So here is my question,
I want to prove that $\ell^1$ is separable. So i need to show that there exists a countable dense subset in $\ell^1$.
Since I am not sure if my idea was right i hoped someone could look over my proof.
My idea was,
Let $Q:=\{(x_n)\subset\mathbb Q:\exists j\in\mathbb N $ s.t $ \forall n\geq j:x_n=0\}$.
Clearly $Q$ is countable so it is left to show that $Q$ is dense in $\ell^1$.
For that we will show that for all $x\in\ell^1\backslash A$ we have,
$$\{y\in\ell^1:\|x-y\|_{\ell^1}=\sum_{n=1}^{\infty}|x_n-y_n|<\epsilon\}=:B_{\epsilon}(x)\bigcap A\neq\emptyset$$
for an arbitrary $\epsilon>0$.
Notice that there exists $n_0\in\mathbb N :\sum_{n=n_0}^{\infty}|x_n|<\epsilon/2$.
Furthermore since $\mathbb Q$ is dense in $\mathbb R$, for every $x_j$ with $j\in\{1,…,n_0-1\}$ we can find $q_j\in\mathbb Q$ s.t $|x_j-q_j|<\frac{\epsilon}{(n_0-1)2}$.
For this $q_j$'s we define,
$$(q_n):=(q_1,…,q_{n_0-1},0,0,…)\in A$$
Since,
$$\|x-q\|=\sum_{n=1}^{\infty}|x_n-q_n|=\sum_{n=1}^{n_0-1}|x_n-q_n|+\sum_{n=n_0}^{\infty}|x_n|<(n_0-1)\frac{\epsilon}{(n_0-1)2}+\epsilon/2=\epsilon$$
it follows that $\overline{A}=\ell^1$ what proves that $\ell^1$ is separble.
Since I am not sure if there is no mistake I wanted to ask if someone could look over my prove and correct it if there are mistakes.
Thanks a lot!
Best Answer
Yes, your proof is good.
In terms of writing, you start using a fixed $x\in\ell^1$ without saying so. Also, the sentence that says "For that we will show..." doesn't really make sense; I understand it because I know how to prove the density, but otherwise it looks hard to understand.