 Completeness of the smooth-function metric

If $(f_n)$ is Cauchy in the smooth-function metric, each derivative sequence $(f_n^{(r)})$ converges uniformly to a continuous function $g_r$. Passing to the limit inshows that $g_r^{\prime }=g_{r+1}$. Hence $g_0$ is smooth and $f_n\to g_0$ in the metric.