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Disjoint convex sets are separable if one has nonempty core and the sets are disjoint.
Let be a real vector space and let be nonempty convex sets. Assume that core(Ω₁) and .
Lemma: The sets and can be separated by a hyperplane.
Context: The proof reduces separation of two sets to separation of a point from a convex set by applying point-vs-set separation to the Minkowski difference .
A vector space over a field is a set equipped with two operations (functions) and such that for all and :
Vector spaces are the basic objects studied via linear maps and invariants such as the determinant and eigenvalues of operators.
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 real vector space and let be nonempty.
We say that and can be separated by a hyperplane if there exists a nonzero linear functional such that
Geometrically, picking any with produces the hyperplane lying between the two sets; see the sup/inf characterization.
Let be a real vector space, let , and let be convex. Assume that core(Ω) is nonempty and .
Theorem: The sets and can be separated by a hyperplane. Equivalently, there exists a nonzero linear functional such that
Context: This is a geometric form of Hahn–Banach. The proof uses the Minkowski gauge of (after translation) to build a sublinear domination bound and then applies Hahn–Banach.