TITLE: Arzela-Ascoli: why is boundedness required?
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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.