![]() In the second version, it may or may not be unique. In the first version of the theorem, evidently the separating hyperplane is never unique. For example, A can be a closed square and B can be an open square that touches A. (Although, by an instance of the second theorem, there is a hyperplane that separates their interiors.) Another type of counterexample has A compact and B open. For example, if A is a closed half plane and B is bounded by one arm of a hyperbola, then there is no strictly separating hyperplane: A more subtle counterexample is one in which A and B are both closed but neither one is compact. For example, A and B could be concentric circles. If one of A or B is not convex, then there are many possible counterexamples. The theorem does not apply if one of the bodies is not convex. The above proof also proves the first version of the theorem mentioned in the lede (to see it, note that K is closed under the hypothesis below.) Consequently K is contained in some hyperplane thus, for all x in K and we finish the proof as before. Finally, if K has empty interior, the affine set that it spans has dimension less than that of the whole space. We have for any x in the interior of K and by continuity the same holds for all x in K. We also normalize them to have length one then the sequence contains a convergent subsequence with limit v, which has length one and is nonzero. Since 0 is not in K, each contains a nonzero vector of minimum length and by the argument in the early part, we have: for any. In general, if the interior of K is nonempty, then it can be exhausted by compact convex subsets. Thus, if v is nonzero, the proof is complete since Hence, for any x in A and y in B, we have. Since is convex, for any x in K, the line segmentĪnd letting gives. By the lemma, the closure of K, which is convex, contains a vector v of minimum norm. Since is convex and the sum of convex sets is convex, K is convex. Proof of theorem: Given disjoint nonempty convex sets A, B, let It is unique since if y is in K and has norm δ, then and x = y. SinceĪs, is a Cauchy sequence and so has limit x in K. Note that is in K since K is convex and so. Proof of lemma: Let Let be a sequence in K such that. Then there exists a unique vector in K of minimum norm (length). Lemma - Let K be a nonempty closed convex subset of R n. See the article on Support Vector Machines for more details. ![]() The margin between the hyperplane and the clouds is maximal. In geometry, a maximum-margin hyperplane is a hyperplane which separates two 'clouds' of points and is at equal distance from the two. The Hahn–Banach separation theorem generalizes the result to topological vector spaces.Ī related result is the supporting hyperplane theorem. The hyperplane separation theorem is due to Hermann Minkowski. An axis which is orthogonal to a separating hyperplane is a separating axis, because the orthogonal projections of the convex bodies onto the axis are disjoint. In the second version, if both disjoint convex sets are open, then there is a hyperplane in between them, but not necessarily any gap. In the first version of the theorem, if both these sets are closed and at least one of them is compact, then there is a hyperplane in between them and even two parallel hyperplanes in between them separated by a gap. In geometry, the hyperplane separation theorem is either of two theorems about disjoint convex sets in n-dimensional Euclidean space. Illustration of the hyperplane separation theorem.
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