Compact space and Bolzano–Weierstrass theorem I have a little question. let $E = \{ f \in C[0,1] : || f ||_{\infty} \leq 1 \}$. How to show that E is not a compact space. I have to find a counterexample like a subsequence that does not respect the Bolzano–Weierstrass theorem ! thanks for the answer and excuse me for my english I am french and I'm not good with this site !
 A: Consider the family of functions $f_n(x)=\sin(nx)\chi_{[0,\frac{\pi}{n}]}$. 
Then $\|f_n\|_{\infty}=1$, whereas we have the pointwise convergence $f_n\to 0$. Suppose $f_{n_k}$ was a convergence subsequence in $\|\cdot\|_{\infty}$. Then its limit would have to agree with the pointwise limit of 0. But by continuity of the norm, we would have $1=\lim_{k\to\infty}\|f_{n_k}\|_{\infty}=\|0\|_{\infty}$, a contradiction.
Hence we have contradicted Bolzano-Weierstrass, so the set is not compact.
A: How about a sequence of orthogonal functions $(f_n)$ with $||f_n||_{\infty} = 1$ and $\int_0^1 f_n(x) f_m(x) \mu(x) dx= 0$ for $n\ne m$. This has no convergent subsequences. For if $f_{n_k} \to g$ then $0 =\int_0^1 f_{n_k}f_{n_{k+1}} \mu(x) dx \to \int_0^1 g^2 \mu(x) dx$ and so $g=0$ but $||f_n||= 1 $ for all $n$, not possible.
Lots of choices: Cebyshev polynomials (suitably normalized), $\sin (\pi n x)$, other orthogonal polynomials,
Obs:  An answer of another user was $\sin nx$;  while not orthogonal on $[0,1]$, it is also an example
A: Not a direct proof-longer than a comment:
compactness of the closed unit ball is actually equivalent to the space being finite dimensional. (One of the reasons we appeal to weaker topologies e.g weak-* etc)
