Let $(C[0,1];d_{\infty})$ the metric space with $C[0,1]$ the continuous functions on $[0,1]$ and $d_{\infty}(f,g)=\max_{x \in[0,1]}|f(x)-g(x)|$
Prove that $$S=\{f:|f(x)| \leq 1\}$$ is closed and bounded but no compact.
My attempt: I've already proved that is bounded using a ball with radius $2$. But I can't figure out how can I prove that is closed and is not compact, I tried with the limit points and the complement but I'm lost.
Best Answer
It is closed because the uniform limit of continuous functions is continuous, and the limit function will belong to S, because it is also the pointwise limit.
For non-compactness, since we are in a metric space, it is enough to find a sequence in S which (you can show) has no convergent subsequence. $\{x^n\}_{n\geq 1}$ will do the trick.