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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.2012-03-14
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    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.2012-03-14
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    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?2012-03-15

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