Let B[0,1] be the set of bounded functions on [0,1 equipped with the supremum metric. Let ($f_n$) be the sequence in B[0,1] defined by,
$f_n(x)$ = $\Bigg\{$ 1 - $nx$ if 0 $\leq$ $x$ $\leq$ $\frac 1n $ or 0 if $\frac 1n$$\leq$ $x$ $\leq$ 1
Prove that ($f_n$) does not converge in {$B$[0, 1], d}.
After drawing what the graph of the function looks like, I can get a picture of it in my head but I am unsure of what steps to take to prove it. Help would be much appreciated!
Observe that $f_n(x) \to 0 $ as $n \to \infty$ for any $x\in (0,1]$ and $f_n(0) = 1$ for all $n\in \mathbb{N}$. Hence, if $\{f_n\}$ converges to some function $f\in B[0,1]$ in sup. metric, it has to be the function $$f_0(x) = \begin{cases} 1, \text{ if } x = 0 ,\\ 0, \text{ if } x\in(0,1] \end{cases}$$ since convergence in supremum metric implies point-wise convergence. But then $$ \sup\limits_{x\in [0,1]} |f_n(x) - f_0(x) | \geq |f_n(\frac{1}{2n})| = \frac 12 , $$ which contradicts to the assumption that $\{f_n\}$ converges to $f_0$ in supremum metric.