I've been asked following question.
The closed unit sphere is not a sequentially compact space. We define the closed unit sphere as $K := \{ f\in T[a,b] \mid \| f \|_\infty \leq 1 \}$
whereas $T[a,b]$ is the set of all step functions of $[a,b] \subset \mathbb{R}$ and $\| f \|_\infty = \sup\{|f(x)|: x\in [a,b]\}$
So I need to find a sequence $(f_n) \subset K$ which has no convergent subsequence that has a limit in K, if I'm correct.
I thought about this problem a lot and I came up with this solution....
$f_n(x) := \begin{cases}1 & \text{for} \quad x=b \\ \frac{1}{r} &\text{for} \quad \frac{b-a}{r}+a\leq x < \frac{b-a}{r-1}+a \quad \text{with} \quad r\in\{2,...,n\}\\0 & \text{for} \quad x=a\end{cases}$
Explanation: $(f_n)$ is obviously a step function in [a,b] and $\|f_n\|\leq 1$. But the limit function $\lim_{n\rightarrow \infty} f_n$ is not a step function since it has infinitely many pieces.
Because the limit of the sequence converges to $f_\infty$, all subsequences converge to $f_\infty$. Therefore no subsequence is in K and therefore K is not a sequentially compact space.
Is that correct or am I entirely on the wrong track? Thanks for helpful answers!