Talk:Arzelà–Ascoli theorem

Simpler Proof?

Does the argument necessarily have to invoke diagonalization on an infinite sequence of sequences?

A family $F=\{f_{n}\}_{n}$ of functions over interval $I$ is uniformly equicontinuous if there exists $\delta (\epsilon )$ for every $\epsilon >0$, such that $|x-y|<\delta \Rightarrow |f_{n}(x)-f_{n}(y)|<\epsilon$. Note that $\delta (\epsilon )$ is independent of $y$ ("uniform") and $n$ ("equi-"). Given such a family $F$, and any fixed $\epsilon >0$, we construct a subsequence terms stay within $3\epsilon$ of each other at all points $t\in I$ (and thereby prove the Arzela-Ascoli theorem) as follows:

Divide the interval $I$ into ${\frac {1}{\delta }}+1=K$ equal overlapping intervals. Pick a point $q_{k}\in I$ in each interval, indexing $k$ from $1$ to $K$. Then construct subsequence $F^{1}=\{f_{n}^{1}\}_{n}$ that converges at $q_{1}$ using the Bolzano Weierstrass theorem, and apply the Bolzano Weierstrass iteratively, to construct subsequence $F^{k+1}=\{f_{n}^{k+1}\}_{n}$ (that converges at $q_{k+1}$) of $F^{k}$ (which converges at $q_{i}$ for $i$ up to $k$). Doing this $K$ times, we have a subsequence of functions that is convergent at all points $q_{k}$.

Convergence implies the Cauchy property - there exists some $N_{k}(\epsilon )$ such that $n,m>N_{k}(\epsilon )\Rightarrow |f_{n}(q_{k})-f_{m}(q_{k})|<\epsilon$. Define $N(\epsilon ):=\max _{k\in \{1...K\}}N_{k}(\epsilon )$. Given arbitrary point $t\in I$, find the interval it lies in (say, the one containing $q_{k}$). We know:

1) That $|f_{n}^{K}(t)-f_{n}^{K}(q_{k})|<\epsilon$, and $|f_{m}^{K}(t)-f_{m}^{K}(q_{k})|<\epsilon$ for any $n,m$

2) That $|f_{m}^{K}(q_{k})-f_{n}^{K}(q_{k})|<\epsilon$ for any $n,m>N(\epsilon )$

Therefore, applying the triangle inequality three times, we conclude that $|f_{n}^{K}(t)-f_{m}^{K}(t)|<3\epsilon$, as claimed. Since this amounts to the Cauchy property, for all $t\in I$.

Uniformly Equicontinuous

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)

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)

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)

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)

For the real line version it is not necessary the uniform equicontinuity, only equicontinuity is enough. By the other hand if you requiry the uniform equicontinuity then the uniform boundness is not necessary, it is only necessary that the sequence is bounded for one single value and the uniform boundness is consequence of it. —Preceding unsigned comment added by G Furtado (talk • contribs) 15:26, 25 October 2008 (UTC)

About the previous paragraph, I believe that over a compact set both equicontinuity and uniform equicontinuity coincide. Nevertheless, I guess that removing uniform boundness in favor of some "uniform bounded at a point" obscures the idea behind the proof and does not offer much advantage. It might be worth to have a "side note", though.André Caldas (talk) 13:19, 17 April 2014 (UTC)

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)

New section

The following was in reply to a message by User:Gsspradlin that the user redacted in which it was asserted that the statement of necessity was incorrect. Sławomir Biały (talk) 19:59, 31 October 2013 (UTC)

The conditions are necessary and sufficient for relative compactness of a subset of C[0,1]. (This is actually true in much greater generality, see Dunford and Schwartz IV.6.5 Theorem 7.) Sławomir Biały (talk) 16:08, 31 October 2013 (UTC)

The article is unclear about whether or not the hypotheses of the theorem are necessary. The first sentence in the article says they are. HOWEVER, the theorem statement given "in simplest terms" is: "Consider a sequence of real-valued continuous functions (ƒn)n∈N defined on a closed and bounded interval [a, b] of the real line. If this sequence is uniformly bounded and equicontinuous, then there exists a subsequence (ƒnk) that converges uniformly." This does NOT state that all the hypotheses are necessary. If indeed they are, the theorem statement "in simplest terms" needs to changed to an "if and only if" type statement. This is important because for theorems of this type, it is rare for all the hypotheses to be necessary. Gsspradlin (talk) 18:53, 31 October 2013 (UTC)

The section on compact metric spaces contains the statement with the necessity result. There is also a section prominently labeled Necessity in the article. So I assume it's difficult to miss this. If you really insist on adding a necessity statement to the "simplest terms" version, I would suggest adding this as a separate statement altogether (since the statement of the theorem then becomes rather less simple, and we should avoid this if possible). Something like the following:

The converse is also true, in the sense that if every subsequence of the (f_n) itself has a subsequence that converges uniformly, then the original sequence is uniformly bounded and equicontinuous.

