Let \((X,d)\) and \((Y,d)\) be metric spaces, let \(f:X\to Y\) be a function and let \(f_n:X\to Y\), \(n\in\mathbb N\) be a sequence of functions.
The sequence \((f_n)_{n\in\mathbb N}\) converges pointwise to \(f\), if for each \(x\in X\) and each \(\epsilon > 0\) there exists an \(N(x,\epsilon)\in\mathbb N\), such that \[d(f_n(x),f(x)) < \epsilon \text{ for all } n\ge N(x,\epsilon),\] i.e. if the sequence \((f_n(x))_{n\in\mathbb N}\) is convergent to \(f(x)\) for each \(x\in X\).
The sequence \((f_n)_{n\in\mathbb N}\) converges uniformly to \(f\), if for each \(\epsilon > 0\) there exists an \(N(\epsilon)\in\mathbb N\), such that \[d(f_n(x),f(x)) < \epsilon \text{ for all } x\in X\text{ and all } n\ge N(\epsilon),\] i.e. if the sequence \((f_n(x))_{n\in\mathbb N}\) is convergent to \(f(x)\) for each \(x\in X\) in such a way that the distance from \(f_n(x)\) to \(f(x)\) for sufficient big index \(n\) does not exceed the threshold value \(\epsilon\) for all \(x\in X\).
Please note that the difference between pointwise and uniform convergence is that in the first case the index \(N\) depends on both, \(x\) and \(\epsilon\), while in the latter case it only depends on \(\epsilon\).
