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For a real dual pair $(E,F)$ and $A\subseteq E$, use the one-sided polar setThen its bipolar is the weak topology closure of the convex hull containing zero:The real-field and one-sided conventions matter. Replacing $f(a)\le 1$ by $|f(a)|\le 1$ instead produces a closed absolutely convex hull. A complex-field one-sided version uses $\mathrm{Re}f(a)$, rather than ordering complex numbers.^[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 $\alpha \ge 0$ with $f(c)\le \alpha <f(x)$ 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.