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.

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