The following is from Real Analysis with Real Applications by Davidson and Donsig
Consider $f_{n}(x) = x^{n}$ for $x \in [0,1]$. It is easy to check that
$$ \lim_{n \to \infty} f_{n}(x)=\lim_{n \to \infty} x^{n} = \begin{cases} 0, & \text{for}\hspace{0.1cm}0 \le x < 1 \\ 1, & \text{for}\hspace{0.1cm}x=1 \end{cases} $$
Thus the pointwise limit is the function $\chi_{\{1\}}$, the characteristic function of the point $\{1\}$. The functions $f_{n}$ are polynomials,, and hence not only continuous but even smooth; while the limit function has a discontinuity at the point $1$.
For each $n \ge 1$, we have $f_{n}(1)=1$ and so
$$\Vert f_{n}-\chi_{\{1\}} \Vert _{\infty} = \sup_{0 \le x < 1} |x^{n}-0|=1.$$
I'm trying to understand how we get the following statement: $$\Vert f_{n}-\chi_{\{1\}} \Vert _{\infty} = \sup_{0 \le x < 1} |x^{n}-0|=1$$
By pointwise convergence, for all $x \in [0,1]$, $\lim_{n \to \infty}f_{n}(x)=f(x)=1$ since $f_{n}(1)=1$ for all $n \ge N$, but I'm not sure how that fits together with $\Vert f_{n}-\chi_{\{1\}} \Vert _{\infty} = \sup_{0 \le x < 1} |x^{n}-0|=1$.
$x^{n} \leq 1$ so $\sup x^{n} \leq 1$. It is greater than or equal to $y^{n}$ for any $y <1$. Taking limit as $y \to 1$ we see that the supremum is also greater than or equal to $1$.
A detailed argument: Let us prove by contradiction that $\sup_{0 \leq x<1} x^{n} \geq 1$. If $\sup_{0 \leq x<1} x^{n} < 1$ then take a number $r$ between $\sup_{0 \leq x<1} x^{n}$ and $1$. We have $x^{n} <r$ for all $x \in [0,1)$. Hence $x<r^{1/n}$ for all $x \in [0,1)$. But this cannot be true since $r^{1/n}$ is less than $1$: just take $x=\frac {1+r^{1/n}} 2$ to get a contradiction.