Bourbaki

Theorem 1 (The bipolar theorem)

Tag 07HD, permanent. Cite it as [Bourbaki, Tag 07HD].

Topological Vector Spaces, II, p. 44, § 6. Weak topologies

Let F, G be two real vector spaces in duality. For every subset M of F the polar set M∘∘M^{\circ \circ} in F of the polar set M∘M^\circ of M in G is the closed convex envelope (for σ(F,G)\sigma(F, G)) of M∪{0}M \cup \{0\}.

We have seen that we need only consider the case where MM is convex and 0∈M0 \in M. Denote the closure, in topology σ(F,G)\sigma(F, G) of MM by M‾\overline{M}, then M‾\overline{M} is a convex set in FF; prop. 4 of II, p. 44 shows that M∘∘⊇M‾M^{\circ\circ} \supseteq \overline{M}. On the other hand if a∈Fa \in F does not belong to M‾\overline{M} then there exists a closed hyperplane HH in FF which separates aa strictly from M‾\overline{M} (II, p. 38, prop. 4); since HH does not contain 0, there exists y∈Gy \in G such that HH has the equation ⟨x,y⟩=−1\langle x, y \rangle = -1 (II, p. 43, prop. 3); thus ⟨x,y⟩>−1\langle x, y \rangle > -1 for all x∈M‾x \in \overline{M} and ⟨a,y⟩<−1\langle a, y \rangle < -1. This implies that y∈M∘y \in M^\circ and a∉M∘∘a \notin M^{\circ\circ}, and the relation M∘∘=M‾M^{\circ\circ} = \overline{M} follows.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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