How can I show that the closed unit ball in $L^1([0,1])$ is not compact?

Question

Consider the following subset of $L^1([0,1])$, $S=\left\{f\in L^1([0,1]):{\|f\|}_1\leq1\right\}$. Prove that $S$ is not compact.

Should I start with an open cover and prove that it has no finite subcover or find a convergent sequence that has no convergent sub-sequence? I am quite confused.

Answer 1

Consider $f_n:=2^n\mathbf{1}_{[0,1/2^n]}$.

Answer 2

Take $f_n(x)$ to be the $n$th binary digit in the expansion of $x \in (0,1)$. Then $\|f_n\|_1 = { 1\over 2}$ and $\|f_n-f_m\|_1 = {1 \over 2}$ for all $m \neq n$.

Let $U_n = B(f_n,{1 \over 4})$, this is an open cover with no finite sub cover.

Addendum: To illustrate $f_n$, suppose $x= {1 \over 2} + {1 \over 8} = 0.101\bar{0}$, then $f_1(x) = 1$, $f_2(x) =0$, $f_3(x) = 1$ and$f_k (x) = 0$ for all $k > 3$. In general ($x \in (0,1)$) $f_n(x) = \lfloor 2^n x \rfloor\pmod 2 $.