Showing that the sum of two sets of sequences is dense in the sequence space $\ell^1$ I'm trying to solve the following problem. It's a part of problem 1.14 in Brezies' book on functional analysis.

Let $E=\ell^1$. Consider $$X=\{x=(x_n)_{n\geq1}\in E:x_{2n}=0\ \forall n\geq1\}$$ and $$Y=\{y=(y_n)_{n\geq1}\in E:y_{2n}=\frac{1}{2^n}y_{2n-1}\ \forall n\geq1\}.$$ Show that $\overline{X+Y}=E$.

Here, $\ell^1=\{x:\|x\|_1<\infty\}$ and $$\|x\|_1:=\sum_{k=1}^\infty|x_k|.$$ We have that $\overline{X+Y}\subseteq E$ as $E$ is the ambient space of $X+Y$, but I'm unsure on how to prove that $E\subseteq\overline{X+Y}$. Maybe an argument by contradiction would be useful. That is, assume that there exists a $z\in E$ such that $z\notin\overline{X+Y}$ and then arrive to some contradiction.
The problem to continue here is that I don't really know what the closure $\overline{X+Y}$ would be. Thus, I don't really know how to proceed with the other part either.
I know that we're working in a sequence space, so every element is a sequence. The sequences may have infinitely many elements, so we're in an infinite dimensional space. I know how the norm $$\|(x_n)_{n\in\Bbb{N}}-(y_n)_{n\in\Bbb{N}}\|:=\sum_{k=1}^\infty|x_k-y_k|$$ works, but not really what the sets of the intersection of closed sets would look like, or how they would influence the outcome of the closure. Here, both $X$ and $Y$ are closed.
I'm also a bit curious on how addition of sets consisting of sequences works. Is the resulting set such that every sequence point in every sequence is the sum of the sequences in the initial sets $X$ and $Y$?
 A: I'll use overlines for the elements of the space and indices for their components, so $\overline{x}\in\ell^1$ is the sequence $x_1,x_2,x_3,\dots$
Let $c_{00}$ be the subspace of $\ell^1$ containing sequences which have only finitely many nonzero terms, this is a dense subspace, if $\overline{x}\in\ell^1$ it can be approximated by the following sequence of elements of $c_{00}$:
$$\begin{align}
\overline{y}_1=x_1&,0,0,0,\dots\\
\overline{y}_2=x_1&,x_2,0,0,\dots\\
\overline{y}_3=x_1&,x_2,x_3,0,\dots\\
\overline{y}_4=x_1&,x_2,x_3,x_4,\dots\\
\end{align}
$$
Now $\|\overline{x}-\overline{y}_i\|_1$ goes to zero since to calculate it you're summing the tails of a converging series.
So we want to show $c_{00}\subseteq X+Y$, this is enough since $\overline{c_{00}}=\ell^1$.
Let $\overline{x}$ be in $c_{00}$ and consider $\overline{y}=x_1,x_1,2x_2,x_2,4x_3,x_3,\dots$ so that $y_{2n}=x_{2n}$ and $y_{2n-1}=2^n x_{2n}$, this is an element of $Y$ since it has only finitely many nonzero terms. But now $\overline{z}=\overline{x}-\overline{y}\in X$ and $\overline{x}=\overline{z}+\overline{y}$, so $\overline{x}\in X+Y$, as we wanted to show.
(Note that the notation in my answer is somewhat awkward, usually $c_{00}$ denotes the space of sequences with finitely many nonzero terms, but as a subspace of $\ell^{\infty}$ rather than of $\ell^1$.)
