# Arzela-Ascoli for $\mathbb R^n$ from the case of $\mathbb R$?

In class, we proved the Arzela-Ascoli theorem for $\mathbb R$. The lecturer said it's also true for $\mathbb R^n$, and this version is deducible from $\mathbb R$. I tried to do this but failed. How does one infer this generalization?

For the record, the theorem I mean is:

Theorem. Let $X$ be a compact metric space. Then a subset of $C(X,\mathbb R)$ is compact iff it's closed, equicontinuous, and totally bounded.

• It looks like the key is the canonical identification $C(X,\mathbb{R}^n) \cong \bigl(C(X,\mathbb{R})\bigr)^n$. – Daniel Fischer Jan 31 '16 at 19:52
• @DanielFischer that much I figured out, but no further :( Could you help me understand how the projections move the desired properties around? – user153312 Jan 31 '16 at 22:14
• By the way, I think you mean "uniformly bounded", not "totally bounded". Since $C(X,\mathbb{R}^n)$ is complete, totally bounded and closed equals compact for subsets of that space. – Daniel Fischer Jan 31 '16 at 22:25

We have a canonical identification $C(X,\mathbb{R}^n) \cong \bigl(C(X,\mathbb{R})\bigr)^n$. Not only as sets, but as topological spaces. By definition of the product topology, the coordinate projections $\pi_k \colon C(X,\mathbb{R}^n) \to C(X,\mathbb{R})$ are continuous. So if $K \subset C(X,\mathbb{R}^n)$ is compact, so is $K_k := \pi_k(K)$ for $1 \leqslant k \leqslant n$. By the known case of $C(X,\mathbb{R})$, it follows that $K_k$ is uniformly bounded and equicontinuous. Now note that that implies that

$$\prod_{k = 1}^n K_k \subset \bigl(C(X,\mathbb{R})\bigr)^n \cong C(X,\mathbb{R}^n)$$

is uniformly bounded and equicontinuous. This is most convenient to see if we use the maximum norm on $\mathbb{R}^n$. Since

$$K \subset \prod_{k = 1}^n K_k$$

it follows that $K$ is uniformly bounded and equicontinuous. As a compact subset of a Hausdorff space, $K$ is also closed.

Conversely, if $S \subset C(X,\mathbb{R}^n)$ is uniformly bounded and equicontinuous, then the sets $S_k := \pi_k(S)$ are uniformly bounded and equicontinuous. Hence $\overline{S_k}$ is a compact subset of $C(X,\mathbb{R})$ for $1 \leqslant k \leqslant n$. By Tíkhonov's theorem (well, a small part of it),

$$K := \prod_{k = 1}^n \overline{S_k} \subset \bigl(C(X,\mathbb{R})\bigr)^n \cong C(X,\mathbb{R}^n)$$

is compact. By construction, $S\subset K$, hence $\overline{S} \subset K$, and $\overline{S}$ is compact as a closed subset of a compact set.

I have used without proof that uniform boundedness and equicontinuity are preserved under projections and products. These things aren't difficult to prove.

I don't know the Arzela-Ascoli proof you have used for a segment $[a,b] \subset \mathbb R$. However if you look at Wikipedia proof, you'll see that the only thing you require is an enumeration of the rationals of the segment $[a,b]$.

You can mimic this proof by using an enumeration of the points of the compact $X \subset \mathbb R^n$ having rational coordinates (as this set is countable) to get the generalization you're looking for.

• Thanks for your input, but it does not answer my question. I want to deduce the theorem for $\mathbb R^n$ using the version for $\mathbb R$ without redoing any work. – user153312 Jan 31 '16 at 22:15