Understanding the Proof of the Arzela-Ascoli Theorem from Carothers

Question

The below is the proof for the Arzela-Ascoli Theorem from Carothers' Real Analysis. I had a few questions regarding some steps in his proof which I have put in blue. If anyone could explain the blue lines, it would be appreciated.

Some definitions that I recently learned:

• Uniformly bounded means $\sup_{f\in\mathcal{F}}||f||_{\infty}<\infty$
• Equicontinuous means $\forall \epsilon>0 \exists \delta:d(x,y)<\delta\implies |f(x)-f(y)|<\epsilon \text{ for all } f\in\mathcal{F}$

Theorem (Arzela-Ascoli)

Let $X$ be a compact metric space, and let $\mathcal{F}\subset C(X)$. Then $\mathcal{F}$ compact if and only if $\mathcal{F}$ is closed, uniformly bounded, and equicontinuous.

(Note that $\mathcal{F}$ is a collection of continuous real-valued functions on $X$, and $C(X)$ is the space of all continuous functions $f:X\to\mathbb{R}$)

Proof.

(I understand the $\implies$ direction, so I omit that part of the proof.)

For $\impliedby$:

Suppose $\mathcal{F}$ is closed, uniformly bounded, and equicontinuous and let $(f_n)\in\mathcal{F}.$ $\color{blue}{\text{We need to show that $(f_n)$ has a uniformly convergent subsequence.}}$

$\color{blue}{\text{First note that $(f_n)$ is equicontinuous.}}$ Thus given $\epsilon>0, \exists \delta >0: d(x,y)<\delta\implies |f_n(x)-f_n(y)|<\epsilon/3, \forall n$.

Next since $X$ is totally bounded, $X$ has a finite $\delta$-net, i.e, there exists $x_1,\ldots, x_k\in X$ such that each $x\in X$ satisfies $d(x,x_i)<\delta$. $\color{blue}{\text{Now since $(f_n)$ is also uniformly bounded,}}$ each of the sequences $(f_n(x_i))_{n=1}^{\infty}$ is bounded in $\mathbb{R}$ for $i=1,2,\ldots, k$.

$\color{blue}{\text{Thus by passing to a subsequence of the $f_n$ (and relabeling), we may suppose that $(f_n(x_i))_{n=1}^{\infty}$ converges for each $i=1,2,\ldots, k$}}$.

In particular, we can find some $N$ such that $|f_m(x_i)-f_n(x_i)|<\epsilon/3$ for any $m,n\geq N$ and any $i=1,2,\ldots, k$.

And now we are done! Given $x\in X$, first find $i$ such that $d(x,x_i)<\delta$ and then whenever $m,n\geq N$, we will have \begin{align*} &|f_m(x)-f_n(x)|\\ &\leq|f_m(x)-f_m(x_i)|+|f_m(x_i)-f_n(x_i)|+|f_n(x_i)-f_n(x)|\\ &<\epsilon/3+\epsilon/3+\epsilon/3\\ &=\epsilon \end{align*}

That is, $f_n$ is uniformy Cauchy, since our choice of $N$ does not depend on $x$. Since $\mathcal{F}$ is closed in $C(X)$ by assumption, it follows that $f_n(x)$ converges to some $f\in\mathcal{F}$, uniformly.

$\blacksquare$

More explictly, my questions are:

• For the first blue line, why are we trying to show that $(f_n)$ has a uniformly convergent subsequence?
• For the second blue line, how do we know $(f_n)$ is equicontinuous?
• For the third blue line, how can we tell that $(f_n)$ is uniformly bounded?
• For the fourth blue line, can someone elaborate on what Carothers is trying to do here? I'm having trouble understanding what is occurring at this step.

Thanks.

Answer 1

Because if you have a uniformly convergent subsequence and the space is closed your subsequence will converge in the space, then you have a that for any sequence you can extract a convergent subsequence in the space, which is one of the characterizations of a compact space.

If the space is equicontinuous any sequence inside it is equicontinuous.

Same as above. Or: $$\{f_n\} \subseteq \{f\in\mathcal{F}\}$$

so

$$\sup_{f_n}\||f_n||_{\infty} \leq \sup_{f\in\mathcal{F}}||f||_{\infty}<\infty $$

For each $x$ he has some subsequence $f_{n_i}(x_i)$ that converges pointwise, taking the intersection of all the $n_i$ and renaming as $n$ he gets a sequence that converges for every $x_i$

Answer 2

I think his proof has a little fail. He says that $f_n$ is uniformly Cauchy, since $N$ does not depend on $x$, but the subsequence he built depends on the $\delta$-net given at the beginning.