Let V be an inner product space, S and $S_0 $be subsets of V, and W be a finite-dimensional subspace of V.
Prove:$S_0 \subset S$ implies $S^{\bot} \subset S_0^{\bot}$
Pf: Let $x \in S^{\bot}$, for all $y \in S$, the inner product $\langle x,y \rangle$ is given $\langle x,y \rangle=0$, which implies $y \in S_0$ Then $S_0 \subset S$, $x \in S_0^{\bot}$.Proved.
Prove: $W=(W^{\bot})^{\bot}$
Pf: Let $x \in W$ and $y \in W^{\bot}$.
By definition, for all $y \in W^{\bot}$, the inner product is given by $\langle x,y \rangle=0$.
Using $x \in W$, this gives $W \subset (W^{\bot})^{\bot}$ How to prove the other side?
Best Answer
Suppose that $w\in (W^\perp)^\perp$ then $\left<w,x \right>=0$ for all $x\in W^\perp$.
If $w\not\in W$, then $w=v+v'$ where $v\in W$ and $v'\in W^\perp\setminus \{0\}$. Then $$0=\left<w,x \right>=\left<v,x \right>+\left<v',x \right>=\left<v',x \right>$$ for all $x\in W^\perp.$ However, $v'\in W^\perp$ and $\left<v',v' \right>\neq 0$ so it is a contradiction.