Dual space whose predual is the closed linear span of $E$ in $X^*$. I want to prove Exercise 5 of the chapter "Weak Topology" of the book J.Diestel "Sequences and Series in Banach Spaces".
"Let $X$ be a Banach Space and $E \subseteq X^*$. Suppose $E$ separates the points of $X$ and $B_X$ is compact in the topology of pointwise convergence in $E$. Then $X$ is a dual space whose predual is the closed linear span of $E$ in $X^*$.
I know that this exercise is solved in "Exercises in Functional Analysis" by C.Costara and D. Popa, but they use the concept of nets. I want to know if there is a way to prove this without using that concept. I tried to follow the proof given on the book, but I'm not sure if I can use sequences instead of nets in this context.
The proofs starts like this, and they use nets only in this part:
Any element $x \in X$ generates an element $\hat{x} \in X^{* *}$. If we restrict $\hat{x}$ to $\operatorname{Sp}(E)$ we obtain the canonical operator $T: X \rightarrow(\overline{\operatorname{Sp}(E)})^{*}, T x=\left.\hat{x}\right|_{\overline{S p}(E)}$. Then $T$ is linear and for any
$x \in X$ we have the fact that
$$
\|T x\|=\left\|\left.\widehat{x}\right|_{\overline{S p}(E)}\right\| \leq\|\widehat{x}\|=\|x\|,
$$
i.e., $T$ is continuous, with $\|T\| \leq 1$. Suppose that $T x=0$. Then $\left.\widehat{x}\right|_{\overline{S p(E)}}=0$, and then $x^{*}(x)=0 \forall x^{*} \in E$, and by hypothesis this implies $x=0$. Therefore $T: X \rightarrow(\overline{\mathrm{Sp}(E)})^{*}$ is a linear bounded injective operator. We shall prove that $T$ is also surjective and for that we will use the compactness hypothesis. We denote by $\tau_{p}$ the topology of pointwise convergence on $E$, i.e., the topology generated by the family of seminorms $\left(p_{x^{*}}\right)_{x^{*} \in E}, p_{x^{*}}: X \rightarrow \mathbb{R}_{+}, p_{x^{*}}(x)=\left|x^{*}(x)\right| \forall x \in X$. We shall prove that $T:\left(B_{X}, \tau_{p}\right) \rightarrow\left((\overline{\mathrm{Sp}(E)})^{*}\right.$, weak $\left.^{*}\right)$ is continuous. Let $x_{0} \in B_{X}$ and $\left(x_{\delta}\right)_{\delta \in \Delta}$ be a net in $B_{X}$ such that $x_{\delta} \rightarrow x_{0}$ in the $\tau_{p}$ topology. Equivalently, $x^{*}\left(x_{\delta}\right) \rightarrow x^{*}\left(x_{0}\right) \forall x^{*} \in E$, and we obtain that $x^{*}\left(x_{\delta}\right) \rightarrow x^{*}\left(x_{0}\right) \forall x^{*} \in \operatorname{Sp}(E)$. Let now $x^{*} \in \overline{\operatorname{Sp}(E)}$. Then for $\varepsilon>0$ there is an $y^{*} \in \operatorname{Sp}(E)$ such that $\left\|y^{*}-x^{*}\right\|_{X^{*}}<\varepsilon / 3$, and therefore $\left|y^{*}\left(x_{\delta}\right)-x^{*}\left(x_{\delta}\right)\right|<\varepsilon / 3$ $\forall \delta \in \Delta\left(\left\|x_{\delta}\right\| \leq 1\right)$. But $y^{*}\left(x_{\delta}\right) \rightarrow y^{*}\left(x_{0}\right)$ and therefore there is $\delta_{\varepsilon} \in \Delta$ such that $\left|y^{*}\left(x_{\delta}\right)-y^{*}\left(x_{0}\right)\right|<\varepsilon / 3 \forall \delta \geq \delta_{\varepsilon}$. Then for any $\delta \geq \delta_{\varepsilon}$ we have
$$
\left|x^{*}\left(x_{\delta}\right)-x^{*}\left(x_{0}\right)\right| \leq\left|x^{*}\left(x_{\delta}\right)-y^{*}\left(x_{\delta}\right)\right|+\left|y^{*}\left(x_{\delta}\right)-y^{*}\left(x_{0}\right)\right|+\left|y^{*}\left(x_{0}\right)-x^{*}\left(x_{0}\right)\right|<\varepsilon
$$
Question: Can I use sequences instead of nets?
