Comparing two $\sigma$-algebras on $B(\ell^1)$

Question

Let us consider $B(\ell^1)$, bounded linear operators on $\ell^1$. We recall the weak operator topology, denoted by $w$, on $B(\ell^1)$ is determined as follow

$$w-\lim T_i=T \Longleftrightarrow \lim\langle (T_i-T)x,f\rangle=0$$ for every $x\in \ell^1 , f\in\ell^{\infty}$.

Let $M$ be the Borel sigma algebra coming from the weak operator topology on $B(\ell^1)$ and let $M_0$ be the $\sigma$-algebra generated by the $w$-basic neighborhoods in $B(\ell^1)$.

Q. Do we have $M=M_0$?

Let us give some information concerning this problem which may be helpful. It is proved by Ju Myung Kim that the weak operator topology on the unit ball of $B(X)$ is relatively second countable if $X^*$ is separable. By an interesting argument in enter link description here, Matthew Daws showed the converse is also valid. So we have that:

$X^*$ is separable if and only if the weak operator topology on the unit ball of $B(X)$ is relatively second countable.

It is clear that, if $X^*$ is separable then $M_0=M$. I feel the converse is not true and $B(\ell^1)$ may serve a counterexample, however I have no clear proof.

With taking account to the comments mentioned in this problem, another approach of this question is:

Is $(B(\ell^1),w)$ hereditary lindelof?

Answer 1

Combination of the following two facts implies that $(B(\ell^1),w)$ is hereditary Lindelof.

1- $(B(\ell^1),sot)$ is hereditary Lindelof.

2- (Kuratowski and Sierpinski [1921]) A regular space $X$ is hereditary Lindelof if and only if every uncountable set $A\subseteq X$ has a condensation point which is contained in $A$