TITLE: An example of space $V$ such that $(V^{\perp})^{\perp} \neq V$
QUESTION [5 upvotes]: I know that if $W$ is a vector space of finite dimension then for any subspace $V$ ,$(V^{\perp})^{\perp} = V$. But I have heard that this is not true for infinite dimensional vector spaces. So I tried to construct a counter example but I could not get any. I tried the vector space formed by infinite tuples but whatever subspace I took it was satisfying $(V^{\perp})^{\perp} = V$. So if any one could give a counter example it would be great.Thanks.

REPLY [4 votes]: In a Hilbert space, $A^{\perp\perp}=\overline{L(A)}$ (= the smallest closed subspace containing $A$) for any set $A$. Thus the counterexamples are exactly the subspaces $V$ that are not closed.
For a concrete example, you can take $H=\ell^2$ and $V$ as the finitely supported sequences. Then it's easy to see directly that $V^{\perp}=\{0\}$, so $V^{\perp\perp}=\ell^2$.