If its possible, I want to know why (I'm not familiarized with the concept).
If is not possible to change nets for sequences, Ideally I would like to have a full solution, but with hints to solve it, its enough for me.
If there is any typo in the solution, tell me and I will edit them.
 A: Let's start with the general case. If we have the set $X$, a family of topological spaces $(Y_s,\mathcal T_s)$, $s\in S$ and functions $f_s\colon X\to Y_s$ then there exists the coarsest topology $\mathcal T_{\{f_s\}}$ on $X$ for which all the functions $f_s$ are continuous. It's defined by the subbase
$$\mathcal S =\{f_s^{-1}(U_s):s\in S,\ U_s\in\mathcal T_s\}.$$ For example this is how we define product topology: it's the coarsest topology for which all the projections are continuous. This is also the case for weak and weak-star topologies.
Denote by $Y$ the space $\overline{\mathrm{Sp}(E)}$. For each $x^*\in Y$ we can define a linear mapping $\widehat {x^*}\colon Y^*\to \Bbb R$ as the evaluation at $x^*$, i.e. $\widehat {x^*}(\varphi)=\varphi(x^*)$ for all $\varphi\in Y^*$.
The space $(Y^*,\mathrm{weak}^*)$, is a space endowed with the coarsest topology for which all the functionals $\widehat {x^*}$ are continuous. This is the weak$^*$ topology.
Similarly, $\tau_p$ is the coarsest topology on $X$ for which all the functionals from $E$ are continuous. Observe that this implies that all the functionals from $\mathrm{Sp}(E)$ are continuous (as a finite linear combinations of continuous functionals). Now we prove that also all $x^*\in Y=\overline{\mathrm{Sp}(E)}$ also are continuous on $B_X$ (be carefull where we use this restriction).
To do this, take any  $x^*\in Y$ and consider any $x_0\in B_X$ and $\varepsilon>0$. We can find $y^*\in\mathrm{Sp}(E)$ such that $\left\|y^{*}-x^{*}\right\|_{X^{*}}<\varepsilon / 3$. From the continuity of $y^*$ there exists an open neighbourhood $U\subset B_X$ of $x_0\in B_X$ such that $|y^*(u)-y^*(x_0)|<\varepsilon/3$ for all $u\in U$. Therefore
$$|x^*(u)-x^*(x_0)|\leq |x^*(u)-y^*(u)|+|y^*(u)-y^*(x_0)|+|y^*(x_0)-x^*(x_0)|\leq $$ $$\|x^*-y^*\|\cdot\|u\|+\frac\varepsilon 3+\|x^*-y^*\|\cdot\|x_0\|\leq 2\cdot\frac \varepsilon 3\cdot 1+\frac\varepsilon 3 = \varepsilon.$$ This finishes the proof of continuity of all $x^*\in Y$ on $B_X$ endowed with topology $\tau_p$.
(Remark: this proof is exactly the same as the proof that the uniform limit of a sequence of continuous functions is continuous and it's not a coincidence, since $x^*$ is close to $y^*$ with respect to the uniform distance on $B_X$).
Now return to the general case. Suppose we have another topological space $(P,\mathcal T)$ and the function $F\colon (P,\mathcal T)\to (X,\mathcal T_{\{f_s\}})$. It is well known that $F$ is continuous if and only if all the functions $$f_s\circ F\colon (P,\mathcal T)\to (Y_s,\mathcal T_s)$$ are continuous.
Therefore, to verify continuity of $T\colon (B_{X}, \tau_{p})\to (Y^*,\mathrm{weak}^*)$ we need to show that all the compositions $\widehat {x^*}\circ T$ for $x^*\in Y$ are continuous.
Let's investigate these maps.
$$\left(\widehat {x^*}\circ T\right)(x) = 
\widehat {x^*}(T(x)) = \widehat {x^*}\left(\hat{x}|_Y\right) = (\hat{x}|_Y)(x^*) = \hat{x}(x^*) = x^*(x).$$
Now observe that the continuity of $x^*\in Y$ was previously shown, so we are done.
