Suppose $(f_n)$ is a sequence of functions and $f_n:[0,1) \to [0,1)$ such that $f_n(x) = x^n$. We need to prove that it converges pointwise but not uniformly.
$f_n$ converges to $f$ pointwise, if there exists an $N(\epsilon , x)$, dependent on both $\epsilon > 0$ and $x \in E$, such that $\mid f_n(x) - f(x)\mid < \epsilon$ when $n \ge N(\epsilon , x)$.
So, for all $n \ge N(\epsilon , x)$ $$\mid x^n \mid = x^n \le x^N < \epsilon$$ Take $N > \dfrac{\ln \epsilon}{\ln x}$. This proves that the sequence converges pointwise. Now, my question is since we are able to find $N$ dependent on both $\epsilon$ and $x$, wouldn't this imply that the sequence does not converge uniformly (because in the statement of uniform convergence $N$ is independent of $x$)?