This is a somewhat stupid question..
In my notes, the Arzela-Ascoli theorem is stated like this:

Let $X$ be a compact space and let $M\subset C(X,\mathbb R)$. Then $M$ is relatively compact in $C(X,\mathbb R)$ if and only if it is uniformly bounded and equicontinuous.

But since every function in $M$ is continuous (belong to $C(X,\mathbb R)$) and bounded (since $X$ is compact), aren't the conditions in the theorem just an interpretation of the fact that $M\subset C(X,\mathbb R)$?

I come up with this question because I was trying to solve questions like: given a subset $M\subset C[0,1]$ with some other conditions, and to prove that $M$ is relatively compact in $C[0,1]$. I tried to use the theorem, but then I ended with:
$M\subset C[0,1] \Rightarrow M$ is uniformly bounded.
$[0,1]$ is compact
$\Rightarrow$ $\forall f\in M$, $f$ is uniformly continuous.
$\Rightarrow \forall \epsilon >0, \exists \delta>0$ such that $|x-y|<\delta \Rightarrow |f(x)-f(y)|< \epsilon$
Then doesn't this imply uniformly bounded and equicontinuous directly without using other condition of M?

Any help is appreciated!


The conditions of Arzela-Ascoli are "uniformly bounded" and "equicontinuous" and not "each function is bounded" and "each function is uniformly continuous". The prefixes "uniformly" and "equi" suggest that those are conditions that are relevant for a set $M$ of functions and cannot be checked for each function $f \in M$ separately. Let us review the relevant definitions:

  1. A set $M \subseteq C(X,\mathbb{R})$ is uniformly bounded if there exists $L > 0$ such that for all $f \in M$ we have $\sup_{x \in X} |f(x)| \leq L$. It is true that each function in $M$ is bounded because it is a continuous function on a compact space but it doesn't mean that all the functions in $M$ can be bounded by the same constant.
  2. A set $M \subseteq C(X,\mathbb{R})$ is equicontinuous if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $f \in C(X,\mathbb{R})$ we have $|x - y| < \delta \implies |f(x) - f(y)| < \varepsilon$. Again, it is true that each function in $M$ is uniformly continuous but it doesn't mean that we can automatically find a $\delta$ that works simultaneously for all the functions in $M$, only for each function separately.
  • $\begingroup$ Thanks for your answer! But why we can't just choose the largest L and smallest delta? $\endgroup$ – J.Y Feb 5 '17 at 3:00
  • $\begingroup$ Well, if $M$ would be finite then you could do that (but then you wouldn't need Arzela-Ascoli to tell you that $M$ is compact). But if $M$ is not finite, the "largest" $L$ might be $+ \infty$ and the "smallest" $\delta$ might be $\0$. Let $X = [0,1]$ and $M = \{ nx \, | \, n \in \mathbb{N} \}$. That is, $M$ contains the functions $f_1(x) = x, f_2(x) = 2x, f_3(x) = 3x, \dots$. Each $f_n(x)$ is bounded by $n$ but there is no $L$ such that $|f_n(x)| \leq L$ for all $x \in [0,1]$ and $n \in \mathbb{N}$. $\endgroup$ – levap Feb 5 '17 at 3:03

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