= Bipolar theorem for a dual pair
{c}
For a real <dual pair> $(E,F)$ and $A\subseteq E$, use the one-sided <polar set>
$$
A^\circ=\{f\in F:f(a)\leq1\text{ for every }a\in A\}.
$$
Then its bipolar is the <weak topology> closure of the <convex hull> containing zero:
$$
A^{\circ\circ}=\overline{\operatorname{conv}}^{\sigma(E,F)}(A\cup\{0\}).
$$
The real-field and one-sided conventions matter. Replacing $f(a)\leq1$ by $|f(a)|\leq1$ instead produces a closed absolutely convex hull. A complex-field one-sided version uses $\operatorname{Re}f(a)$, rather than ordering complex numbers.<Appendix D: The bipolar theorem (ETH Zürich lecture notes)>[Theorem D.1 and following separation remark, for the real L-infinity/L1 dual pair]{ref}
Write $C$ for the right-hand closed <convex set>. The bipolar is an intersection of weakly closed half-spaces containing $A$ and zero, so it contains $C$. If $x\notin C$, the <Hahn-Banach separation theorem> supplies $f\in F$ and $\alpha\geq0$ with $f(c)\leq\alpha<f(x)$ for every $c\in C$. Choose $r>0$ between $\alpha$ and $f(x)$. Then $f/r\in A^\circ$ but $(f/r)(x)>1$, excluding $x$ from the bipolar. This proves the reverse inclusion.
For $E=F=\mathbb R$ with the usual multiplication pairing and $A=\{2\}$, the one-sided polar is $(-\infty,1/2]$ and the bipolar is $[0,2]$. The absolute-value polar would give the symmetric interval $[-2,2]$ instead.
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