Does Arzelà-Ascoli theorem hold for pointwise convergence? Consider the space $C(\left[0,1\right])$ of continuous, real-valued functions on the interval, equipped with the topology of pointwise convergence.
Is it true that a subset $S\subseteq C(\left[0,1\right])$ is compact if and only if it is bounded and equicontinuous? 
I guess that the answer is no, as this seems to be a weakened version of Arzelà-Ascoli theorem, which guarantees the validity of the statement when the topology is given by uniform convergence instead. I was trying to find a counterexample but it's not easy to check that a certain subset is or is not compact, especially because pointwise convergence topology is not metrisable, so we cannot use sequential compactness arguments. Can you help me with that? Thank you.
 A: As a starting point we need to take the correct version of Arzela-Ascoli for the compact-open (or in this case also uniform) topology: $S$ is relatively compact (the closure is compact) iff it is pointwise relatively compact, and equicontinuous.
In the weaker pointwise topology, we have more compact sets, and there the criterion is simply: $S$ is relatively compact iff it is pointwise relatively compact: this makes it a subset of a product space of compact intervals of the reals, and thus has a compact closure due to the Tychonoff theorem.
In both cases, to characterize compactness, add closed to the list of conditions.
A: The answer is no. Take a sequence of continuous functions $f_n$ that pointwise converges to a continuous function $f$ but that does not uniformly converge. The set $S=\{f_n\}\cup\{f\}$ is not equicontinuous but is compact in the topology of pointwise convergence. 
For instance, let $f_n(x) = 0$ if $x \ge 1/n$ and $f_n(x) = 4n^2x(x-1/n)$ otherwise.  The sequence pointwise converges to $0$ but not uniformly. Clearly, $S=\{f_n\}\cup\{f\}$ is not equicontinuous.
A: It might be helpful if you think of C([0,1]) as a subspace of the (compact,hausdorff) product space $R^{[0,1]}$. 
- Apologies, this is not correct!
