Annihilators and closedness

Question

Let $E$ be a normed space. Let $A\subset E$ and $B \subset E^*$ be arbitrary subsets. Then the annihilators $A^\perp$ and $B_\perp$ are closed subspaces of $E^*$ resp. $E$. Why is that?

Answer 1

Let $x\in A$, define $h_x:E^*\rightarrow E^*$ by $h_x(y)=f(x)$, $h_x$ is continuous, so $Ker(h_x)$ is closed. The anihilator of $A$ is $\cap_{x\in A}ker(h_x)$ is closed since it is the intersection of closed subsets.