Weak* and product topology are the same on $B_1((\ell^1)^*) = D^N$ (p19 (C))

Let \mathbb{D} = \{ x \in \mathbb{C} : |x| \le 1 \}. Show that B_1(\ell^\infty = (\ell^1)^*) = \mathbb{D}^\mathbb{N} and that the weak*-topology on B_1(\ell^\infty) is the same as the product topology on \mathbb{D}^\mathbb{N}.

Remarks
Without further comment, I will interchange between S \in (\ell^1)^* and (s_n) \in \ell^\infty.  Also, I will sometimes abbreviate B_1( (\ell^1)^* ) to simply B_1.  A basic open neighborhood of the weak*-topology on B_1 is
V_{a_1, \ldots, a_n ; \epsilon ; T} = \{ S \in B_1( (\ell^1)^* ) : |S(a_1) - T(a_1)| < \epsilon, \ldots,|S(a_n) - T(a_n)| < \epsilon\} ,
for some a_i = (a_{in}) \in \ell^1, \epsilon > 0, and T \in B_1. Also, e_n \in \ell^\infty is defined by m \neq n \Rightarrow e_n(m) = 0, and e_n(n) = 1.

Proof.
B_1(\ell^\infty) = \{x \in \ell^\infty : \| x \| \le 1 \} = \mathbb{D}^\mathbb{N}.

Let U be a basic open neighborhood of (t_n) in the product topology on \mathbb{D}^\mathbb{N}. That is, U = \prod_{i = 0}^\infty U_i, such that U_i = \mathbb{D} if i \notin \{i_1, \ldots, i_n \} and
U_{i_j} = \{ x \in \mathbb{D} : | x - t_{i_j} | < \epsilon_j \}.
Note that |(S - T)(e_{i_j})| = |s_{i_j} - t_{i_j}|. Let \epsilon = \min \{ \epsilon_j \}.  Then V_{e_{i_1}, \ldots, e_{i_n} ; \epsilon ; T } \subset U.

Then, let V be a basic open neighborhood of B_1 in the weak*-topology, that is V = V_{a_1, \ldots, a_n ; \epsilon ; T}. Now, \exists M > 0, such that \forall i, \sum_{n = M}^\infty |a_{in}| < \epsilon / 4.  Let K > 0, such that K \ge \max \{ \| a_1 \|_1, \ldots, \| a_n \|_1 \}.  Let \eta = \frac{\epsilon}{2MK}. Choose (s_n) \in \mathbb{D}^\mathbb{N}, such that n < M \Rightarrow |s_n - t_n| < \eta.
Then
|(S - T)(a_i)| = \left| \sum_{n = 0}^\infty (s_n - t_n)a_{in} \right| \le \sum_{n = 0}^\infty |(s_n - t_n)a_{in}|
= \sum_{n = 0}^{M - 1} |(s_n - t_n)a_{in}| + \sum_{n=M}^\infty |(s_n - t_n)a_{in}|
< \epsilon / 2 + \epsilon / 2 = \epsilon.

Thus, if U = \prod_{i = 0}^\infty U_i, where i < M \Rightarrow U_i = \{x \in \mathbb{D} : |x - t_i| < \eta \} and i \ge M \Rightarrow U_i = \mathbb{D}, then U \subset V.

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