Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 22H / c / Solution

The finite truncations of $x=(1,1/2,1/3,\dots )$ lie in $c_{00}$ and converge in the supremum norm to $x\notin c_{00}$. Thus $c_{00}$ is not closed. Every element of $c_0$ is the norm limit of its finite truncations, while $c_0$ is closed and contains $c_{00}$. Hence $\boxed{\overline{c_{00}}=c_0}$.