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.

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    $\begingroup$ It looks like the key is the canonical identification $C(X,\mathbb{R}^n) \cong \bigl(C(X,\mathbb{R})\bigr)^n$. $\endgroup$ Commented Jan 31, 2016 at 19:52
  • $\begingroup$ @DanielFischer that much I figured out, but no further :( Could you help me understand how the projections move the desired properties around? $\endgroup$
    – user153312
    Commented Jan 31, 2016 at 22:14
  • $\begingroup$ 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. $\endgroup$ Commented Jan 31, 2016 at 22:25

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – user153312
    Commented Jan 31, 2016 at 22:15

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.


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