Question about compactness of C[0,1]

Question

I am trying to prove that subset of $C[0,1]$ given as $\{h\in C[0,1]:||h||\leq 1\}$ in supremum metric is not compact. As I know, in that set lays sequence $g_n(x)=x^n$ which converges to non-continuous function (0 everywhere else despite 1). And I came up after some research that this implies that such a subset of $C[0,1]$ can't be compact, be I couldn't find out why. Maybe it's simple but I don't see that. So... I will be really grateful if someone could explain that. Or show other, simple way to prove that statement.

Thank you in advance!

Answer 1

In any metric space, every compact set is closed (see for example the first answer of A compact set, which is not closed.). So if your subset doesn't contain all its limits, it is not closed and therefore cannot be compact.

Answer 2

Let $B = \{f \in C[0,1] \:|\: \|f\| \leq 1\} \subseteq C[0,1]$ denote the set under consideration. If $B$ were compact, then every sequence in $B$ would have to have a convergent subsequence (convergent with respect to $\|\cdot\|$) converging to an element in $B$. However the sequence $(g_n)_n$ you found fails to have such a subsequence, since the $\| \cdot \|$-limit would in particular have to be the pointwise limit which is not continuous as you already noticed.