Is $\ell^1(\mathbb{N})^*$ separable? If so, why?

Question

I have reason to believe that the dual of $\ell^1$ is separable but I cannot prove it easily since $\ell^1$ is not reflexive. I have also tried searching around but have not had any success. If anyone would be able to shed a light on this I would be very grateful.

Answer 1

If $A\subset\Bbb N$ define $f_A\in\ell_\infty=(\ell_1)^*$ by $$f_A=\chi_A.$$What is $||f_A-f_B||_\infty$ if $A\ne B$? How many subsets does $\Bbb N$ have?

Revised version of the same answer, not assuming we know the dual is $\ell_\infty$: if $A\subset\Bbb N$ define $\lambda_A\in(\ell_1)^*$ by $$\lambda_A x=\sum_{j\in A}x_j.$$What is $||\lambda_A-\lambda_B||$ if $A\ne B$? How many subsets does $\Bbb N$ have?

Details: First, the answer is that $||\lambda_A-\lambda_B||=1$ if $A\ne B$. Because if $C=A\setminus B$ and $D=B\setminus A$ then $$\lambda_A x-\lambda_Bx =\sum_{j\in C}x_j-\sum_{j\in D}x_j;$$since $C\cap D=\emptyset$ this shows that $|\lambda_A x-\lambda_Bx|\le\sum|x_j|=||x||$, so $||\lambda_A-\lambda_B||\le 1$. On the other hand, say $k$ is in one of $A$ or $B$ but not the other. Let $x_j=1$ if $j=k$, $0$ if $j\ne k$. Then $|\lambda_A x-\lambda_B x|=1=||x||$, so $||\lambda_A-\lambda_B||\ge1$.

Hence if $A\ne B$ then $B(\lambda_A,1/2)\cap B(\lambda_B,1/2)=\emptyset$, because if not then if $\lambda$ is in the intersection the triangle inequality shows that $||\lambda_A-\lambda_B||\le||\lambda_A-\lambda||+||\lambda-\lambda_B||<1/2+1/2=2/2=1$.

So $(B(\lambda_A,1/2))_{A\subset\Bbb N}$ is an uncountable family of disjoint nonempty open sets. It follows that any dense set is uncountable: Say $D$ is dense. The definition of the word "dense" shows that for every $A\subset\Bbb N$ there exists $x_A\in D$ such that $x_A\in B(\lambda_A,1/2)$. Since the $B(\lambda_A,1/2)$ are disjoint we must have $x_A\ne x_B$ for $A\ne B$. So $A\mapsto x_A$ is an injective mapping from the power set of $\Bbb N$ to $D$, and since the power set of $\Bbb N$ is uncountable this shows $D$ is uncountable.