Definition 1. Suppose that $\left(f_{n}\right)$ is a sequence of functions $f_{n}: A \rightarrow \mathbb{R}$ and $f: A \rightarrow \mathbb{R}$. Then $f_{n} \rightarrow f$ pointwise on $A$ if, for every $\epsilon>0$, there exists $N_{x,\epsilon} \in \mathbb{N}$ such that
$n>N_{x,\epsilon}$ implies that $\left|f_{n}(x)-f(x)\right|<\epsilon$ for all $x \in A$.
Definition 2. Suppose that $\left(f_{n}\right)$ is a sequence of functions $f_{n}: A \rightarrow \mathbb{R}$ and $f: A \rightarrow \mathbb{R}$. Then $f_{n} \rightarrow f$ uniformly on $A$ if, for every $\epsilon>0$, there exists $N_{\epsilon} \in \mathbb{N}$ such that
$n>N_{\epsilon}$ implies that $\left|f_{n}(x)-f(x)\right|<\epsilon$ for all $x \in A$.
Suppose in definition $2$ we take $N_{\epsilon}=\max(N_{x,\epsilon})$ this does not means that pointwise convergence implies uniform convergence?
Best Answer
Because $\{N_{x,\varepsilon}\mid x\in A\}$ is an infinite set (in general), and therefore you cannot be sure that it has a maximum (or even that it has an upper bound).
If, for instance, $A=[0,1]$, if$$f_n(x)=\begin{cases}1&\text{ if }x=1/n\\0&\text{ otherwise,}\end{cases}$$and if $\varepsilon=\frac12$, you can take$$N_{x,\varepsilon}=\begin{cases}n+1&\text{ if }x=1/n\text{ for some }n\in\Bbb N\\1&\text{ otherwise.}\end{cases}$$It is then clear that $\{N_{x,\varepsilon}\mid x\in A\}$ has no upper bound.