Exercise 3.F.25

Proof. Let $B=v\in V:\phi (v)=0\text{ for every }\phi \in U^0$.

Suppose that $u\in U$. By definition, $\phi (u)=0$ for all $\phi \in U^0$. Thus $u\in B$ and, therefore, $U\subset B$.

For inclusion in the other direction, we will prove the contrapositive. Suppose that $v\notin U$. Since $0\in U$, it follows that $v\ne 0$. Let $u_1,\dots ,u_n$ be a basis of $U$. We have that $v,u_1,\dots ,u_n$ is a linearly indepedent list in $V$. Extend it to a basis $v,u_1,\dots ,u_n,v_1,\dots ,v_m$ of $V$ and let $\phi ,{\psi }_1,\dots ,{\psi }_n,{\phi }_1,\dots ,{\phi }_m$ be its dual basis. It is easy to see that $\phi ,{\phi }_1,\dots ,{\phi }_m$ is a basis of $U^0$. But $\phi (v)=1$, thus $v\notin B$.

By modus tollens, $v\in B$ implies $v\in U$. Therefore $B\subset U$. □