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.

  • $\begingroup$ 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. $\endgroup$ – Davide Giraudo Mar 14 '12 at 17:00
  • $\begingroup$ 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. $\endgroup$ – Sam Mar 14 '12 at 17:39
  • $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – fatoddsun Mar 16 '12 at 11:32
  • $\begingroup$ 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 $1_{\{1\}}$. $\endgroup$ – Nate Eldredge Apr 7 '13 at 13:46
  • $\begingroup$ @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.) $\endgroup$ – 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.


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!

  • $\begingroup$ 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... $\endgroup$ – Sam Mar 14 '12 at 21:00
  • 1
    $\begingroup$ @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. $\endgroup$ – t.b. Mar 14 '12 at 23:50
  • $\begingroup$ @t.b.: Ah, I see. I didn't really register the part in paranthesis. That clears everything up, then. Thanks. =) $\endgroup$ – Sam Mar 15 '12 at 0:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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