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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!

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  • $\begingroup$ The general statement about $V \not\cong V^*$ when $V$ is infinite-dimensional needs the axiom of choice. That does not mean a specific example needs the axiom of choice. See the recent post math.stackexchange.com/questions/3757844/…. Your vector space $F^\infty$ (sequences with finite support) is essentially the same as the example $F[X]$ on that page, with $F = \mathbf R$. $\endgroup$
    – KCd
    Jul 20, 2020 at 2:51
  • $\begingroup$ That statement is regarding the 'single' dual space. I was interested in the double dual, so I can't really see the connection to that question. $\endgroup$
    – learning_linalg
    Jul 20, 2020 at 3:00
  • $\begingroup$ Ah, I see. Why do you think an example has to be expressible in a concrete way when you intend to involve the axiom of choice? That is a nonconstructive axiom. For instance, in the countable direct product of fields $\prod_{n \geq 1} F$, the proper ideal $\bigoplus_{n \geq 1} F$ of sequences with finite support has to be contained in a maximal ideal by the axiom of choice, but it is unrealistic to expect to see an actual down-to-earth example of such a maximal ideal. $\endgroup$
    – KCd
    Jul 20, 2020 at 3:07

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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.

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  • $\begingroup$ This example is really interesting! I certainly would not be able to come up with this. Would there perhaps be a simpler argument to show $E$ (the evaluation map defined in the question) is not surjective? $\endgroup$
    – learning_linalg
    Jul 20, 2020 at 6:53

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