Proof of $\text{dim}(E)\leq\text{dim}(E^*)$ without using a basis Let $E$ be a vector space over a (not necessarily commutative) field $K$. How can we show that $\text{dim}(E)\leq\text{dim}(E^*)$ without using a basis?
My proof uses a basis. Let $(b_i)_{i\in I}$ be a basis of $E$. We have a family $(b^*_i)_{i\in I}$ of elements of $E^*$ characterized by $\langle b_i,b^*_j\rangle=0$ for $i\ne j$ and $\langle b_i,b^*_i\rangle=0$. Clearly this family is free. Let $H$ be the subspace of $E^*$ generated by $(b^*_i)_{i\in I}$. Then
$$\text{dim}(E)=|I|=\dim(H)\leq\dim(E^*).$$
Is there a way to prove this without using a basis?
 A: It's a result due to Blass that "every vector space has a basis" is equivalent to the axiom of choice. Without AC this result is not necessarily true; in fact it's consistent with ZF that there exist infinite-dimensional vector spaces $E$ such that $E^{\ast} = 0$. See this MO question for some details, giving the example of $E = \mathbb{R}$ as a vector space over $\mathbb{Q}$.
Of course without bases / AC we have the further problem that it's no longer even clear what $\dim E$ means. We could take it to mean either the smallest size of a generating subset or the largest size of a linearly independent subset and it's no longer clear that these coincide. But in any case it's clear that with either of these definitions $\dim_{\mathbb{Q}} \mathbb{R}$ is infinite (and with a little effort uncountable; this, requires explicitly constructing an uncountable linearly independent subset which is done here), whereas it's consistent that $\text{Hom}_{\mathbb{Q}}(\mathbb{R}, \mathbb{Q}) = 0$ which certainly has dimension $0$.
