If $X$ is closed, then $X=(X^0)^{\dagger}$

Question

We define the nullifier of a subspace $X$ of a normed space $\mathcal{N}$ as

$$X^0:=\{f\in\mathcal{N}^*:f(\xi)=0,\forall\xi\in X\}$$

and the nullifier of a subspace $\Lambda$ of $\mathcal{N}^*$ as

$$\Lambda^{\dagger}:=\{\xi\in\mathcal{N}:f(\xi)=0,\forall f\in \Lambda\}.$$

They are closed subspaces of $\mathcal{N}^*,\mathcal{N}$, respc.

I proved that $X\subset (X^0)^{\dagger}$, but I need to prove also that

If $X$ is closed, then $X=(X^0)^{\dagger}$.

Well, if $\mu \in (X^0)^{\dagger}\setminus X$, so $\mu=\xi+\gamma$ with $\xi\in X$ and $\gamma\not\in X$. We may have

$$f(\gamma)=f (\mu)-f (\xi)=0\forall f\in X^0$$...

I don't know how I can prove that $\mu$ is a limit of a sequence in $X$.

Many thanks for any clue. Also, I'd like to know if the term "nullifier" is correct, once I've translated from another idiom.

Answer 1

There's a result from Hahn-Banach Theorem that states that $\exists f\in\mathcal{N}^*$ s.t. $||f||=1$, $f(\mu)=d(\mu,X)>0$ and $f\mid_X=0$.

So, $f\in X^0$, then $\mu\not\in (X^0)^{\dagger}$ once $(X^0)^{\dagger}=\{\xi\in\mathcal{N}: f(\xi)=0,\forall f\in X^0\}$. This is a contradiction.