Hello my lecturer gave a proof that $\displaystyle f_n(x)=x^n$ is not uniformly convergent to $\displaystyle f(x)=0\ \mbox{if} \ x<1\ \mbox{and}\ 1\ \mbox{if}\ x=1$ on $\displaystyle [0,1]$. Which went as follows:

Given $\displaystyle \epsilon\in(0,\frac{1}{3})$ and suppose we have uniform convergence then:

$\displaystyle \exists N s.t. \forall n\geq N\forall x\in[0,1]:\ |f_n(x)-f(x)|<\epsilon$

Given such an N take n=N $\displaystyle x=(1-\epsilon)^{(\frac{1}{n})})<1$

then $\displaystyle |f_n(x)-f(x)|=|((1-\epsilon)^{(\frac{1}{n})^n)}=1-\epsilon>epsilon$ which is a contradiction and we are done.

However I am confused why this breaks down if the function is only defined on $\displaystyle [0,a)$ where $\displaystyle a\in(0,1)$ in which case it is uniformly convergent?

Thanks for any help