Although the converse should be largely parenthetical, even just a footnote. Sławomir Biały (talk) 19:05, 31 October 2013 (UTC)

Thanks for the rapid response. Again, the very first sentence of the article says the theorem gives "necessary and sufficient conditions" for something to happen, but then in "simplest terms", the hypotheses are described merely as being sufficient. I find this very confusing. I don't think a reader should have to have to search for a separate "Necessity" section to find that a converse of sorts is true. It is far from obvious (at least to me) that a uniformly convergent sequence of functions is equicontinuous (it's more obvious that it's uniformly bounded). It's also an interesting fact (at least to me). I approve of your "converse" wording. I don't think it should be a footnote, because that would make it hard to find. If it were in parentheses, that would be fine with me. I wouldn't worry about making the theorem less simple. At present, it is only two sentences. You propose adding one sentence, which make a grand total of three sentences. Still pretty simple. The "converse" may be less important than the "forward" direction, but the first sentence of the article says the theorem gives "necessary and sufficient conditions", so a statement of the theorem should explain what "necessity" means here. A good theorem statement should agree with the description of the theorem that begins the article.

Is your proposed "converse" statement equivalent to "if (f_n) converges uniformly, then (f_n) is uniformly bounded and equicontinuous?" It it is, that would be simpler than what you proposed. I'm not sure right now if it is equivalent - it's confusing. By the way, I posted something earlier that was simply incorrect, so I deleted it. I hope it confused no one. Gsspradlin (talk) 20:06, 31 October 2013 (UTC)

(I changed my wording above from "logically equivalent" to merely "equivalent" because I think establishing the equivalence requires a little metric space topology in addition to logic)Gsspradlin (talk) 20:54, 31 October 2013 (UTC)

As you say, it's equivalent modulo some rather trivial details. Actually, the proof that a uniformly convergent sequence on a compact metric space is equicontinuous is only a few lines long and could easily be added to the article. Sławomir Biały (talk) 21:52, 31 October 2013 (UTC)

Thanks. I had no idea the "converse" was so easy to prove. It would be nice if the proof (a few lines long) could be added to the article, but I think that it is more important that this fact be easy to find in the article. I suspect that the proof of the "converse", and indeed the "converse" itself, is not that well-known even among mathematicians. I'm sure it is less well-known than the more difficult direction. Indeed, I'm not even sure if the "converse" property is considered to be part of the "official" Arzela-Ascoli theorem. I did some Googling and the first statement of the AA theorem I found described the hypotheses as "sufficient" conditions. If the AA theorem is generally stated in terms of certain assumptions being sufficient for a certain subsequence to exist, then maybe the word "necessary" should simply be removed from the first sentence of the article. If this is the case and this is change is made, I still think the question of necessity is one that will occur to many readers, that its answer will be surprising to many, and an answer to that question should be easy to find. I think the necessity question should be answered in the theorem statement (it takes only a sentence), or else there should be a link to the Necessity section of the article (I think one can do this) or a suggestion to refer to the Necessity section (the link or suggestion should be in the Statement and First Consequences section in my opinion and the reader shouldn't have to hunt for it). Currently, the Necessity section is hard to use. Unlike the main theorem, it is given in a very general case (when most people, probably even most mathematicians, will be interested only in the case of functions on a compact interval), and the necessity result seems to be mixed with its own proof, making both hard to read (I would rather see a statement of what is proved first, followed by a proof). The proof contains the notation "osc", which is not explained or defined, and which I, at least, have never seen. Gsspradlin (talk) 23:12, 31 October 2013 (UTC)

