A problem with annihilators in infinite-dimensional vector spaces It's a known fact that $\mathrm{Ann}(S^\circ)=S$, where $S$ is a subspace of a finite dimensional vector space $V$. I'll include the definitions for the sake of completeness, since $\mathrm{Ann}(S)$ (sometimes notated $^\circ S$ and called the pre-annihilator or joint kernel of $S$) hasn't got a more or less standard definition (for more on this, check this MO thread).
Given a vector space $V$ and a subspace $S\subseteq V$, we define:
$S^\circ=\{\varphi\in V^\star:\varphi(S)=0\}$
Similarly, given a subspace $S^\star\subseteq V^\star$, we define:
$\mathrm{Ann}(S^*)=\{v\in V:\varphi(v)=0\,\,\forall\varphi\in S^\star\}$
On the other hand, if we're dealing with an infinite dimensional vector space, we can only affirm that $S\subseteq \mathrm{Ann}(S^\circ)$. For instance, the inclusion is strict if we work under a normed vector space and choose $S\neq V$ to be any dense subspace (it follows from continuity that any linear functional that vanishes at $S$ must be identically $0$, and therefore $S\subseteq\mathrm{Ann}(S^\circ)=\mathrm{Ann}(\{0\})=V$).
However, a friend of mine sketched a proof that demonstrates that the equality always holds (regardless of the dimension of $V$), and I can't seem to find a hole in neither his argument or the one showed above. It goes like this:
Given $S$, pick any basis $B$ of $S$ and extend it to a basis of $V$ by adding linearly independent vectors outside of $S$, thus forming a basis $B'$ of V. Then, we can define a linear functional $\varphi$ such that $\varphi(v)=0$ for every $v\in B$ and $\varphi(w)=1$ for every $w\in B'\setminus B$. Then, since $\varphi\in S^\circ$, there's no vector $v\in V\setminus S$ such that $v\in\mathrm{Ann}(S^\circ)$. Therefore $S\supseteq\mathrm{Ann}(S^\circ)$, and the other inclusion is easily proved; thus, the equality holds.
 A: $\newcommand\Ann{\operatorname{Ann}}$Let $V$ be a vector space and let $S\subseteq V$ be a subspace. Suppose there is a vector $v\in\Ann(S^\circ)\setminus S$. There exists a linear map $\phi:V\to\mathbb R$ such that $\phi|_S=0$ and $\phi(v)=1$: to construct it, let $B$ be a basis of $S$; then $B'=B\cup\{v\}$ is linearly independent, and we can find a basis $B''$ of $V$ which contains $B'$. There is a linear map $\phi:V\to\mathbb R$ such that $\phi(b)=0$ for all $b\in B''\setminus\{v\}$ and $\phi(v)=1$. But then $\phi\in S^\circ$ and this is absurd, because then $v\in\Ann(S^\circ)\subseteq\ker\phi$.
A: Questions to ask yourself about the sentence "Then, since $\varphi \in S^{\circ}$ , there's no vector $v \in V \setminus S$ such that $v \in \operatorname{Ann}(S^{\circ})$":


*

*How is the "there's no vector..." part of this sentence specific to the functional $\varphi$ just mentioned?  (It seems completely independent of it?)

*How is the assertion "there's no vector $v \in V \setminus S$ such that $v \in \operatorname{Ann}(S^{\circ})$" any different from the statement you are trying to prove?  Why is it true?
It may help your intuition to note that, in general, the kernel of the functional $\varphi$ you have just mentioned need not be $S$.  (If $B' \setminus B$ has two distinct elements $w$ and $w'$, then $w - w'$ is in $\ker(\varphi)$ also.)
