Some subspaces of $l^1$ are closed

Question

Let $l^1$ be the $\mathbb{R}$-Banach space of all sequences such that $\|x\|=\sum\limits_{i=1}^\infty |x_i|<\infty$ and consider the subset $$X=\{x_n\in l^1: x_{2n}=0\quad n>0\},\qquad Y=\{y_n\in l^1: y_{2n}=\frac{1}{2^n}y_{2n-1}\quad n>0\}$$

1. Show that $X,Y$ are closed subspaces of $l^1$

2. Assume that $\omega=(w_k)_{k\geq 1}\in l^1$ is defined by $$w_{2n-1}=0,\quad w_{2n}=\frac{1}{2^n}\quad n>0$$ verify that $w\not\in X+Y$

My attempt was the following, but I don't know if it is true:

I. $X$ is closed.

Since we have $\|x\|=\sum\limits_{i=1}^\infty |x_i|<\infty$ for all sequences in $l^1$, we also have $\|t\|=\sum\limits_{i=1}^\infty |t_i|<\infty$ for all $t_i\in X$

Since they are absolut convergent, we can use any summation order and $\|t\|=\sum\limits_{q=1}^\infty |t_q|<\infty$ will still be true.

So we choose a decreasing order $$t_1\geq t_2 \geq \dots \geq 0 \geq 0 \geq \dots$$

Now we can rebuild ever sequence into a cauchy sequence $$\|t_n-t_{n+1}\|\geq \|t_{n+1}-t_{n+2}\|\geq \dots \geq \|0-0\|$$ And since we know $x_{2n}=0\in X$ we proved that $X$ is a closed subspace.

II. $Y$ is closed.

So $$\sum\limits_{n=0}^\infty \frac{C}{2^n}=2\cdot C$$ and since $C=y_n$ we know that $C<\infty$ hence $$\sum\limits_{n=0}^\infty \frac{C}{2^n}=2\cdot C<\infty$$

But how do I show that the limit of every cauchy sequence is in $Y$?

2. $w\not\in X+Y$

Now I'm not sure if $X+Y$ means that every sequence $q_n\in X+Y$ looks like this $q_n=x_n+y_n$ or if $w_n=x_n+y_m$ for $n\neq m$ is also included. I assume only sequences of the form $q_n$ are included. Then clearly $q_{2n}=y_{2n}$ $$q_{2n+2}=y_{2n+2}=\frac{1}{2^n}y_{2n-1}=\frac{1}{2^{n+2}}y_{2n}=\frac{1}{2^{n+2}}q_{2n}$$ and

So clearly $q_{2n+2}< q_{2n}$ thus $z_n= w$ only if $y_{n}=1\quad \forall n$

But since then $\|y\|=\sum\limits_{i=1}^\infty |y_i|=\infty$ we have that $y\not\in l^1$, hence $w\neq z_n$ and $w\not\in X+Y$

Answer 1

There are easier ways. For instance, $$ X=\bigcap_{k=1}^\infty\{(x_n)\in\ell^1:x_{2k}=0\}, $$ and each $\{(x_n)\in\ell^1:x_{2k}=0\}$ is closed because it is the kernel of the continuous linear functional $(x_n)\mapsto x_{2k}$.

You can use a similar argument for $Y$.