Closed subspace $M=(M^{\perp})^{\perp}$ in PRE hilbert spaces. [duplicate]

Is it true that $M=(M^{\perp})^{\perp}$ if $M$ is a closed subspace of a PRE hilbert space (a space with a scalar product, but that is not complete)?

The proof of the analog fact for hilbert spaces uses the projection on the subspaces $M^{\perp}$ and $(M^{\perp})^{\perp}$, but if the space is not complete that projection may not exists.

marked as duplicate by Daniel Fischer, Andrew D. Hwang, Moishe Kohan, apnorton, user147263 Jul 7 '14 at 19:36

Consider the pre-hilbert space $H = \ell^2_{f}$ consisting of sequences of finite support in $\ell^2$. Let $x \in \ell^2$ be the sequence $x_{n} = 1/n$ and $M = \{y \in H \mid \langle x,y\rangle = 0\}$. Then $M \subsetneqq H$ is closed and one can check that $M^{\perp\perp} = H$.