Canonical Isomorphism Between $\mathbf{V}$ and $(\mathbf{V}^*)^*$ For the finite-dimensional case, we have a canonical isomorphism between $\mathbf{V}$, a vector space with the usual addition and scalar multiplication, and $(\mathbf{V}^*)^*$, the "dual of the dual of $\mathbf{V}$." This canonical isomorphism means that the isomorphism is always the same, independent of additional choices. 
We can define a map $I : \mathbf{V} \to (\mathbf{V}^*)^*$ by $$x \mapsto I(x) \in (\mathbf{V}^*)^* \ \text{ where } \ I(x)(f) = f(x) \ \text{for any } \  f \in \mathbf{V}^*$$
My Question: what can go wrong in the infinite-dimensional case? The notes I am studying remark that if $\mathbf{V}$ is finite-dimensional, then $I$ is an isomorphism, but in the infinite-dimensional case we can go wrong? How?

(This is just a bit too longer to be a comment)
It is important to remark that the fact that for an infinite dimensional space $\bf V$ is not isomorphic to $\bf V^{**}$ requires the axiom of choice. Furthermore, the counterexample is not some pathological space.
In some models without the axiom of choice a peculiar thing happens: every linear operator from a Banach space into a normed space is continuous. One example of such model is Solovay's model in which all sets of reals are Lebesgue measurable. We remark that in this model the principle of Dependent Choice holds, which is enough to develop most classical analysis.
In such model $\ell_2$, a Banach space, has only continuous linear functionals, so the algebraic dual is the same as the topological dual. The fact that $\ell_2$ is a self-dual (in the topological sense) does not require much of the axiom of choice, not more than we have in Solovay's model anyway.
Now we need to verify that the evaluation map is a linear isomorphism, it is that and more. It is an isometry. This is not a very hard exercise in applying basic functional analysis theorems.
