# 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.

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You have to ask for a compact closure. Indeed, if you take $\{r_n\}$ a sequence of rationals which converges to $2^{-1/2}$ the sequence $f_n(x):=|x-r_n|$ converges pointwise to $f(x)=|x-2^{-1/2}|$, so the set $\{f_n\}$ is bounded, equicontinuous but not closed. Since the topology of pointwise convergence is Hausdorff, it cannot be compact. –  Davide Giraudo Mar 14 '12 at 17:00
Don't you want to assume $S$ to be closed in the topology of pointwise convergence as well? Otherwise the family of constant functions $S = \{ x\mapsto c \mid c\in [0,1) \}$ is a trivial counterexample. –  Sam Mar 14 '12 at 17:39
I don't want to assume $S$ to be closed as what I want to do is actually find a counter-example, and I think yours is perfect and easy! But why this should fail in the uniform topology? I mean: $S$ is equicontinuous and bounded, but $\left\{x\mapsto 1-1/n\right\}$ converges to the constant function $f(x)=1\notin S$ both pointwisely and uniformly, so $S$ is not closed, and thus not compact as a subspace of an Hausdorff set. How is it possible? And also: what does "bounded" mean in a non-metrisable space? –  fatoddsun Mar 15 '12 at 17:00

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.

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I see... but what is an example of a closed subset $S$ which is compact in pointwise topology but not equicontinuous or not bounded for example? That would solve any problem. –  fatoddsun Mar 16 '12 at 11:32
I don't think the second paragraph here is right. If $S$ is pointwise relatively compact, then its closure in $[0,1]^\mathbb{R}$ is compact. The intersection of this with $C([0,1])$ need not be compact anymore. For instance, consider $f_n(x) = x^n$; $\{f_n\}$ is pointwise relatively compact, but is not relatively compact in the pointwise topology on $C([0,1])$, as the only possible limit point is $1_{\{1\}}$. –  Nate Eldredge Apr 7 '13 at 13:46
@fatoddsun: Consider a tent function $f_n$ which has a spike of height $n$ in the interval $(0,1/n)$, and is 0 elsewhere. Then $\{f_n\} \cup \{0\}$ is compact in the pointwise topology but neither uniformly bounded nor equicontinuous. (Any compact set $S$ must be pointwise bounded, since the evaluation maps are continuous and hence $\{f(x) : f \in S\}$ must be compact and hence bounded.) –  Nate Eldredge Apr 7 '13 at 13:49

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.

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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!

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Why is it not correct? I would say the hint is fine: It would indeed be very helpful to see that the topology of pointwise convergence is the same as the product topology, as the answer by Henno shows! So I really don't understand the downvotes... –  Sam Mar 14 '12 at 21:00
@Sam: I voted down for two reasons: 1. (compact,hausdorff) and 2. This answer had two votes while Henno's answer had zero when I saw the thread. If Polynomial edits that bad goof out I will remove the vote. –  t.b. Mar 14 '12 at 23:50
@t.b.: Ah, I see. I didn't really register the part in paranthesis. That clears everything up, then. Thanks. =) –  Sam Mar 15 '12 at 0:39