Show annihilator is closed subspace of dual space. Let $X$ be normed vector space and $M$ be subspace. Let $M^{0} = \{ \lambda \in X^{*} : \lambda(x)= 0, \forall x \in M\}$, where $X^{*}$ is dual space of $X$. I want to show that $M^{0}$ is closed subspace. How can do I show that? I'm tried to show that let $\{\lambda_{n}\}_{n=1}^{\infty}$ be cauchy sequence and it converges to some element in $M^{0}$, but failed. Anyone give me some hint?
Update : I think I found the answer. For fixed $x \in i(M)$, where $i: M \to M^{**}$ by natural map, intersection of all $x^{-1}(\{0 \})$ is $M^{0}$. Hence it is closed. Is it right?
 A: The null space of $x^{\star}\in X^{\star}$ is closed because $x^{\star}$ is continuous and the null space is the inverse image of $\{ 0\}$ under $x^{\star}$. Annihilators are intersections of such closed subspaces, which makes them closed.
For example, let $Jx(x^{\star})=x^{\star}(x)$. $Jx$ is a continuous linear functional on $X^{\star}$. So $\mathcal{N}(Jx)=\{ x^{\star} \in X^{\star} : Jx(x^{\star})=x^{\star}(x)=0 \}$ is a closed subspace of $X^{\star}$ and $M^{0}=\bigcap_{x\in M}\mathcal{N}(Jx)$.
A: I'm studying an introductory course in functional analysis and maybe you can use a sequence like this:
Let $\lambda\in\overline{M^0}$ so there exists a sequence $\langle \lambda_n : n\in\omega\rangle$ in $M^0$ such that $\lambda_n \longrightarrow \lambda$. In other symbols:
$$\lim_{n\to\infty}\lambda_n=\lambda$$
Then, it's enough to show that $\lambda\in M^0$. But that is easy since you know that $(\forall n\in\omega)(\lambda_n=0)$, using the limit above you get $\lambda=0$. Hence $M^0$ is closed.
