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,


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!

  • $\begingroup$ After reading your proof, I can guess your $A$ is $Q$. But when I was going through your proof, such sudden change in symbol without explicit explanation really confused me. $\endgroup$ – Sam Wong May 29 '18 at 13:47

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.

  • $\begingroup$ Thanks for the comment! As you see or read english is not my first language. I used the sentence "For that we will show..." because I know that there are different techniques to prove that something is dense and I wanted to make sure that the reader knows how I want to show the density. I am trying to improve my "math-english" as good as it is possible, so I am really happy about comments like yours! $\endgroup$ – Thorben Apr 8 '14 at 23:19

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