A problem regarding a sequence in $C[0;1]$ 
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.
 A: You were nearly there. Suppose that $(x_n)_{n\in\mathbb N}$ converges in $C\bigl([0,1]\bigr)$ to some function $x$. Then, for each $t\in[0,1]$,$$x(t)=\lim_{n\to\infty}x_n(t)=\begin{cases}1&\text{ if }x\in(0,1]\\0&\text{ otherwise.}\end{cases}$$But there is no such function in $C\bigl([0,1]\bigr)$.
Note that the equality $x(t)=\lim_{n\to\infty}x_n(t)$ follows from the fact that you're using the norm that you are using. If, say, $\|x\|=\int_0^1\bigl|x(t)\bigr|\,\mathrm dt$, this would not work.
A: 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 $f$.
A: You can prove it by contradiction:
If there exist $\xi$ such that $x_n \to \xi$ then:


*

*As uniform convergence implies point-wise convergence for all $t$, $\xi(t)=x(t)$, and $\xi=x$ is not continuous.

*As a uniform limit of continuous function is continuous, $\xi$ is continuous.
So there is a contradiction.
