TITLE: A problem regarding a sequence in $C[0;1]$
QUESTION [5 upvotes]: Given normed space $C[0;1]$ (the space of continous functions defined on $[0;1]$) with the norm $||.||$ defined as following
$$||x|| = \max\limits_{x \in [0;1]} |f(x)| $$
  Prove that the sequence $\{x_n\}_n$ doesn't converge in $(C[0;1],||.||)$ with 
  $$x_n(t) = \left\{ \begin{array}{ll}
 nt,& 0 \le t \le \dfrac{1}{n} \\ 
 1,&  \dfrac{1}{n} < t \le 1
\end{array} \right. $$

In the attempt of solving this problem, I found out that the sequence converges by individual points to the function $x$ (means for every $t \in [0;1]$, $\lim\limits_{n \to \infty} x_n(t) = x(t)$) with
$$x(t) = \left\{ \begin{array}{ll}
 0,& t = 0 \\ 
 1,& 0 < t \le 1
\end{array} \right. $$
which isn't an element of $C[0;1]$.
This might be a point in the solution. However, I don't know how to rigorously proceed with the correct definition of the coverage in a normed space (as $x_n - x$ isn't an element of the normed space). Please help me. Thank you for reading.

REPLY [3 votes]: Here is an alternative formal approach. Take any function $f\in C[0,1]$, and prove that
$$
\lim_{n\to\infty}\|f-x_n\|\geq \frac12
$$
by dividing into two cases: either $f(0)<\frac12$, or $f(0)\geq\frac12$. Thus $x_n$ cannot converge to $f$.