Bourbaki

Proposition 17

Tag 07C5, permanent. Cite it as [Bourbaki, Tag 07C5].

Topological Vector Spaces, chapter II, § 2. Convex sets

— In a topological vector space E, let A be a set with at least one interior point, and such that all its points lie on the same side of an hyperplane H. Then H is closed, the interior points of A lie strictly on the same side of H, and the cluster points of A lie on the same side of H. In particular open (resp. closed) half spaces are determined by closed hyperplanes.

In fact suppose that H contains the origin and that f(x)=0f(x) = 0 is an equation of H; suppose, for definiteness, that f(x)⩾0f(x) \geqslant 0 for all x∈Ax \in A. The half space formed by the points y such that f(y)>−1f(y) > -1 contains at least one interior point, and, by translation, the same is true of the half space of points y such that f(y)>0f(y) > 0; this shows that H is closed (I, p. 11, corollary). Then we know that ff is a strict morphism of E on R\mathbf{R} (I, p. 13, corollary), therefore f(A˙)f(\dot{A}) is an open set in R\mathbf{R}. This set cannot contain 0 or it would contain numbers <0< 0 contrary to hypothesis; it is thus contained in the open interval ]0,+∞[]0, +\infty[. On the other hand, the half space of those y for which f(y)⩾0f(y) \geqslant 0 is closed and contains A, therefore it contains A‾\overline{A}.

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

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.