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?

In the infinite dimensional case, $V^{**}$ will always have dimension strictly greater than that of $V.$ Let $\mathcal{A}$ be a basis for $V$ over $F.$ By the universal property for direct products, there exists a linear map $f: \hat{V} \rightarrow \displaystyle\prod_{a\in \mathcal{A}} F_a $ satisfying $\pi_A \circ f (v) = \epsilon_a (v)$ for all $v\in V^*,$ where $\epsilon_a $ is the evaluation at $a$ map and $\pi_a $ projects the range of $f$ onto its $a$th coordinate. It isn't too difficult to show that $f$ is an isomorphism. All that is left to show is that $\displaystyle\prod_{a\in \mathcal{A}} F_a$ has dimension stirctly larger than that of $V$ (for this, note that $V \cong \bigoplus_{a\in \mathcal{A}} F_a$ and give a cardinality-based argument) 
