Theorem 4

The dimension of the dual $E^*$ of a vector space E is at least equal to the dimension of E. For $E^*$ to be finite-dimensional, it is necessary and sufficient that E be so, and then $\dim E^* = \dim E$.

If K is the field of scalars of E, E is isomorphic to a space $K_s^{(I)}$ and therefore $E^*$ is isomorphic to $K_d^I$ (§ 2, no. 6, Proposition 10). As $K_s^{(I)}$ is a subspace of $K_d^I$, $\dim E = \mathrm{Card}(I) \leq \dim E^*$ (no. 3, Corollary 4 to Proposition 4); further, if I is finite, $K_d^I = K_d^{(I)}$ (cf. Exercise 3(d)).