TITLE: If $M=M^{\perp\perp}$ for every closed subspace $M$ of a pre-Hilbert space then $H$ is complete
QUESTION [7 upvotes]: It is well-known that if $H$ is a Hilbert space then for every closed subspace $M$ we have $M=M^{\perp\perp}$ see here
I would like to prove the converse by showing that: if $M=M^{\perp\perp}$ for every closed subspace $M$ of a pre-Hilbert space  then $H$ is complete

REPLY [10 votes]: We will show that the embedding $i:H\hookrightarrow \hat H$ of $H$ into its completion is onto. Given $z\in \hat H\setminus \{0\}$, the subspace $M=\{x\in H:\langle i(x),z\rangle =0\}$ is closed in $H$ and its orthogonal complement $M^\perp$ is different from $\{0\}$ since otherwise $M=M^{\perp\perp}=\{0\}^\perp=H$ which implies $z=0$. For fixed $y\in M^\perp\setminus \{0\}$ and the linear functionals $\varphi(x)= \langle x,y\rangle$ and $\psi(x)=\langle i(x),z\rangle$ we have kern$(\psi)=M\subseteq$ kern$(\varphi)$ which implies $\varphi=a\psi$ for some scalar $a$ (indeed, if $e\in H$ satisfies $\psi(e)=1$ we get $x-\psi(x)e\in$ kern$(\psi)$ so that $0=\varphi(x-\psi(x)e)=\varphi(x)-a\psi(x)$ for all $x\in H$ with $a=\varphi(e)$). Now, the functionals on $\hat H$ given by $\langle\cdot,i(y)\rangle$ and $\langle\cdot, az\rangle$ coincide on the dense subspace $i(H)$ and hence on $\hat H$ which implies $i(y)=az$. Since $a\neq 0$ this gives $z\in i(H)$.