TITLE: Sequence in $C[0,1]$ with no convergent subsequence
QUESTION [19 upvotes]: 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.

REPLY [7 votes]: This is (more-or-less) a modification of the hint from t.b's comment.
It suffices to show that any subsequence of $(f_n)$ is not Cauchy. (Since every convergent sequence is a Cauchy sequence.)
To do this, notice that
$$\lVert f_{2n}-f_n \rVert = \sup_{x\in[0,1]} (x^n - x^{2n}) = \sup_{x\in[0,1]} (x^n - (x^{n})^2) = \sup_{t\in[0,1]} (t-t^2)=\frac14.$$
(It is easy to find maximum of the quadratic function $f(t)=t-t^2=t(1-t)$.)
In addition, if we use the monotonicity of the sequence $(f_n)$, we see that
$$k\ge 2n \qquad \Rightarrow \qquad \lVert f_{k}-f_n \rVert \ge \frac14.$$
Now, if we have any subsequence of $(f_n)$ then the above estimate shows that this subsequence is not Cauchy. For any given $k_0$ we can find $k'>k_0$ such that $n_{k'}>2n_{k_0}$ and $\lVert f_{n_{k'}}-f_{n_{k_0}} \rVert \ge \frac 14$.