## The dual basis of an infinite dimensional vector space does not span the dual space

Let $F$ be a field and let $V$ be an infinite dimensional vector space over $F$; say $V$ has basis $B = \{v_i\}_I$. Prove that the dual basis $\{\widehat{v}_i\}_I$ does not span the dual space $\widehat{V} = \mathsf{Hom}_F(V,F)$.

Define a linear transformation $\varphi$ on $V$ by taking $v_i$ to 1 for all $i \in I$. Note that for all $i \in I$, $\varphi(v_i) \neq 0$. Suppose now that $\varphi = \sum_{i \in K} \alpha_i\widehat{v}_i$ where $K \subseteq I$ is finite; for any $j \notin K$, we have $(\sum \alpha_i \widehat{v}_i)(v_j) = 0$, while $\varphi(v_j) = 1$. Thus $\varphi$ is not in $\mathsf{span}\ \{\widehat{v}_i\}_I$.

So the dual basis does not span $\widehat{V}$.