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Let be $A\subset H$ a subset of $H$ Hilbert space. I know that $A^{\perp\perp}$ is the smallest closed subspace of $H$, such that $A\subset A^{\perp\perp}$.

But if $H$ is a inner product space (or pre-Hilbert, i.e., not necessarily complete), can I say something about this? $A^{\perp\perp}$ is the smallest closed subspace of $H$, such that $A\subset A^{\perp\perp}$?

Anyone has a counterexample?

With help of @user1952009, I tried something. Let be $H^{'}$ the completed Hilbert space of $H$ and $F\subset H^{'}$ closed subspace, such that $A\subset F$. Then, $A^{\perp\perp}\subset F^{\perp\perp}=F\subset H^{'}$.

So I need to show that $F\cap H$ is closed subspace in $H$ to conclude, cause the hypothesis is smallest closed subspace of $H$.

Ok so far?

Irddo
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  • closed $\implies$ complete so $A^{\perp}$ and $A^{\perp \perp}$ are Hilbert spaces. now considering the Hilbert space obtained from $H$ by completing it, you retrieve in the 1st situation ? – reuns Jan 20 '16 at 20:15
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    @user1952009: closed only implies complete if the "surrounding space" (in which we consider closedness) is itself complete. – PhoemueX Jan 20 '16 at 21:24
  • @user1952009 to (closed $\implies$ complete) I need that $H$ be complete. That doesn't work. – Irddo Jan 20 '16 at 23:02
  • ok $A^\perp = { x \in H | \forall a \in A : \langle x, a \rangle = 0 } $ ? so consider $B = A^{\perp \perp}$ and $B' = A^{\perp \perp}$ in the completed hilbert space, and write that $B = B' \cap H$ ? – reuns Jan 20 '16 at 23:16
  • @user1952009 I'm not sure that I understood what you say, let me try here. Let be $F\subset H^{'}$ (Complement of $H$) closed subspace, such that $A\subset F$. Then, $A^{\perp\perp}\subset F^{\perp\perp}=F\subset H^{'}$. But I need that $F\subset H$ to conclude, or not? (Hypothesis: smallest closed subspace of $H$ – Irddo Jan 20 '16 at 23:34

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