Evaluation map infinite dimensional vector space

Question

In a finite-dimensional vector space, we have an isomorphism ev: $ V \to V^{**}$, where ev(v)$: V^* \to F$ and $f \mapsto f(v)$. But what exactly happens in the infinite-dimensional case? Why is ev$(v)$ not surjective?

Answer 1

If you follow the argument you see that there is a dual basis in $V^*$. So if $v_1, ...,v_n$ is a basis of $V$, then $f_i$ where $f_i(v_j)=\delta_{ij}$ is a dual basis. But in the case of infinite dimensions this is not a basis, because there are many functions which take nonzero values on infinitly many of the basis $\{v_i\}$ of $V$. In any case they dont even have the same cardinality, $|V^*|=2^{|V|}$.