Let $V$ a finite dimensional vector space over $F$, $S \subset V$ and $\Phi: V \to V^{**}$ the isomorphism given by $\Phi(v) = \phi_v$, where $\phi_v:V \to F$ is given by $\phi_v(f) = f(v).$ Show that $S$ is a generator of $\Phi^{-1}((S^{0})^{0})$.
Addictional information: $S^0= \{f:V \to F: f(s) = 0, \forall s \in S\}$ and therefore $(S^{0})^{0} = \{g:V^{*} \to F: g(f) = 0, \forall f \in S^{0} \}$.
So, I have no Idea how to prove that.
I know that if $v \in \Phi^{-1}((S^{0})^{0})$, then $\Phi(v) \in (S^0)^0$ and $\Phi(v)=\phi_v$ where $\phi_v: V^{*} \to F$ is given by $\phi(f) = f(v)$. Since $\phi_v \in (S^{0})^{0}$ we have $\phi_v(f) = 0$, for all $f \in S^0$.
I can't see how it implies that $v$ can be written as a linear combination of elements in $S$.
Can you help me?
Note that $\operatorname{span} S = \Phi^{-1}((S^0)^0)$ if and only if $\Phi(\operatorname{span} S) = (S^0)^0$. Now, as $\Phi(S) \subseteq \Phi(\operatorname{span} S)$ and $\Phi(\operatorname{span} S)$ is a subspace of $V^{**}$ that contains $\Phi(S)$, it follows that $\operatorname{span}(\Phi(S)) \subseteq \Phi(\operatorname{span} S)$. Also, the inclusion $\Phi(\operatorname{span} S) \subseteq \operatorname{span}(\Phi(S))$ it is easy to check.
So, it is enough to prove that $(S^0)^0 = \operatorname{span}(\Phi(S))$. You can check my proof of this fact right here.