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In geometry, the hyperplane separation theorem is either of two theorems about disjoint convex sets in ''n''-dimensional Euclidean space. 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 the second version, if both disjoint convex sets are open, then there is a hyperplane in between them, but not necessarily any gap. 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. The hyperplane separation theorem is due to Hermann Minkowski. The Hahn–Banach separation theorem generalizes the result to topological vector spaces. A related result is the supporting hyperplane theorem. In geometry, a maximum-margin hyperplane is a hyperplane which separates two 'clouds' of points and is at equal distance from the two. The margin between the hyperplane and the clouds is maximal. See the article on Support Vector Machines for more details. == Statements and proof == The proof is based on the following lemma: Proof of lemma: Let Let be a sequence in ''K'' such that . Note that is in ''K'' since ''K'' is convex and so . Since : as , is a Cauchy sequence and so has limit ''x'' in ''K''. It is unique since if ''y'' is in ''K'' and has norm δ, thenand ''x'' = ''y''. Proof of theorem: Given disjoint nonempty convex sets ''A'', ''B'', let : Since is convex and the sum of convex sets is convex, ''K'' is convex. By the lemma, the closure of ''K'', which is convex, contains a vector ''v'' of minimum norm. Since is convex, for any ''x'' in ''K'', the line segment : lies in and so :. For , we thus have: : and letting gives: . Hence, for any ''x'' in ''A'' and ''y'' in ''B'', we have: . Thus, if ''v'' is nonzero, the proof is complete since : In general, if the interior of ''K'' is nonempty, then it can be exhausted by compact convex subsets . 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 . 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. We have for any ''x'' in the interior of ''K'' and by continuity the same holds for all ''x'' in ''K''. We now 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. Consequently ''K'' is contained in some hyperplane ; thus, for all ''x'' in ''K'' and we finish the proof as before. 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.) Here, the compactness in the hypothesis cannot be relaxed; see an example in the next section. Also, This follows from the standard version since the separating hyperplane cannot intersect the interiors of the convex sets. 抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)』 ■ウィキペディアで「hyperplane separation theorem」の詳細全文を読む スポンサード リンク
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