Bipolartheorem for canonical pair (X,X*) Let $X$ be a vector space and $X^*$ its algebraic dual. For $A \subseteq X$ we define $A' = \lbrace f \in X^* | f(u) = 0, \hspace{2mm} \forall u \in A  \rbrace$ and $A'' = \lbrace x \in X | \langle f,x\rangle = 0, \hspace{2mm} \forall f \in A' \rbrace$. I want to show $A'' = \mbox{span} A$. For finite dimensional vector spaces this is clear. If $X$ is an infinite dimensional vector space then first note that $A' = \sideset{^\perp}{}A$. By definition we have $A'' = (\sideset{^\perp}{}(\mbox{span}A))^{\perp}$. By the bipolar theorem one gets $(\sideset{^\perp}{}(A))^{\perp} = \overline{\mbox{span} A}^{\sigma(X^*,X)}$. Now I am not sure if the linear span is already $\sigma(X^*,X)$ closed. I know that $X^*$ is $\sigma(X^*,X)$-complete.
 A: Note that this holds for vector spaces without any topology over every field.
$A \subset A''$. Indeed, take $a \in A$. Let $f \in A'$, this means that $f(u)=0$ for all $u \in A$, in particular $f(a)=0$. As this holds for all $f$, we get $a \in A''$. Now it's easy to show that $A''$ is a vector space, so we get $\operatorname{span} A \subset A''$. Set $Y:=\operatorname{span} A$
Now suppose that $x \in X \setminus Y$. Now consider the algebraic dual space $(X/Y)^*$. Because $x \notin Y$, we get that $x+Y \neq Y$, so there exists a linear functional $f:X/Y \to \Bbb k$ ($\Bbb k$ being the ground field) such that $f(x+Y)\neq 0$ (start with a linear functional on the subspace spanned by $x+Y$ that doesn't vanish on $x+Y$ and use that $\Bbb k$ is an injective object in the category of all $\Bbb k$-vector spaces (with no topology) to extend this to $X/Y$. If you don't like using the notion of injective modules, use Zorn's lemma directly.). Compose this with the projection $X \to X/Y$ to get a linear functional $f \in X^*$ that vanishes on $Y$ and that is non-zero on $x$. As $f$ vanishes on $Y$, we get $f \in A'$, but we have $\langle f,x\rangle = f(x) \neq 0$, so $x \notin A''$.
