[Paper Review] On hyperplanes and semispaces in max-min convex geometry
This paper investigates separation by hyperplanes in max-min convex geometry, establishing that only specific semispaces—those corresponding to the diagonal points in the unit cube—have closures that are hyperplanes. The key result is a characterization of when semispaces are hyperplanes, resolving the question of when classical hyperplane separation is possible in this setting, with implications for convexity and fuzzy algebra.
The concept of separation by hyperplanes is fundamental for convex geometry and its tropical (max-plus) analogue. However, analogous separation results in max-min convex geometry are based on semispaces. This paper answers the question which semispaces are hyperplanes and when it is possible to classically separate by hyperplanes in max-min convex geometry.
Motivation & Objective
- To clarify under what conditions hyperplanes can separate points from closed max-min convex sets.
- To determine which semispaces in max-min convex geometry have closures that are hyperplanes.
- To resolve the contrast between tropical convexity (where semispace closures are always hyperplanes) and max-min convexity.
- To provide a geometric and algebraic characterization of hyperplanes and semispaces in the max-min semiring over [0,1]^n.
- To establish conditions under which classical hyperplane separation is possible in max-min convex geometry.
Proposed method
- Analyzes the structure of semispaces at a point x⁰ ∈ [0,1]^n by ordering coordinates and partitioning them into equal or strictly decreasing blocks.
- Defines n+1 maximal max-min convex sets avoiding a given point, called semispaces, and studies their closures.
- Introduces the concept of closure of a semispace S_i(x⁰) and characterizes when it forms a hyperplane using componentwise max-min operations.
- Uses a case analysis based on the position of x⁰ relative to the diagonal D_n, distinguishing between diagonal points (x_i = x_j for all i,j) and non-diagonal points.
- Applies a contradiction argument to show that if a hyperplane contains two specific points y and z derived from x⁰, it must also contain a third point v, proving closure is a hyperplane.
- Employs compactness arguments and inequalities involving α and β to show that strict inequalities in the semispace closure imply hyperplane membership.
Experimental results
Research questions
- RQ1Which semispaces in max-min convex geometry have closures that are hyperplanes?
- RQ2Under what conditions can a point be separated from a closed max-min convex set by a hyperplane?
- RQ3Why does the classical separation by hyperplanes fail in max-min convex geometry, unlike in tropical convexity?
- RQ4What is the role of diagonal points in enabling hyperplane separation in max-min convex sets?
- RQ5Can the closure of any semispace be a hyperplane, or are there structural obstructions?
Key findings
- The closure of a semispace S_i(x⁰) is a hyperplane if and only if i = 0 or x⁰ lies on the diagonal D_n, i.e., all coordinates are equal.
- For non-diagonal points, the closure of any semispace is not a hyperplane, and such semispaces cannot be separated from the point by a hyperplane.
- A counterexample exists in dimension 2 where a point x = z₁ ∧ z₂ cannot be separated from the segment [z₁, z₂] by any hyperplane.
- The closure of S₀(x⁰) is a hyperplane if x⁰ is not on the diagonal, but only when x⁰ is on the diagonal does S₀(x⁰) itself become a hyperplane.
- For any point x ∈ D_n (the diagonal), any closed max-min convex set not containing x can be separated from x by a hyperplane.
- The proof relies on constructing points y, z, and v such that y and z lie in the closure of a semispace but v does not, and showing that if a hyperplane contains y and z, it must also contain v, leading to a contradiction unless the semispace is a hyperplane.
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This review was created by AI and reviewed by human editors.