I can't speak for the majority of mathematicians, but Arzela definitely stated the theorem as a necessary and sufficient condition. Presumably this is the relevant criterion in assessing what the "official" version of the theorem ought to be. Sławomir Biały (talk) 17:19, 5 November 2013 (UTC)

I've written a lot here, but the issue is simple: the first sentence of the article doesn't really match the theorem statement. These are probably the two most important elements of the article, and if they don't go together well, the reader may be confused, no matter how good the rest of the article is. I've suggested a couple alternatives (changing one or the other - I'm not sure which one should be changed), and you've come up with a good fix for the theorem statement. Gsspradlin (talk) 00:56, 1 November 2013 (UTC)

Metric spaces

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)

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)

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 {{citation needed}} 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)

Counterexamples

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)

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)

You are right, stupid question... :| --Pokipsy76 (talk) 06:49, 6 June 2008 (UTC)

Toppology on C(X, Y)

The formulation of the theorem for metric spaces contains the phrase

`Then a subset F of C(X,Y) is compact if and only if [conditions]'.

However this only has a meaning when one specifies what is the topology on C(X, Y). Does anyone know the answer? (My first guess would be compact open topology.) It would also be nice to show how the version for the interval is a special case of the version for metric spaces.

Uniform equicontinuity

The version just restored by Silly Rabbit has several deficiencies.

One is that I could not think of a reference in modern times that would use Zorn for justifying the diagonal procedure. I think this is just his personal point of view. To my knowledge, nobody else does it. To the most, you use the Axiom of Dependent Choice, that everybody in Analysis uses without quoting.

The second is that if you know (end of the proof) that there is a delta for which equicontinuity holds, then you just need to cut the interval in pieces smaller than delta. Using Borel-Lebesgue makes a really bizarre impression, to say the least. Bdmy (talk) 17:28, 25 December 2008 (UTC)

Problems have been solved, to the satisfaction of both (I hope) parts. Bdmy (talk) 14:22, 27 December 2008 (UTC)

It may seem needless in the special case of the unit interval, but the various generalizations do require the explicit use of Borel-Lebesgue. The primary purpose in having a proof in the very special case of an interval is to illustrate the features that the proofs of all the various generalizations have in common. So, while I definitely agree that the result can be proven more efficiently for intervals, the goal is not to provide the most efficient proof for this special case. siℓℓy rabbit (talk) 18:06, 25 December 2008 (UTC)

I still have two objections. First, your claim of generality is relative; the case of compact Hausdorff does not follow this scheme of proof, simply because the space need not be separable. Second, the fact that the proof generalizes is not a sufficient excuse to doing (without a warning to the reader) something that the competent reader will find absurd: apply Borel-Lebesgue to a covering of a compact interval by intervals of equal length.

There is a simple way out of this: change only a few words in order to get the proof for compact metric spaces (it is almost already there). Bdmy (talk) 19:11, 25 December 2008 (UTC)

There is also a less simple way: do not use uniform equicontinuity, but just equicontinuity. Using uniform continuity first and Borel-Lebesgue after is overkill. Bdmy (talk) 19:35, 25 December 2008 (UTC)

More or less any space where limit makes sense

"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)

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)

Looking at Frechet's paper, with my weak command of French, he seems to be talking about real-valued functions on compact subsets of R^n. I accessed the paper through Springerlink -- I'd be grateful if someone more accustomed to reading 19th century French mathematics could weigh in on this. Since the passage in our article is a garbled section from a rather extemporaneous section of Dunford and Schwartz, we likely want a different tone. I've edited it to correspond to the rigorous statement closest to that made by Dunford and Schwartz, though I personally suspect that's still too strong a statement for what Frechet in fact proved. Magnus the Blind (talk) 23:38, 8 October 2010 (UTC)

Most General Formulation

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)

Necessary properties of the family of functions

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)

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)

Proof

The statement in the proof that says:

The collection of intervals U_x, x ∈ I, forms an open cover of I.

I am not sure needs to be true. It seems the U_x could fail to cover I. The size of union of U_x'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)

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)

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 U_x could be balls of e/2^n. The union of all the U_x 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 U_x's more carefully. Oliverks1 (talk) 22:36, 16 November 2014 (UTC)

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)

OK I am with you now. I thought the cover was for the x_k not all x in I. Sorry for the confusion.

Oliverks1 (talk) 09:19, 17 November 2014 (UTC)

Hausdorff condition not needed

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)

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)

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)

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)