TITLE: What's the reason why this sequence of function doesn't converge uniformly to $f$?
QUESTION [7 upvotes]: Consider $f_n = \sqrt[n]{x}$ on $[0,1]$
So it converges to the step function $f = 0$ if $x = 0$ and $f=1$ otherwise
I could see why it doesn't converge if i draw an epsilon rectangle over one part since for each $n$, $f_n$ lies completely outside of the function. 
If I draw an epsilon rectangle over the whole $f$, I don't see why this isn't uniform convergence
EDIT: I got this from Spivak, so keep it at that level please...
Added question: if $f_n\nrightarrow f$ for some $x \in \mathbb{R}$, can I conclude that it is not uniformly convergent over $\mathbb{R}$?

REPLY [3 votes]: @sizz : It does look as if you don't know the difference between pointwise convergence and uniform convergence.
If for every $x\in\mathbb R$, $\lim_{n\to\infty} f_n(x)=f(x)$, that is pointwise convergence of $f_n$ to $f$.  That's what you've got here (although in order for the statement to be true just as you've stated it, you ought to have $\sqrt[n]{|x|}$).
Now look at the supremum over all $x\in\mathbb R$ of the distance between $f_n(x)$ and $f(x)$.  If that goes to $0$ as $n\to\infty$, then that is uniform convergence of $f_n$ to $f$.  That doesn't happen here.  You have
$$
f(x) = \begin{cases} 1 & \text{if }x\ne 0, \\ 0 & \text{if }x=0 \end{cases}
$$
and $f_n(x)=\sqrt[n]{|x|}$.  As $x\to0$, you have the distance between $f_n(x)$ and $f(x)$ approaching $1$.  So the "uniform distance" between $f_n$ and $f$ is $1$.  And that doesn't go to $0$ as $n\to\infty$.  So you don't have uniform convergence.