If $V$ is infinite dimensional with basis $A$, prove that $V^*$ is isomorphic to the direct product of copies of $F$ indexed by $A$. Deduce that dim $V^*$ > dim $V$.
I've shown in the past that the direct product of copies of $F$ indexed by $A$ is a vector space over $F$ and it has strictly larger dimension than the dimension of $V$. This fact would easily imply that dim $V^*$ > dim $V$ if I can show the isomorphism in the first part of the problem.
I think for the isomorphism, I have to find a bijection between the bases? I don't know how I would get a good map between $V^*$ and the direct product so that seems like a better idea. I'm having trouble getting off the ground with this one so any help would be appreciated.
Given any $f\in V^*$, you can map $f\longmapsto \{f(a)\}_{a\in A}$. This map is clearly linear. It is one-to-one: if $f(a)=g(a)$ for all $a\in A$, then $f=g$. And it is onto: for any $\{\lambda_a\}_{a\in A}$, the map given by $f(a)=\lambda_a$ is mapped to $\{\lambda_a\}$. So that's your isomorphism.