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Let $A=\{ x,y,z: x,y,z\in[0,1] \}$ and $B=\{(x-2)^{2}+(y-2)^{2}+(z-2)^{2}\le 1\}$. Show if the sets $A$ and $B$ can be properly or strictly separated. Does anyone know the solution of this problem?

Alex Ravsky
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Laura
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1 Answers1

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The set a $A$ is a cube with side $1$, the set $B$ is a ball with radius $1$. The sets $A$ and $B$ are closed compacts. The sets $A$ are $B$ disjoint because if $(x,y,z)\in A$ then $(x-2)^2\ge 1$, $(y-2)^2\ge 1$, and $(z-2)^2\ge 1$, therefore $(x,y,z)\not\in B$. Thus, by Hahn-Banach separation Theorem, the sets $A$ and $B$ can be separated.

Alex Ravsky
  • 90,434
  • I understand, but I am supposed to find also the separating hyperplane, that is, to find the corresponding normal vector. – Laura Dec 22 '15 at 21:48
  • @Laura This should be done with a bit of geometric imagination. In general it should be a hyperplane going through an interior point of a segment realizing the minimal distance between the sets $A$ and $B$ and perpendicular to the segment. In this partial case it seems that the segment is $$[(1,1,1), (2-\frac 1{\sqrt{3}},2-\frac 1{\sqrt{3}},2-\frac 1{\sqrt{3}})],$$ so it suffices to take a hyperplane $x+y+z=c$ for any $c\in (3, 6-\sqrt{3})$. – Alex Ravsky Dec 23 '15 at 03:41
  • Thanks for your help. Could I ask you what about in case that we have the sets $A=[0,1]\times [0,1]$ and $B={(x,y)\in\mathbb{R}^{2}_{+}:xy\ge 2 }$. What would be the normal vector of the separating hyperplane? – Laura Oct 13 '20 at 15:46
  • @Laura This becomes clear when we draw a picture of $A$ and $B$. As a separating hyperplane we can take a line $\ell$ described by the equation $x+y=c$ (so with the normal vector $(1,1)$) for any $c\in (2, 2\sqrt{2})$. We can easily verify analytically that $\ell$ separates $A$ and $B$. Indeed, if $(x,y)\in A$ then $x,y\le 1$ so $x+y\le 2<c$. If $(x,y)\in B$ then $xy\ge 2$ and by the inequality between arithmetic and geometric means we have $x+y\ge 2\sqrt{xy}\ge 2\sqrt{2}>c$. – Alex Ravsky Oct 13 '20 at 18:56
  • It makes sense, however, I read that for the normal vector $\alpha$ must be fulfilled $\inf <\alpha, a> \ge \sup <\alpha,b>, a\in A, b\in B$. By $<,>$ I mean a scalar product. It does not work for $\alpha =(1,1)$. – Laura Oct 13 '20 at 20:39
  • @Laura It works the converse way, $\sup_{a\in A} \langle a, \alpha\rangle=2<2\sqrt{2}=\inf_{b\in B} \langle b, \alpha\rangle$. – Alex Ravsky Oct 13 '20 at 21:13
  • Ok, I understand now. And last thing, sorry for bothering you, would you know how to describe the closed ball in $\mathbb{R}^{3}$ as a intersection of supporting halfspaces? – Laura Oct 14 '20 at 07:38
  • @Laura Let $B$ be a closed ball of in $\Bbb R^n$ of positive radius $R$ centered at a point $x_0\in\Bbb R^n$. Then $B=\bigcap_{x\in S} H_x$, where $S$ is the sphere bounding $B$, and $H_x={y\in\Bbb R^n: \langle x-x_0,y-x_0\rangle\le R^2}$ is a half-space supporting $B$ at $x$. – Alex Ravsky Oct 14 '20 at 10:28
  • Could you explain in more details why the half-space supporting $B$ at $x$ is the $H_{x}$ set? I would say that a tangent plane to the sphere at point $x$ is the equation $<x-x_{0},y-x>=0$ so $H_{x}={ y\in\mathbb {R}^{n}:<x-x_{0},y-x>\le 0}$. Is it incorrect? – Laura Oct 15 '20 at 07:35
  • @Laura This condition defines the same set $H_x$, because $\langle x−x_0,y−x_0\rangle \le R^2$ iff $\langle x−x_0,y−x\rangle \le 0$ (since $\langle x−x_0,x−x_0\rangle=R^2$. – Alex Ravsky Oct 15 '20 at 08:18
  • Ok, it's really true. Just one more question about the sign. The hyperplane is divided in two half spaces and how we know which of the sign $\leq$ or $\geq$ is correct for the general point $x$? I mean the hyperplane $\langle x-x_{0},y-x\rangle =0$ is the same as $\langle x_{0}-x,y-x\rangle=0$ so we are supposed to write that the half-space supporting $B$ at $x$ is either $H_{x}={y\in\mathbb R^{n}:\langle x-x_{0},y-x\rangle\le 0}$ or $H_{x}={y\in\mathbb R^{n}:\langle x_{0}-x,y-x\rangle\ge 0 }$. Why not for example $H_{x}={y\in\mathbb R^{n}:\langle x-x_{0},y-x\rangle\ge 0}$? – Laura Oct 16 '20 at 11:25
  • @Laura I guess you mean “The space $\Bbb R^n$ is divided into two half spaces”. We choose the sign for the expression determining $H_x$ to assure that $H_x$ contains $x_0$. Since $\langle x−x_0,x_0−x\rangle=\langle x−x_0,x-x_0\rangle=-R^2\le 0$, we should impose the condition $\langle x−x_0,y−x\rangle \le 0$. – Alex Ravsky Oct 16 '20 at 11:40