Show that $X$ and $Y$ are closed linear spaces and $\overline{X+Y}=E$. Question: Let $E=l^{1}$ (the space of real, absolutely convergent series) and define
$$X=\{x=(x_{n})_{n\geq1}\in E:x_{2n}=0 \ \forall n\geq 1\}$$
and
$$Y=\{y=(y_{n})_{n\geq1}\in E:y_{2n}=\frac {1}{2^{n}}y_{2n-1} \ \forall n\geq 1\}.$$
Show that $X$ and $Y$ are closed linear spaces and $\overline{X+Y}=E$.
My attempt: I've proved that $\overline{X}=X\subset E$ and also $\overline{Y}=Y\subset E$. I also proved that $X$ and $Y$ are convex spaces. I defined $Z=\{z=x+y: x\in  X \text{ and } y\in Y\}$. And I tried to build new subsequences in order to prove the equality required. However I'm struggling with this. Would someone help me?
P.S. My doubt is different from this other question I found here
 A: Try to write a general $z \in \ell^1$ as $z = x+y$ with $x \in X$ and $y \in Y$. Then since $x_{2n} = 0$, it is easy to see that the only candidates are:
$$\begin{cases}
y_{2n} = z_{2n},\\
y_{2n-1} = 2^nz_{2n},
\end{cases} \qquad\qquad \begin{cases}
x_{2n} = 0,\\
x_{2n-1} = z_{2n-1}-2^nz_{2n},
\end{cases}$$
Now, if $z$ is a finitely-supported sequence, then the above relations yield well-defined elements $x\in X$ and $y \in Y$ such that $z =x+y$.
Therefore
$$\text{finitely-supported sequences} \subseteq X+Y \implies \ell^1 = \overline{\text{finitely-supported sequences}} \subseteq \overline{X+Y} \subseteq \ell^1$$
so $\overline{X+Y} = \ell^1$.
This exercise is interesting because $X+Y \ne \ell^1$, which means that $X+Y$ is a sum of two closed subspaces but it is not closed.
Namely, if we take $z = \left(\frac1{n^2}\right)_n \in \ell^1$ then the above relations yield
$$\begin{cases}
y_{2n} = \frac1{4n^2},\\
y_{2n-1} = \frac{2^{n-2}}{n^2},
\end{cases} \qquad\qquad \begin{cases}
x_{2n} = 0,\\
x_{2n-1} = \frac1{(2n-1)^2} - \frac{2^{n-2}}{n^2},
\end{cases}$$
which are not $\ell^1$-sequences so $z \notin X+Y$.
