer (talk) 07:35, 28 January 2013 (UTC)


 * What I said was: Please leave well alone if it's not clear to you what needs to be done. --prime mover (talk) 08:55, 28 January 2013 (UTC)

To put matters at rest, an inspection of the history will almost surely show that it grew so hysterically (i.e. the refactor predates the multiple def paradigm). --Lord_Farin (talk) 09:57, 28 January 2013 (UTC)


 * The equivalence proof might require the axiom of countable choice; unless this is determined not to be the case, I don't think it would be a good idea to use one definition as definition $1$ and the other as definition $2$. --abcxyz (talk) 16:37, 28 January 2013 (UTC)


 * Ah, that could be. If it is so, do you know if the closed ball version corresponds to the notion of a complete uniform space? --Dfeuer (talk) 17:14, 28 January 2013 (UTC)


 * I think ACC can be avoided by taking the supremum of the distances $d(p_n, p_m), m > n$ plus some fixed epsilon as radius (it requires some remark to establish existence of this supremum) for the $n$th ball in the one direction (or some suitable adaptation to ensure all conditions are satisfied); the other direction should be simply taking the centres of the balls. --Lord_Farin (talk) 17:32, 28 January 2013 (UTC)


 * How would you choose the centers of the balls without ACC? --abcxyz (talk) 19:05, 28 January 2013 (UTC)


 * Simply as the terms from the sequence, of course. --Lord_Farin (talk) 19:11, 28 January 2013 (UTC)


 * What sequence? --abcxyz (talk) 19:14, 28 January 2013 (UTC)

Okay, so it is to be shown that "Sphere condition" implies "Cauchy condition" (the converse is dealt with by the new Nested Sphere Theorem). So we are given a Cauchy sequence. That's the sequence I mean. --Lord_Farin (talk) 19:17, 28 January 2013 (UTC)


 * Sorry, I meant that to prove "Cauchy" $\implies$ "Sphere" using Nested Sphere Theorem might need ACC to choose the centers of the spheres. --abcxyz (talk) 19:26, 28 January 2013 (UTC)


 * So we are given a sequence of spheres. A sphere has associated a centre, by definition (also, characterised by having the greatest distance to the complement of the sphere). These centres will form a Cauchy sequence because the spheres are nested. --Lord_Farin (talk) 19:31, 28 January 2013 (UTC)


 * Also, I fail to see what the condition that the radii tend to zero has to do with it. It seems superfluous. --Lord_Farin (talk) 19:34, 28 January 2013 (UTC)


 * But isn't it true that a sphere might not have a unique center? --abcxyz (talk) 19:35, 28 January 2013 (UTC)


 * I finally discern your point. To me, a ball always has associated a centre. Dropping that condition and only insisting that it is a "ballable set", so to speak, proves your point accurate. We could for example take a suitable sequence in the Cantor set, which I think provides an example of non-unique radii, and probably also of non-unique centres. Thanks for that new insight; precaution added to my system. --Lord_Farin (talk) 20:04, 28 January 2013 (UTC)


 * Just to come in here: might it be demonstrable that a ball in a complete metric space does only have one centre? I feel that intuitively it ought to have, and I would expect it to be a consequence of the Cauchy condition, and/or that a sequence can have at most one limit. --prime mover (talk) 20:34, 28 January 2013 (UTC)


 * A two-point space provides a simple counterexample. --abcxyz (talk) 20:42, 28 January 2013 (UTC)


 * But is a two-point space a complete metric space? --prime mover (talk) 21:25, 28 January 2013 (UTC)


 * Yes. Any Cauchy sequence is eventually constant. --Lord_Farin (talk) 21:27, 28 January 2013 (UTC)

I think we really should check into whether the closed ball def gives complete uniform spaces. I don't know enough to check myself at the moment, but I'm pretty sure the Cauchy sequence definition does not (without ACC), and if we can use one that does as the primary def, that should give a cleaner hierarchy. IF that's possible. --Dfeuer (talk) 21:03, 28 January 2013 (UTC)