# Compactness of $C([0,1])$

I have to verify if the $C([0,1])$, space of all continuous functions defined on interval $[0,1]$ with supremum metric is compact.

As I know, we have to check if every sequence of functions $f_{n}(x)$ has subsequence that $f_{n_{k}}(x)$ is convergent. In this metric of course conervgence implies uniform convergence, so there won't be a problem of showing continuity of the limit function. But I really don't know if we can make subsequence convergent.

The sequence $$f_n(x)=x^n$$ does not have a convergent subsequence since $$f_n$$ converges pointwise towards $$f(x)=0$$ if $$x\neq 1$$ and $$f(1)=1$$ which is not continuous. Hence, the space is not compact.

• Of course! Thank you very much, I was blind or something :) May 24, 2016 at 11:15
• ...is not *continuous. May 24, 2016 at 11:15
• While this is a good example, notably they were asking about all of $C[0,1]$, which is not even bounded. Any compact set in a metric space must be bounded.
– Ian
May 24, 2016 at 11:16
• @Tsemk Arostide the solution you have done is it for the supremum metric asked in the question ? Feb 26, 2022 at 15:02

There is a general result:

Theorem The unit ball in a normed space is compact if and only if the space is finite dimensional.

An immediate consequence of this is that the unit ball in $$C([0,1]$$ is not compact and hence (as a closed subset of a space being non-compact implies the space itself is not compact) the whole of $$C([0,1])$$ is non-compact.

• "as a subset of a space being non-compact implies the space itself is not compact" This is not true. The set $(0,1)$ is not compact with Euclidean metric, but $[0,1]$ is. Dec 21, 2021 at 15:26
• oh yeah - probably I meant to say "as a closed subset of a space being non-compact implies the space itself is not compact" Dec 31, 2021 at 0:02
• I see, it makes sense then. Dec 31, 2021 at 2:25

Let be $$f_n:[0,1]\to\mathbb{R}$$ with $$f_n(x):=n$$ for all $$n\in\mathbb{N}$$. Clearly, $$f_n\in C([0,1])$$ for all $$n\in\mathbb{N}$$ but $$(f_n)_{n\in\mathbb{N}}$$ is unbounded and can't contain any convergent subsequence. Hence, $$C([0,1])$$ is not compact.

A Banach space is finite-dimensional if (and only if) it is locally compact. It's easy to see that $C[0,1]$ is a Banach space with respect to the supreme norm. And, trivially, the property of compactness implies local compactness. To sum it up, there is not infinite-dimensional Banach space (such as $C[0,1]$ ) that is also compact.

• I do not understand your third sentence at all (perhaps you meant to switch compactness and local compactness in that sentence.) This is also a higher level view than I think is appropriate for the OP.
– Ian
May 24, 2016 at 11:17
• @Ian Absolutely!
– user335721
May 24, 2016 at 11:27