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Disjoint convex sets can be separated strictly on the side of an open set.
Let be a real normed space. Let be nonempty, disjoint convex sets, and assume that is open.
Corollary. There exist a nonzero (see dual space) and such that
This follows from closed hyperplane separation under an interior condition together with the openness of , which makes the inequality strict on the -side.
Let be a vector space over (or ). A norm on is a function such that for all and scalars :
A pair is called a normed vector space.
Let be a vector space (typically over ). A set is convex if for all and all we have
Context. Equivalently, is convex iff it contains every line segment joining any two of its points. Convexity is the core geometric notion underlying convex analysis and optimization.
Let be a metric space and let .
The set is open if for every there exists such that
where is the open ball in .
Open sets are stable under arbitrary unions and finite intersections (see basic properties of open sets). Complements of open sets are closed.
Let be a normed space over .
The (continuous) dual space is the set of all continuous linear functionals (equivalently, all bounded linear functionals on ).
For and , the duality pairing is
Let be a real normed space and let be nonempty convex sets. Assume that int(Ω₁) and
Theorem: The sets and can be separated by a closed hyperplane; i.e., there exists such that
Context: In vector spaces, the analogous separation uses possibly discontinuous linear functionals (see core-based separation). The interior assumption ensures the separating functional is continuous, using closed level-set continuity.