# Arzela-Ascoli Theorem on metric spaces

I've been looking for a proof of one particular direction of this theorem for metric spaces. I've looked online, but everyone seems to use different terminology/notation to state the theorem, so I'd like to ask for an outline of a proof specific to my text's version, which my text "leaves to the reader".

Let $A \subset M$ where $A$ is compact and $M$ is a metric space.
Let $B \subset C_{b}(A,N)$ where $N$ is a metric space and $C_{b}(A,N)$ is the set of all bounded continuous functions from $A$ to $N$.

Then $B$ is compact if and only if $B$ is equicontinuous, closed, and pointwise compact.

I'm trying to prove the forward direction. The reverse direction (showing compactness) is based on the diagonalization argument, which is described well in the textbook, but the text makes no remarks on the forward direction. I already managed to prove pointwise compactness, and closure, which were trivial, but equicontinuity seems difficult. Could someone provide an outline of the proof, or link me to the proof of this particular version of the theorem?

Remarks:
- The metric on $B$ is defined by $d_{B}(f,g) = sup(d_{N}(f(x),g(x)|x \in A)$ where $d_{N}$ is the metric on $N$. So compactness of $B$ is defined relative to this metric $d_{B}$. (Showing $d_{B}$ is a metric is trivial).

• An observation (I might be wrong, though): since $A$ is a compact metric space and $N$ is a metric space, doesn't it hold $C_b(A,N)=C(A,N)$?
– user239203
Commented Mar 28, 2016 at 17:48
• That is actually correct! Commented Mar 28, 2016 at 18:28
• In the "if and only if"-statement above, on the right hand side "pt.wise compact" should really be "pt.wise relatively compact" (i.e. having a compact closure), or am I mistaken? Commented May 14, 2019 at 10:47

This is a standard application of compactness. Fix $\epsilon>0$ and $x_0\in A$. By continuity, for each $f\in B$ there exists $\delta_f>0$ such that $d(f(x),f(x_0))<\epsilon$ whenever $d(x,x_0)<\delta_f$.

Our aim is to show that $\delta_f$ may be chosen independently of $f$. For each $\delta>0$, let $B_\delta=\{f\in B\;\colon\;d(f(x),f(x_0))<\epsilon\textrm{ whenever }d(x,x_0)<\delta\}$. By what we have said above, the $B_\delta$ cover $B$.

You can show yourself that each $B_\delta$ is open. Then compactness of $B$ gives us a finite subcover $(B_{\delta_j})_{j=1}^n$. But since clearly $B_\delta\subset B_{\delta'}$ whenever $\delta'\le\delta$, we may conclude that $B=B_\delta$, where $\delta=\min\{\delta_1,\dots,\delta_n\}$. This gives us equicontinuity.

If you would rather use the sequential definition of compactness, then you can do that too. We suppose that $B$ is not equicontinuous. Then there exist $\epsilon>0,x_0\in A$ such that for all $\delta>0$ there is some $f\in B,x\in A$ such that $d(x_0,x)<\delta$ and $d(f(x_0),f(x))\ge\epsilon$.

We may therefore choose sequences $x_1,x_2,x_3,\dots$ and $f_1,f_2,f_3,\dots$ such that $d(x_0,x_n)<1/n$ and $d(f_n(x_0),f_n(x_n))\ge\epsilon$ for each $n$.

By compactness of $B$, the sequence $f_1,f_2,\dots$ must have a convergent subsequence. After passing to this subsequence, we may assume without loss of generality that $(f_n)$ is a convergent sequence converging uniformly to some $f\in B$.

We claim that $f$ is not continuous at $x_0$. Indeed, let $\delta>0$. We find $x\in A$ such that $d(x_0,x)<\delta$ but $d(f(x_0),f(x))\ge\epsilon/3$.

Let $n$ be chosen so that

• $1/n<\delta$
• $d(f,f_n)<\epsilon/3$

Then we have $d(x_0,x_n)<\delta$. Lastly, observe that: \begin{align} \epsilon\le d(f_n(x_0),f_n(x_n))&\le d(f_n(x_0),f(x_0)) + d(f(x_0),f(x_n)) + d(f(x_n),f_n(x_n))\\ &\le d(f_n,f) + d(f(x_0),f(x_n)) + d(f_n,f)\\ &\le d(f(x_0),f(x_n))+2\epsilon/3 \end{align}

We may conclude that $d(f(x_0),f(x_n))\ge\epsilon/3$. Since $\delta>0$ was arbitrary, this means that $f$ is not continuous at $x_0$.

• For equicontinuity, do you not need to also show that $\delta$ can be chosen independent of $x$, i.e. $\delta$ depends only on $\epsilon$? Commented Mar 28, 2016 at 18:31
• @user308485 I call that uniform equicontinuity. But it's not hard to show that $\delta$ can be chosen independent of $x_0$; indeed, by compactness of $A$, each $f\in B$ is uniformly continuous, so we may choose $\delta_f$ independent of $x_0$ for each $f$. Then change the definition of $B_\delta$ to $\{f\in B\;\colon\;d(f(x),f(y))<\epsilon\textrm{ whenever }d(x,y)<\delta\}$ and continue as before. Commented Mar 28, 2016 at 18:38
• How doyou know that each $f$ is uniformly continuous? Commented Mar 31, 2016 at 5:22
• @user308485 A continuous function $f$ from a compact metric space $A$ to a metric space $B$ is always uniformly continuous. Sketch proof: Fix $\epsilon>0$. Then for each $\delta>0$ let $U_\delta$ be the set of points $x\in A$ such that $d(f(x),f(y))<\varepsilon$ whenever $d(x,y)<\delta$. Show that each $U_\delta$ is open; then, by continuity of $f$, $(U_\delta)$ is an open cover of $A$, so has a finite subcover as $A$ is compact. Take the minimal $\delta$ such that $U_\delta$ appears in this finite subcover and show that $d(f(x),f(y))<\epsilon$ if $d(x,y)<\delta$. Commented Mar 31, 2016 at 9:56
• Alternatively, copy the proof from real analysis that a continuous function on $[0,1]$ is uniformly continuous, replacing the Bolzano-Weierstrass theorem with the definition of sequential compactness. Commented Mar 31, 2016 at 9:57