Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

One of the ways the Arzela-Ascoli Theorem is stated is as follows:

Given a compact space $X$ and a set $ M \subset C(X) := \{f: X \rightarrow \mathbb{R}, \| . \|_{\infty}\}$, the following are equivalent:

1) $M$ is bounded, closed and uniformly equicontinuous.
2) $M$ compact.

Why is the condition on boundedness required, and does it not follow from uniform equicontinuity? It appears to me that uniform equicontinuity implies (pointwise) continuity of any function $f$ in $M$, and since $X$ is compact and $f$ is continuous, $f(X)$ is compact so $f$ takes its maximum and minimum on its image. In particular, it would follow that $\| f\|_{\infty} < \infty$, for every $f \in M$, so $M$ is bounded. Since the theorem is well-established, it seems to me that there must be some mistake in this argument.

share|cite|improve this question
Of course $f$ is bounded, if $f \in M$. Here you nedd a stronger assumption: there exists $C>0$ such that $\|f\|_\infty \leq C$ for every $f \in M$. $C$ is the same for every $f \in M$. – Siminore Nov 22 '12 at 17:38
I think I prefer the statement of Arzela-Ascoli where one drops "closed" from "1)" and changes "compact" in "2)" to "totally bounded" (or "pre-compact"). It makes it clear that it's the other properties that are essential to making the proof work. – kahen Nov 22 '12 at 18:10
It does appears that if $V$ is connected, and there is a point $x \in V$ such that $\{ f(x) | f\in M \}$ is bounded, then M is bounded. This is an exercise I am currently stuck on. – Yiteng Nov 22 '12 at 20:45
up vote 5 down vote accepted

You are right: If $f \in C(X)$, then $f$ is bounded. But: $M$ bounded means

$$\sup_{f \in M} \|f\|_{\infty}<\infty$$

... and this condition is not necessarily fulfulled if every function in $M$ is bounded.

Example: Let $f_n(x) := n$, then $f_n$ is continuous and bounded, but $M := \{f_n;n \in \mathbb{N}\}$ is not bounded.

share|cite|improve this answer

The subset of constant functions is closed and equicontiuous, but not compact. Your mistake was to conclude from bounded functions to a bounded set of bounded functions...

share|cite|improve this answer

Generally, compactness of $M$ is proven by showing that it is sequentially compact. In proving this, you need to take a sequence of functions in $M$, and although each one may be bounded, they might not be bounded by the same bound. Read any proof and you will see that it is important that they are uniformly bounded.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.