# Elements of double dual that are not evaluation maps

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!

• 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 $F = \mathbf R$.
– KCd
Jul 20, 2020 at 2:51
• 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. Jul 20, 2020 at 3:00
• 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.
– KCd
Jul 20, 2020 at 3:07

We use AC in the following way: for $$V_1 $$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.
• 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? Jul 20, 2020 at 6:53