Closed ball is not compact Show that the closed ball in $C([0,1])$ of center $0$ and radius $1$ is not compact.
I thought it will be compact since every closed and bounded set in $\mathbb{R}$ is compact?
Why is it not compact and how can I prove it? 
 A: Let $f_n$ be zero except on $[\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n}), \frac{1}{2}(\frac{1}{n}+\frac{1}{n-1})]$, where the graph is described by joining the points
$(\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n}),0), (\frac{1}{n},1), (\frac{1}{2}(\frac{1}{n}+\frac{1}{n-1}),0)$. Then $\operatorname{supp} f_n = [\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n}), \frac{1}{2}(\frac{1}{n}+\frac{1}{n-1})]$, and $f_n(\frac{1}{n}) = 1$. Hence $\|f_n-f_m\| = \delta_{mn}$.
The collection $\{ B(x, \frac{1}{2}) \}_x$ is an open cover of $C[0,1]$. If we take any finite sub-collection, then at most one of the $f_k$ can be contained in each one, so the finite sub-collection can contain only a finite number of $f_n$. It follows that $C[0,1]$ is not compact.
A: This answer is for posterity, and I hope someone appreciates it.  $C([0,1])$ is a metric space, so it suffices to show it has a bounded sequence with no convergent subsequence.  Such a sequence is $(f_n)$ where $f_n(x)=x^n$.  The boundedness is obvious.  The sequence converges pointwise to a noncontinuous function.  No subsequence can converge (in the metric of $C([0,1])$, that is, uniformly) because if a sequence of continuous functions converges uniformly to a function, that limiting function must be continuous. 
