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enter link description here I was working on the proof of Ascoli-Arzela Theorem (10.3) at the link above. However two points in there are not clear for me.

1st For necessity, we take $\mathcal F$ as a collection of functions from $C(X)$. Why do we prove elements of $\mathcal F$ are uniformly continuous i.e. $\mathcal F$ is equicontinuous? I think since $X$ is compact every functions of $C(X)$ must be uniformly continuous thus every subset of $C(X)$ is equicontinuous. I’m in I miss somewhere.

2nd For sufficiency, I’m okay with the parts until defining $\varphi$ function. However I cannot get remainings. We’re trying to show that $\mathcal F$ is totally bounded since it will be sufficient for the compactness of $\overline {\mathcal F}$. However I cannot understand how we get totally boundedness. What is our $\varepsilon$-net and open balls?

If anyone can explain them to me, I will be very very happy. I appreciate any help. Thanks

P.S. : My questions can be so silly. I’m sorry in advance.

user519955
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  • "Every functions of $C(X)$ must be uniformly continuous thus every subset of $\mathcal{F}$ is equicontinuous." Why? –  Oct 22 '19 at 21:04
  • @Jack I’m sorry I have miswritten. I thought it because equicontinuity means every element (function) of set must be uniformly continuous but all elements of $C(X)$ are already uniformly continuous? – user519955 Oct 22 '19 at 21:08
  • Equicontinuity of a family of functions is a different thing from uniform continuity of a function. For example if $X=[0,1]$ and $f_n(x)=x^n$ and $F={f_n: n\in \Bbb N}$ then $F$ is not an equicontinuous family. In fact, in $C[0,1]$ the set $F$ is an infinite closed bounded discrete subspace. – DanielWainfleet Oct 23 '19 at 03:26

2 Answers2

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  • It is true that if a subset $\mathcal{F}\subset C(X)$ is equicontinuous, then every $f\in\mathcal{F}$ is uniformly continuous. But the converse is not true. Namely, in order to show that a given collection $\mathcal{F}\subset C(X)$ is equicontinuous, it does not suffice to show that each $f\in\mathcal{F}$ is uniformly continuous. See, for instance, this example.

  • Note that $\varphi$ is a map from $\{1,\cdots,N\}$ to $\{1,\cdots,K\}$. There are only finitely many such maps. Denote the set of all these maps as $J$. Then $$ \mathcal{F}=\bigcup_{\varphi\in J}\mathcal{F}_{\varphi} $$ where $J$ is a finite set. It is shown in your notes that $\operatorname{diam} F_{\varphi}\le \varepsilon$. This shows that $\mathcal{F}$ is totally bounded. See this definition. The language of "open balls" is not used in this context.

  • I got my 2nd point thank you. I think my definition of equicontinuity is inadequate from my lecture. For saying A function family is equicontinuous they have to be continuous uniformly with same balls. Am I right? Thanks in advance – user519955 Oct 23 '19 at 08:22
  • @user519955: the definition is on page 60 of your linked notes. –  Oct 23 '19 at 11:28
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Consider the set $\{f_n : n \ge 1\}$ where $$ f_n(x) = \begin{cases} 1 - nx & 0 \le x \le \frac1n \\ 0 & \frac1n < x \le 1 \end{cases}$$

This set is not compact because the sequence $(f_n)$ has no convergent subsequence. This is because the limit $\lim_{n\to\infty} f_n$ is not continuous (so there is no limit in $C[0,1]$).

What you can see at $x = 0$ is that for $0 < \varepsilon < 1$, there is no single $\delta$ such that $$ \sup_{|x - 0| < \delta} |f_n(x) - f_n(0)| = |f_n(\delta) - f_n(0)| = \min\{1, n\delta\} \le \varepsilon.$$ These functions are each uniformly continuous (take $\delta = \varepsilon/n$) but this delta depends on $n$ and there is no single $\delta$ that will do.

Sera Gunn
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