I'm revising some of my notes and I stumbled across this question. Let $V = F^\infty$, the set of sequences in a field with finite support, and let the evaluation map $E : V \rightarrow V^{**}$ be defined as $E(v)(f) = f(v)$ with $ f \in V^*$. It is easy to show that $E$ is injective and I know that it is not surjective, but I cannot seem to find any examples of elements in $V^{**}$ that are not mapped to by $E$ for some element in the domain.
I know that this would need some kind of the axiom of choice, but am lost as to how to proceed.
I would appreciate any and help in finding an example of this!
We use AC in the following way: for $V_1<V_2$ $F$-vectorspaces, any element of $V_1^\vee$ can be extended to an element of $V_2^\vee$. (Proof: partially order the extensions by domain inclusion and pick a maximal $f$ by Zorn. If $\operatorname{dom}f\neq V_2$ you can clearly make an extension by pick some $v\notin\operatorname{dom}f$ and send it to say $0\in F$, extending $f$ to a bigger domain $\operatorname{dom}f\oplus Fv$, contradiction.)
So we can declare what our element $\beta\in V^{\vee\vee}$ does: it sends $e_i^*$ to $0$ for every $i\in\mathbb{N}$, and it sends the element $\phi\in V^{\vee}$, $\phi((a_i))=\sum a_i$ (note $(a_i)\in V$ has finite support, so this sum makes sense) to $1\in F$. This $\beta$ defines an element of $[(1,1,1,\dots)F\oplus\bigoplus^\infty F]^\vee$ and so we can extend to some $\tilde\beta\in V^{\vee\vee}$. This $\tilde\beta$ is not in the image of the natural embedding $j\colon V\to V^{\vee\vee}$ because any element of $j(V)$ which vanishes on $e_i^*$ must be zero.