Requirement of closed and bounded set $[a,b]$ in the Ascoli theorem

In Wikipedia, the Ascoli theorem requires the functions to be continuous on the closed and bounded interval $[a,b]$. However, in the proof given in the book "Theory of Ordinary Differential Equations" by Coddington and Levinson (Tata MaGraw-Hill Edition 1972) (given at page 5, Ascoli Lemma) the authors only require that the interval be bounded. The lemma is:-

On a bounded interval $I$, let $F=\{f\}$ be an infinite, uniformly bounded, equicontinuous set of functions. Then $F$ contains a sequence $\{f_n\}$, $n = 1,2,\cdots,$ which is uniformly convergent on $I$.

(The authors mention in the text before the lemma that $I$ denotes an open interval.)

Reading through the article on Compact Space on wikipedia, I get a feeling that the interval should be compact. However, the proof in the above book also seems to be correct.

-
Perhaps you could provide a quotation of the theorem, as it appears in "Theory of Ordinary Differential Equations", in your post. If you did, you would be more likely to get an answer. – matt Jan 4 '12 at 9:53
The statement you wrote talks only about convergence - but you didn't state which convergence. What you will not get without compactness is uniform convergence (and it is easy to find a counter example). But there are other definitions - you can get point convergence, or - a definition which is popular for the use of ODEs - "uniform convergence on every compact subset". – yaakov Jan 4 '12 at 10:23

The authors define "equicontinuous on $I$" to mean "uniformly equicontinuous on $I$". That is, for any $\epsilon>0$ there is a $\delta>0$ such that for any $f\in \cal F$ and for any $x$, $y$ in $I$,

$$|f(x)-f(y)|<\epsilon\quad\text{ whenever }\quad|x-y|<\delta.$$

The theorem you are referring to does give uniform convergence (you omitted this in your post). And, indeed, with the assumption of uniform equicontinuty and uniform boundedness, the assumption that $I$ be closed is not needed.

What is needed is that the interval be totally bounded. If you examine the proof in Coddington and Levinson , you'll see that's what they are using.

See here for a different proof.

Incidentally, some authors say $\cal F$ is equicontinuous over $I$ to mean that given $x\in I$ and $\epsilon>0$, there is a $\delta_{x,\epsilon}$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta_{x,\epsilon}$ and $f\in\cal F$ . One can show that if $I$ is compact, then this notion of equicontinuity implies the notion of uniform equicontinuity above.

-
Thanks for finding the precise mistake and the help. I have not understood the definition of equicontinuity and uniform equicontinuity clearly before. – jpv Jan 5 '12 at 6:17
The Wikipedia article on this theorem assumes uniform equicontinuity (as the dependence of $\delta$ on $x$ has not been made clear) and a closed and bounded interval. From your answer I do not think that both these assumptions are necessary. Is it a good idea to edit the Wiki page? – jpv Jan 5 '12 at 6:27