$\displaystyle e_m(t)=e^{imt}, f_n=e_{-n} + n e_n (m \in Z, n \in N)$ and $\displaystyle A = \overline{Lin \{ e_n| n \in N\}}, B = \overline{Lin \{ f_n| n \in N\}}$. (Clousures in $\displaystyle L^2(-\pi, \pi))$. Are A+B closed? What is $\displaystyle \overline{A+B}$ ?
$\displaystyle e_m(t)=e^{imt}, f_n=e_{-n} + n e_n (m \in Z, n \in N)$ and $\displaystyle A = \overline{Lin \{ e_n| n \in N\}}, B = \overline{Lin \{ f_n| n \in N\}}$. (Clousures in $\displaystyle L^2(-\pi, \pi))$. Are A+B closed? What is $\displaystyle \overline{A+B}$ ?
If you include 0 in the natural numbers then A+B will contain all the trigonometric polynomials and therefore $\displaystyle \overline{A+B}= L^2(-\pi, \pi)$. If you do not allow 0 as a natural number then $\displaystyle \overline{A+B}= \{x\in L^2(-\pi, \pi): \int_{-\pi}^\pi\!\!\!x(t)\,dt=0\}$ (the orthogonal complement of the constant functions in $\displaystyle L^2(-\pi, \pi)$).
To see that A+B is not closed, you only need to find some element $\displaystyle x\in L^2(-\pi, \pi)$ such that $\displaystyle x\notin A+B$. For that, notice that every x in A+B must be of the form $\displaystyle \sum_{m\in\mathbb{Z}}x_me_m$ with $\displaystyle \sum_{n=1}^\infty n^2|x_{-n}|^2 < \infty$ (and of course $\displaystyle \sum_{m\in\mathbb{Z}}|x_m|^2<\infty$ to ensure that $\displaystyle x\in L^2(-\pi, \pi)$). So for example $\displaystyle x = \sum_{n=1}^\infty \frac{e_{-n}}n \notin A+B$.