Exercise 3.F.22

Proof. Suppose $\phi \in {(U+W)}^0$. Let $u\in U$ and $w\in W$. By definition of sum of subspaces, we have that $u+w\in U+W$. Thus

Taking $u=0$ implies that $\phi \in W^0$. Similarly, taking $w=0$ implies that $\phi \in U^0$. Hence $\phi \in U^0\cap W^0$ and ${(U+W)}^0\subset U^0\cap W^0$.

To prove the inclusion in the other direction, suppose $\phi \in U^0\cap W^0$. Let $v\in U+W$. There are $u\in U$ and $w\in W$ such that $v=u+w$. We have

Since $v$ was arbitrary, $\phi \in {(U+W)}^0$. Hence $U^0\cap W^0\subset {(U+W)}^0$, as desired. □