Talk:Arzelà–Ascoli theorem

message: The article defines uniform equicontinuity, but the concept is not mentioned later. In the proof for the real line, the concepts of equicontinuity and uniform equicontinuity are used as if they were the same thing. Of course, they coincide because $latex I$ is compact, but this fact is not mentioned.André Caldas (talk) 13:19, 17 April 2014 (UTC)
replies: text: I have just changed it from "uniformly equicontinuous" to "equicontinuous".  These notions coincide on a compact domain, and anyway for most authors and readers "equicontinuous" is already regarded as the uniform version.  (Thus "pointwise equicontinuous" is a separate notion.). Sławomir Biały (talk) 14:49, 17 April 2014 (UTC)
replies: text:  This is a pitty... in my opinion, in a set with one function only, "equicontinuous" should reduce to continuous, and "uniformly equicontinuous" should reduce do "uniformly continuous". The "pointwise" comes from the idea of "at a point". Since equicontinuity can be defined at a point, the word "pointwise" makes sense to use as an optional way to emphasise the fact that it is "equicontinuous" at every point, just like we do for continuity. I guess Wikipedia is not a place for this kind of discussion... but maybe "most authors" might not be the best... André Caldas (talk) 03:04, 2 May 2014 (UTC)
replies: text:  Well, not really.  Equicontinuity of a family of functions F (equipped with the discrete metric) on a metric space X is just uniform continuity of the function u:F\times X\to \mathbb R given by u(f,x) = f(x).  So in this sense, the "natural" definition of equicontinuity already extends uniform continuity.  From that perspective, it makes more sense to regard pointwise equicontinuity as a separate notion.  Sławomir Biały (talk) 10:45, 2 May 2014 (UTC)

text: I'm not really sure what point that post is trying to make.  On a compact space, uniform equicontinuity and equicontinuity coincide (fix an ε, and cover the space with finitely many balls on which the oscillation of each function in the sequence is less than ε).  Also for an equicontinuous family of functions on a compact space, boundedness at a single point is equivalent to uniform boundedness.  Obviously, neither of these statements are true for non-compact spaces:  does the "real line version" that this post refers to have something to do with a version of the theorem true for the whole real line?  It's unclear to me, but if so the remark doesn't seem to have any bearing on the article as currently written.  Sławomir Biały (talk) 20:59, 17 April 2014 (UTC)
thread_title: Uniformly Equicontinuous

message: I could not find the stated general theorem for metric spaces in the literature. 
The Hausdorff version is not in Dunford & Schwarz. The quotation of IV.6.7 is absolutely wrong! This theorem is about C(S), "the set of all bounded continuous real or complex functions defined on S." (page 261 in Dunford).    —Preceding unsigned comment added by Tigili12 (talk • contribs) 12:06, 21 August 2008 (UTC)
replies: text: The set of continuous functions on a compact Hausdorff space is coextensive with the set of bounded continuous functions, so this is precisely the version in Dunford and Schwartz.  This does hold when the target space is a metric space, although you are right that it isn't mentioned in the cited section of Dunford and Schwartz (it appears in the Notes and remarks, IV.16).   siℓℓy rabbit (talk) 17:02, 25 October 2008 (UTC)

text: I have restored the section.  The version *was* correct as stated contrary to your contention that it was "Absolutely wrong".  In the future, if you feel that the source doesn't completely support the statement of the theorem, it may be better to add a  template instead of simply removing the material.  The editor who added it may have intended to reference the "Notes and remarks" section of D&S as well, but failed to.  This is then a courtesy to indicate that you would like to see more specific references.  I have added a number of other references to the disputed content.  siℓℓy rabbit (talk) 17:02, 25 October 2008 (UTC)
thread_title: Metric spaces 

message: It would be nice to insert counterexamples in case the equicontinuity hypothesis is not satisfied. Does someone know any?--Pokipsy76 (talk) 14:27, 29 April 2008 (UTC)
replies: text: Since equicontinuity is the main hypothesis, more or less anything will work. How about f(x)=xn on the closed unit interval? Algebraist 11:03, 5 June 2008 (UTC)
replies: text: You are right, stupid question... :| --Pokipsy76 (talk) 06:49, 6 June 2008 (UTC)
thread_title:  Counterexamples 

message: "A further generalization of the theorem was proven by Fréchet (1906) for any space in which the notion of a limit makes sense (such as a metric space or Hausdorff space)." 
Can this statement in the introduction be tightened up a little bit? It seems unlikely that the theorem was generalised to any space where the notion of limit makes sense. Using filters, limits can be defined for any topological space, even non-T0 ones. 202.36.179.66 (talk) 03:52, 9 July 2009 (UTC)
replies: text: The attribution is taken directly from Dunford and Schwartz, although the "metric space or Hausdorff space" part seems to have been added after the fact.  Anyway, without having checked exactly what Frechet proved, we have to go with what reliable secondary sources have said about it. Sławomir Biały (talk) 20:43, 10 December 2009 (UTC)
thread_title:  More or less any space where limit makes sense 

