Roughly speaking, the difference between the pointwise convergence and the uniform convergence is the N's dependence on x. So for pointwise convergence in domain $E$:
$$ \exists N \text{ s.t. } |f_n(x)-f(x)|<\epsilon \text{ for } n>N, x \in E, \epsilon >0$$
For uniform convergence to work, we add the condition $\forall x$. ie:
$$ \exists N \text{ s.t. } |f_n(x)-f(x)|<\epsilon \text{ for } n>N, \forall x \in E, \epsilon >0$$
In the former, N can depend on x, while in the latter, it can only depend on epsilon. The thing is, can't we choose the greatest of all Ns in the former? E.g: if:
$$ |f_n(x_1)-f(x_1)|<\epsilon \implies N\geq 2 $$
$$ |f_n(x_2)-f(x_2)|<\epsilon \implies N\geq 6 $$
$$ |f_n(x_3)-f(x_3)|<\epsilon \implies N\geq 3 $$
can't we choose $N \geq 6$, such that all of the following statements are true? Hence implying uniform convergence?
Best Answer
If $E$ consists of finitely many points, then what you're saying works. But generally you will be interested in infinite domains, in which case the maximum you're considering need not exist.
For example, if $E=(0,1]$, and
$$ |f_n(1)-f(1)|<\epsilon \implies N\geq 1 $$ $$ |f_n(1/2)-f(1/2)|<\epsilon \implies N\geq 2 $$ $$ |f_n(1/3)-f(1/3)|<\epsilon \implies N\geq 3 $$ $$ \dots$$ $$ |f_n(1/k)-f(1/k)|<\epsilon \implies N \geq k $$
then there is no single $N$ you can choose that will work for all $x$.