Summable complex sequence

Question

Let $V = \{ v \in \ell^2 \setminus \lvert \sum_{n=1}^\infty v_n \rvert < \infty \}$ be the subspace of $\ell^2$ over $\Bbb C$ of summable complex sequences

Let $U=\overline{V}$ and be $u \in U-V$

I would like to know if $u$ is summable $\lvert \sum_{n=1}^\infty u_n \rvert < \infty$

Answer 1

since the zero sequence is in V, if every $u\in U-V$ is summable (in your sense) then $\bar{V}=V$. And reciprocally.

Answer 2

It seems that, in this context "minus" is the Minkowski operation ($-$ is different from $\setminus$). Then, unless I missed something, $U=\bar{U}=V$ (these are vector subspaces), one has $U−V=U$ (otherwise $U\setminus V=\emptyset$ and this should be meaningless). All this holds with the stronger requirement $\sum_{n\geq 1}|u_n|<+\infty$ (and $U=\ell^2\cap \ell^1=\ell^1$ in this case).