## If V is an infinite dimensional vector space, then its dual space has strictly larger dimension

Let $V$ be an infinite dimensional vector space over a field $F$, say with basis $A$. Prove that the dual space $\widehat{V} = \mathsf{Hom}_F(V,F)$ has strictly larger dimension than does $V$.

We claim that $\widehat{V} \cong_F \prod_A F$. To prove this, for each $a \in A$ let $F_a$ be a copy of $F$. Now define $\varepsilon_a : \widehat{V} \rightarrow F_a$ by $\varepsilon_a(\widehat{v}) = \widehat{v}(a)$. By the universal property of direct products, there exists a unique $F$-linear transformation $\theta : \widehat{V} \rightarrow \prod_A F_a$ such that $\pi_a \circ \theta = \varepsilon_a$ for all $a \in A$. We claim that $\theta$ is an isomorphism. To see surjectivity, let $(v_a) \in \prod_A F_a$. Now define $\varphi \in \widehat{V}$ by letting $\varphi(a) = v_a$ and extending linearly; certianly $\theta(\varphi) = (v_a)$. To see injectivity, suppose $\varphi \in \mathsf{ker}\ \theta$. Then $\theta(\varphi) = 0$, so that $(\pi_a \circ \theta)(\varphi) = 0$, and thus $\varepsilon_a(\varphi) = 0$ for all $a$. Thus $\varepsilon(a) = 0$ for all $a \in A$. Since $A$ is a basis of $V$, we have $\varphi = 0$. Thus $\theta$ is an isomorphism, and we have $\widehat{V} \cong_F \prod_A F$.

By this previous exercise, $\widehat{V}$ has strictly larger dimension than does $V$.