message: I couldn't speak to whether or not there exists a 'most general formulation' of AA, but I myself have read a number of formulations that don't require X be compact Hausdorff -- e.g., Willard's General Topology '70, page 287 requires only that X be a compactly generated Hausdorff space, and deals with compact open C(X,Y) and Y a uniform space, thus is strictly more general than the version stated here.  Kelley's General Topology '91, page 234, has an almost identical version.  Even Munkres' Topology '00, page 290, though it assumes Y is a metric space, assumes X is locally compact Hausdorff -- and then only for one direction!  Thus I've reworded the statement in the introduction to bring it in closer agreement to the literature on this point.  Magnus the Blind (talk) 23:38, 8 October 2010 (UTC)
thread_title:  Most General Formulation 

message: The conditions in the article:
* Uniformly bounded
* Equicontinuous
Except that the equicontinuity as defined here seems (to my untrained eye) to be uniform equicontinuity (the delta doesn't depend on the point).
Also, Rudin ("Real and Complex Analysis") only requires pointwise boundedness. The proof here also only uses pointwise boundedness.
Is there an advantage in changing the conditions to pointwise boundedness and pointwise equicontinuity? I believe both conditions are equivalent to their uniform version in a compact space. I think pointwise boundedness would cause a minor change in the proof, but pointwise equicontinuity might be a bigger change. --72.226.86.106 (talk) 23:23, 18 May 2014 (UTC)
replies: text: For most authors (including Rudin) "equicontinuous" actually means "uniformly equicontinuous".  We have also adopted this convention in the article.  The pointwise notion is of comparatively little use, and anyway the two notions are equivalent on a compact space by a straightforward argument.  It's true that boundedness is only really needed at a single point.  For an equicontinuous family of functions on a compact space, this implies uniform boundedness easily.  Sławomir Biały (talk) 10:57, 19 May 2014 (UTC)
thread_title:  Necessary properties of the family of functions 

message: The statement in the proof that says:
The collection of intervals Ux, x ∈ I, forms an open cover of I.
I am not sure needs to be true.  It seems the Ux could fail to cover I. The size of union of Ux's might be significantly less than the measure of I as one example, even though the x's are dense.
Oliverks1 (talk) 20:20, 16 November 2014 (UTC)
replies: text: Every point of I is one of the x's, though.  So this family tautologically covers I.  Sławomir Biały (talk) 21:17, 16 November 2014 (UTC)
replies: text: I am not sure I follow that.  Assume I is [0,1] and chose my x's to be the rational numbers inside I.  If we imagine they are enumerated by n, then Ux could be balls of e/2^n.  The union of all the Ux would be less than or equal to e.  If I let e be less than 1 then the union can not cover I.  I think you need to select the Ux's more carefully. Oliverks1 (talk) 22:36, 16 November 2014 (UTC)
replies: text: The open cover is \{ U_x | x\in I\}.  Since every x is in the corresponding U_x, it is a cover.  The example you are thinking of is of the form \{ U_r | r\in \mathbb Q\cap I\}, which may indeed fail to be a cover.  (It will, however, obviously cover every rational point of the interval, for the same reason that the original cover includes every point of the interval.)  23:06, 16 November 2014 (UTC)
replies: text: OK I am with you now.  I thought the cover was for the xk not all x in I.  Sorry for the confusion.
Oliverks1 (talk) 09:19, 17 November 2014 (UTC)
thread_title:  Proof 

message: The section on generalizations mentions that the domain can be generalized to a compact Hausdorff space. But the Hausdorff assumption is not actually required, see Theorem 45.4 in Munkres's Topology. In case this is OK with others I will edit the article to reflect this, changing the reference from Dunford & Schwartz to Munkres.
Jyotirmoyb (talk) 07:13, 22 February 2015 (UTC)
replies: text: It looks like Munkres requires Hausdorff for the converse to hold (though this is probably not as essential as the assumption of local compactness).  See also Dugundji's "Topology", where a similar statement is given.  I would just add another statement, rather than replace any of the current ones, at least for now (start a new section: something like "General topological spaces").  This at least solves the problem of how to handle the converse.  Sławomir Biały (talk) 12:49, 22 February 2015 (UTC)
replies: text: How if we have two versions stated: one for compact (not necessarily Hausdorff) spaces and one for locally compact Hausdorff spaces? Stating a theorem with extraneous assumptions may leave readers confused (as I was). Also see for example this Math.SE question. Jyotirmoyb (talk) 13:12, 23 February 2015 (UTC)
replies: text:  I guess that would make sense, and probably would make the section easier to follow.  Also, I think metric spaces should be done as a separate section (together with some sketchy proof about how the "diagonalization" argument still works there), but that's rather a larger undertaking.  The version for topological spaces should follow that, with the more high-brow topological proof sketched (as currently in the text).  I don't know to what extent the sketched proof there depends on the separation axiom, but it should probably be checked carefully.  Sławomir Biały (talk) 22:25, 23 February 2015 (UTC)
thread_title:  Hausdorff condition not needed