Dimension of $(l^\infty)^*$ mod $l^1$.

Question

We can consider $l^1$ as a subspace of $(l^\infty)^*$ because for every $x \in l^1$, there exists a bounded linear functional $T_x: l^\infty \to \mathbb{R}$ defined by $T_x(y) = \sum_n x_n y_n$. Furthermore $l^1$ is closed in $(l^\infty)^*$. I am interested in the dimension of the quotient space $(l^\infty)^*$ mod $l^1$ but I don't know where to even start. I solved a few problems about dimensions of quotient spaces for sequence spaces by considering linearly independent vectors, but here the notion of dual confuses me.

Answer 1

Suppose that $\ell_\infty^*/\ell_1$ is finite dimensional, and let $\{x_1,...,x_n\}$ be a finite subset of $\ell_\infty^*$ such that the set $$\big\{x_1+\ell_1,...,x_n+\ell_n\bigr\}$$ is basis for the quotient $\ell_\infty^*/\ell_1$. Now, let $(e_n)$ be the standard Schauder basis of $\ell_1$. I.e. $e_n(k)=1$ when $k=n$ and $e_n(k)=0$ when $k\neq n$. Being a Schauder basis means that for every $x=(x(k))_{k=1}^{\infty}\in \ell_1$ we have that $$\biggl|\biggl|x-\sum_{n=1}^{k}x(n)e_n\biggr|\biggr|_1\to 0$$ as $k\to \infty$. In other words $\overline{Y}=\ell_1$ where $Y=span(e_m:\,m\in \mathbb{N})$ (This shows that $\ell_1$ is separable). Now, its easy to check that $$\tag{*}\ell_\infty^* = \overline{Y}\oplus Z$$ with $Z=span(x_k:\,1\leq k\leq n)$. Now, since both $\overline{Y}$ and $Z$ are separable by $(*)$ it follows that $\ell_\infty^*$ is separable, hence $\ell_\infty$ must be separable which cant be true. So, the quotient $\ell_\infty^*/\ell_1$ has infinite dimension.