Annihilator of an annihilator 
Let X be a normed space and X' the dual space of X. The annihilator of a vector subspace $M\subset X$ is defined by:
  $$M^{\perp}:=\{f\in X'|f(y)=0 \forall y\in M\}\subset X'$$ Is $N\subset X'$ a vector subset of a dual space, then the annihilator of N in X is defined by:
  $$N^\perp:=\{x\in X|f(x)=0 \forall \in N\}\subset X$$
  Show:
  i)$(M^\perp)^\perp=\overline M$ ii) $\overline N\subset (N^\perp)^\perp$

I found the following proof online, but I have some difficulties understanding the notation:

1) $(A^\perp)^\perp$ is a closed span of A
2)$(B^\perp)^\perp$ is a weak* closed span of B
1)We know $A\subset (A^\perp)^\perp$, and $(A^\perp)^\perp$ is a closed subspace. Since $\overline{span}$ A is the smallest closed subspace containing A, it suffices to show that $(A^\perp)^\perp \subset\overline{span}A$. Suppose not, and pick $x\in (A^\perp)^\perp/\overline{span}A$. Then, since $\overline{span}A$ is a closed subspace, by the Hahn-Banach Theorem we can choose $f\in X'$ such that $f(x)\neq0$ and $f(y)=0$ for all $y\in\overline{span}A$. In particular, $f(y)=0$ for all $y\in A$, so $f\in A^\perp$. Since $f(x)\neq0$ we have $x\notin(A^\perp)^\perp$, and this is a contradiction.

What exactly is the closed span of A? I tried to google it but I didn't find much. Same with the weak* closed span. Is $\overline{span}$ the closed span?. How exactly did we use the Hahn-Banach Theorem to find this f? Can someone help me?
 A: To see that $(A^\perp)^\perp$ is a closed subspace, notice that if $A \subset X$, then $A^\perp \subset X'$ and
$(A^\perp)^\perp = \bigcap_{f \in A^\perp} f^{-1}(0)$. Because any $f \in X'$ is continuous, we have that $f^{-1}(0)$ closed. Because $(A^\perp)^\perp$ is an intersection of closed sets, $(A^\perp)^\perp$ is closed too.
Let $M = \overline{\text{span}}$ and suppose $x \not \in M $. Now consider the normed space $X/M$ and the natural projection $\pi: X \to X/M$. Then take $V = \{ \alpha (x+M) \mid \alpha \in \mathbb{C}\} \subset X/M$. Note that $\|x+M\| >0$ so we can define a functional $g:V \to \mathbb{C}: \alpha(x+M) \mapsto \alpha \|x+M\|$. Note that $\|g\| =1$. Now extend $g$ by the Hahn Banach theorem to $g:X/M \to \mathbb{C}$. Finally we can define $f = \frac{1}{\|x+M\|} g \circ \pi$. Notice that $\pi(y) = 0$ for all $y \in M$ so $f(y) = 0$ for all $y \in M$. But $f(x) = 1$ by construction. 
Notice that this is really the proof of the following nice corollary of the Hahn Banach theorem:

If $X$ is a normed space, $M$ a closed linear subspace of $X$, $x \in X\setminus M$ and $d = \text{dist}(x_0, M)$ then there's an $f \in X^*$ such that $f(x)=1$, $f(y) =0$ for all $y \in M$ and $\|f\| = d^{-1}$.

If by any chance you haven't encountered this corollary, I advise you to look it up. It's pretty useful.
