I am trying to show that $C[0,1],$ the space of all real – valued continuous functions with the sup metric is not sequentially compact with the sup metric by showing that the sequence $f_n = x^n$ has no convergent subsequence. The sup metric $\|\cdot\|$ is defined as
$$\|f – g \| = \sup_{x \in [0,1]} |f(x) – g(x)|$$
where $|\cdot|$ is the ordinary Euclidean metric. Now I know that $f_n \rightarrow f$ pointwise, where
$$f = \begin{cases} 0, & 0 \leq x < 1 \\ 1, & x = 1.\end{cases}$$
However $f \notin C[0,1]$ so this means by theorem 7.12 of Baby Rudin that $f_n$ cannot converge to $f$ uniformly. However how does this tell me that no subsequence of $f_n$ can converge to something in $C[0,1]$?
Thanks.
Best Answer
Since $f_n$ converges to $f$ pointwise, all subsequences $f_{n_k}$ also converge pointwise to $f$. So $f_{n_k}$ can not converge uniformly to continuous $g$, because uniform convergence implies pointwise convergence as well, and $f$ is not continuous.