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This question already has an answer here:

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.

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

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  • $\begingroup$ It generally doesn't hold. $\endgroup$ – Daniel Fischer Jul 7 '14 at 15:03
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No, it is not true. In the edit to this question on mathoverflow there is a simple and explicit counterexample.

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 $M^{\perp\perp} = H$.

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  • $\begingroup$ Why M is closed?@giovanna lamberti $\endgroup$ – C.Ding Apr 17 '16 at 12:02

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