Bi-annihilator of a subspace of the dual of an infinite-dimensional vector space Let $V$ be an infinite-dimensional vector space and $V^*$ its dual.
For a linear subspace $W\subset V$ define $W^ \circ\subset V^*$ as the subspace of linear forms on $V$ vanishing on $W$.
Dually, for $\Gamma\subset V^*$ define $\Gamma^\diamond \subset V$ as the set of vectors $v\in V$ such that $\gamma(v)=0$ for all linear forms $\gamma\in \Gamma$.
It is slightly surprising but not too difficult to show that that we  have for all subspaces $W\subset V$ the equality $(W^\circ) ^\diamond=W$.
But is it true that for all $\Gamma\subset V^*$ we have $(\Gamma^\diamond)^\circ=\Gamma$ ?
And is there a reference (article, book, lecture notes,...) where this problem is mentioned?
 A: No, $(\Gamma^\diamond)^\circ$ need not always equal $\Gamma$.  Let $\mathcal B$ be a basis for $V$, and let $\Gamma$ be the span of the 'dual' set $\{e_b \mathrel: b \in \mathcal B\}$, so $e_b(c)$ is the Iverson bracket $[b = c]$ for all $b, c \in \mathcal B$.  Then $\Gamma^\diamond$ is $0$, so $(\Gamma^\diamond)^\circ$ is all of $V^*$; but $\Gamma$ itself does not contain, for example, the element $\sum_{b \in \mathcal B} e_b$ of $V^*$.
A: The equality is false in general.
Here is a counterexample: fix a basis $v_i, i\in I$ of $V$ and consider the set of coordinate linear forms $v^*_i, i\in I$.
These forms are linearly independant but never form a basis since $V$ is infinite-dimensional.
So complete these forms to a basis $(v^*_j), j\in J$ with $J\setminus I\neq\emptyset$.
Choose $l\in J\setminus I$ and put $J'=J\setminus \{l\}$
If you define $\Gamma \subset V^*$ as the vector space generated by the $v_j^*, j\in J'$, then $\Gamma^\diamond =0$ (since already the subspace of $V^*$ generated by the $v_i^*, i\in I$ kill all vectors in $V$) so that $\Gamma\subsetneq (\Gamma^\diamond)^\circ=\{0\}^\circ=V^*$ yielding the required counterexample.
