Prove that $S^{\bot \bot} = \operatorname{Span}S$ Let $V$ be a vector space and let $S$ be a set of vectors in $V$.
If $k\in\operatorname{Span}S$ then $k = a_1v_1+\cdots+a_nv_n$ where $v_i \in S$ and $i\in\{1,2,\ldots,n\}$.
Now $\forall v_i \in S \implies v_i\cdot w=0, \forall w \in S^{\bot}$
so $$k\cdot w = \sum_{i = 1}^na_iv_i\cdot w = 0$$
so by definition $k \in S^{\bot \bot}$.
If $k \in S^{\bot \bot}$ then $k\cdot w = 0 , \forall w\in S^{\bot}:= \{ w \in V:v\cdot w = 0, \,\,\forall v \in S\}$ 
This implies that $k$ is a linear combination of vectors in $S$ and hence $k$ is in the span of $S$.
 A: We need the condition $\operatorname{Span}S$ is finite-dimensional, which is used in the Lemma.
Suppose that $x\notin\operatorname{Span}S$, by the Lemma there exists
$y\in V$ such that $y\in(\operatorname{Span}S)^\perp$, but 
$$x\cdot y\ne0.\tag{1}$$
Also, we have $(\operatorname{Span}S)^\perp\subseteq S^\perp$ because given $v\in(\operatorname{Span}S)^\perp$, then
$$v\cdot w=0\quad\mbox{for all }w\in S\subseteq\operatorname{Span}S\quad\Longrightarrow\quad v\in S^\perp.$$
Therefore $y\in S^\perp$, and hence by $(1)$ we have 
$x\notin S^{\perp\perp}$. 


Lemma. Let $V$ be an inner product space, and $W$ be a finite-dimensional subspace of $V$. If $x\notin W$, then there exists $y\in V$
  such that $y\in W^\perp$, but $x\cdot y\ne 0$.

Proof. 
Note that there exist unique vectors $x_1\in W$ and $x_2\in W^\perp$ such that
$x=x_1+x_2$. Also, we have $x_2\ne {\it 0}$ because otherwise $x=x_1\in W$, which contradicts to the assumption that $x\notin W$. Thus by taking 
$y=x_2\in W^\perp$, and using the fact that $x_2\ne{\it 0}$,
$$x\cdot y=(x_1+x_2)\cdot x_2=x_1\cdot x_2+x_2\cdot x_2=\Vert x_2\Vert^2>0.$$
