A question on pointwise convergence. The function $f_n(x):[-1,1] \to \mathbb{R}, \, \, \,f_n(x) = x^{2n-1}$ tends pointwise to the function $$f(x) = \left\{\begin{array}{l l}1&\textrm{if} \quad x=1\\0&\textrm{if} \quad -1<x<1\\-1&\textrm{if} \quad x=-1\end{array}\right.$$ but not uniformly (for obvious reasons as $f(x)$ isn't continuous). But then surely in this case $||f_n(x) - f(x)||_\infty \to 0$ as for any $x \in (-1,1)$ and any $\epsilon > 0$ you can make $n$ large enough so that the max distance between $f_n(x)$ and $f(x)$ at that particular $x$ is less than $\epsilon$? What am I doing wrong? 
Thanks! 
 A: Your argument


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*for any $x \in [-1,1]$ and any $\epsilon > 0$ you can make $n$ large enough so that the max distance between $f_n(x)$ and $f(x)$ at that particular $x$ is less than $\epsilon$


is a great proof that for any $x$, the sequence $f_n(x)$ tends to $f(x)$. In other words, it's a proof that $f_n$ tends to $f$ pointwise.
But the assertion $||f_n(x) - f(x)||_\infty \to 0$ is the assertion that $f_n$ tends to $f$ uniformly on $[-1,1]$, which as you've noted is false.
The difference in the two statements (hence in how you'd need to prove them) is in the order of quantifiers. For pointwise convergence, given any $\epsilon>0$, it's "for all $x$, there exists $n$ such that..." - in other words, $n$ can depends on $x$ as well as $\epsilon$. For uniform convergence, given any $\epsilon>0$, it's "there exists $n$ such that for all $x$, ..." - in other words, $n$ cannot depend on $x$.
If you try reorganizing your proof so that you have to choose $n$ before $x$ is given, then you'll see the proof break down.
