TITLE: Does the fact that this vector space is not isomorphic to its double-dual require choice?
QUESTION [44 upvotes]: Let $V$ denote the vector space of sequences of real numbers that are eventually 0, and let $W$ denote the vector space of sequences of real numbers.  Given $w \in W$ and $v \in V$, we can take their "dot product" $w \cdot v$, and for any give $w \in W$, this defines a linear functional $V \to \mathbb{R}$.  In fact, under this association, we see that $W \cong V^{\ast}$.  Assuming choice, $W$ has a basis and has uncountable dimension, whereas $V$ has countable dimension.  So $\mathrm{dim} (V) < \mathrm{dim} (V^{\ast}) \leq \dim(V^{\ast \ast})$ so $V$ is not isomorphic to its double-dual.  In particular, the canonical map $\widehat{\cdot} : V \to V^{\ast \ast}$ defined by $\widehat{x} (f) = f(x)$ is not an isomorphism.
Question 1: Can you explicitly write down an element of $V^{\ast \ast}$ that isn't of the form $\widehat{x}$?
Question 2: What's the situation if we don't assume choice? (i.e. might the canonical map be an isomorphism, might $W$ not even have a basis, might $V$ be isomorphic to its double-dual but not via the canonical map, etc?)

In light of Daniel's and Yemon's comments, and Konrad's related question here, I'd like to reorganize my question(s).  So consider the following statements:

The canonical map $\widehat{\cdot} : V \to V^{\ast \ast}$ is injective but not surjective.
The canonical map is not an isomorphism.
There is no isomorphism from $V$ to its double-dual.
The double-dual is non-trivial.
$V^{\ast}$ has a basis.

This is the question I'm really interested in, and which clarifies and makes more precise my original Question 1:
Revised Question 1: Are there models without choice where (1) holds?  If so, can we find some $x \in V^{\ast \ast}$ witnessing non-surjectivity to describe a witness to non-surjectivity in choice models?  Are there models of choice where (1) fails?  
The following question is a more precise version of Question 2.  
Revised Question 2: Under AC, all 5 statements above are true.  There are $2^5 = 32$ ways to assign true-false values to the 5 sentences above that might hold in some model of ZF where choice fails.  Not all 32 are legitimate possibilites, some of them are incompatible with ZF since, for instance, (1) implies (2), (5) implies (4), and the negation of (4) implies (1) through (3), etc.  Which of the legitimate possibilites actually obtains in some model of ZF where choice fails?
This second question is rather broad, and breaks down into something on the order of 32 cases, so seeing an answer to any one of the legitimate cases would be cool.

REPLY [6 votes]: I have recently noticed$^1$ that in models of ZF+DC+"All sets of real numbers have the Baire property" (e.g. Solovay's model or Shelah's model mentioned by Andreas) there is a very interesting property for Banach spaces:

If $V$ is a Banach space, $W$ is a normed space and $T\colon V\to W$ is linear then $T$ is continuous.

In particular this means that if $V$ is a Banach space over $\mathbb R$ then every functional is continuous. This means that the algebraic dual of a Banach space is the same as the continuous dual of the space.
For example, $\ell_p$ for $p\in(1,\infty)$ have this property, using DC we can develop the basic tools of functional analysis just fine (except Hahn-Banach, though). This implies that $\ell_p$ is reflexive in the algebraic sense, not only in the topological sense.
We further remark$^2$ that the assertion $\ell_1\subsetneq\ell_\infty^\prime$ (where the $\prime$ denotes all the continuous functionals, but in our model these are all the functionals) implies the negation of "All sets of real numbers have the Baire property". The result is that $\ell_1$ is also reflexive in the algebraic sense and in the topological sense.
So with this in mind we have $V_p=\ell_p\oplus\ell_q$ (where $\frac1p+\frac1q=1$) is a self-dual as well algebraically reflexive space, and for distinct $p_1,p_2\in[1,2]$ we have $V_{p_1}\ncong V_{p_2}$ as well, so there are many non-isomorphic examples for this.

Notes:

After sitting to write all the details of the above I found out it is mentioned in Eric Schechter's book, Handbook of Analysis and its Foundations in the last section of chapter 27.
The above reference does not discuss the case of $p=1$, which is discussed later in the end of chapter 29.