Let \(E\) be a set, and let \(f_n\colon E\to\CC\) for \(n\geq 1\) and \(f\colon E\to\CC\) be functions. The sequence \((f_n)\) converges uniformly to \(f\) on \(E\) if for every \(\varepsilon>0\) there exists \(N\in\NN\) such that \(\abs{f_n(x) - f(x)} < \varepsilon\) for all \(n\geq N\) and all \(x\in E\).
Let \(E\), \((f_n)\), and \(f\) be as in the definition, and put \(\norm{f_n - f}_\infty \coloneqq \sup_{x\in E}\abs{f_n(x) - f(x)}\in[0,\infty]\). Then \((f_n)\) converges uniformly to \(f\) on \(E\) if and only if \(\norm{f_n - f}_\infty\to 0\) as \(n\to\infty\).
If \((f_n)\) converges uniformly to \(f\), then for \(\varepsilon>0\) and \(N\) as in the definition, \(\norm{f_n - f}_\infty \leq \varepsilon\) for all \(n\geq N\), so \(\norm{f_n - f}_\infty\to 0\). Conversely, if \(\norm{f_n - f}_\infty\to 0\), then for \(\varepsilon>0\) there exists \(N\) with \(\norm{f_n - f}_\infty < \varepsilon\) for all \(n\geq N\), and then \(\abs{f_n(x) - f(x)}\leq\norm{f_n - f}_\infty<\varepsilon\) for all \(n\geq N\) and \(x\in E\).