Clarification on this corollary of the Arzela-Ascoli Theorem I am given the following corollary without proof:

A family of continuous functions on a compact metric space into
  $\mathbb R^m$ is compact iff it is closed, equicontinuous and bounded.

Does the last word above mean pointwise bounded or uniformly bounded?
(I'm pretty sure equicontinuous above means uniformly equicontinuous.)
 A: The adjective `bounded' refers here to the family of continuous functions, i.e. there is a universal constant $M$ so that for each function $f$ in the family and every $x$ in the domain, $$|f(x)| \leq M.$$ You can get counterexamples if you don't assume uniform boundedness. Let your compact metric space be an arbitrary nonempty metric space, and consider the family $$\mathcal F := \{f_c \textrm{ : } c \in \mathbb R\},$$ where $f_c$ is the function defined by $$c \mapsto (c,0,\ldots,0).$$ Clearly each element of the family is uniformly continuous (and the family itself is equicontinuous), pointwise bounded, and the family itself contains its limit points, so it is closed. However $\mathcal F$  is not sequentially compact, take for instance the sequence $$n \mapsto f_n$$ which has no convergent subsequence.
EDIT: In light of the comment below, I may have misinterpreted the OP's meaning of pointwise boundedness. 
However, let me point out that if we define as in the comment a pointwise bound on $\mathcal F$ to mean that for each $x$ in the domain, there exists $M$ so that $|f(x)| \leq M$ for all $f \in \mathcal F,$ then equicontinuity of $\mathcal F,$ pointwise boudnedness of $\mathcal F$, and compactness of the underlying metric space implies that (without appealing to the Arzela-Ascoli thm) $\mathcal F$ is uniformly bounded. In particular, the hypotheses in the OP's question imply that $\mathcal F$ is uniformly bounded even if $\mathcal F$ is only assumed at first to be pointwise bounded. Here is the proof. 
If $\mathcal F$ is pointwise but not uniformly bounded, then there exists a sequence of points $x_n$ in the (compact) domain and functions $f_n \in \mathcal F$ so that $f_n(x_n) \to \infty$. Thus, in order to prove that pointwise boudnedness implies uniform boundedness, it suffices to prove that for any such sequence $f_n(x_n)$, there exists a bounded subsequence. Since the domain is compact, we may choose a subsequence of $x_n$ (which we still denote $x_n$) that converges to a limit, say $x_n \to x$. Then
$$|f_n(x_n)| = |f_n(x_n) - f_n(x) + f_n(x) - f(x) + f(x)| \leq$$ 
$$|f_n(x_n) - f_n(x)| + |f_n(x) - f(x)| + |f(x)|.$$ The term  $|f_n(x_n) - f_n(x)|$ is bounded since the sequence $f_n$ is in particular equicontinuous and since if $n$ is large, $x_n$ is close to $x$. The term $|f_n(x) - f(x)|$ is bounded since $\mathcal F$ is pointwise bounded, and $f(x)$ is of course independent of $n$. Thus the subsequence $f_n(x_n)$ is uniformly bounded, as desired.
This shows that it `doesn't matter' whether or not you interpret the hypothesis to mean pointwise bounded (in the sense defined in this edit) or uniformly bounded. However, since pointwise bounded is a weaker concept, the theorem is stronger if you only assume pointwise boudnedness in the hypothesis.
