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In problem 2.23 of Boyd & Vandenberghe's Convex Optimization, it is said that the following two sets can not be separated by a hyperplane

$$\begin{aligned} C &:= \left\{ x \mid x \in \Bbb R^2, x_2 \leq 0 \right\} \\ D &:= \left\{ x \mid x \in \Bbb R^2, x_1 x_2 \geq 1 \right\} \end{aligned}$$

Why isn't the horizontal line a separating hyperplane for these sets?

Rodrigo de Azevedo
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Frank Moses
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  • Are there any other restrictions? Seems like $x_1=-1$ and $x_2=-1$ would be in set $C$ and $D$. The title of your question makes me think you want disjoint sets though – Casey May 13 '18 at 17:04
  • Looking at the problem, it seems it wants strict separability. Indeed you are right (if set $D$ is restricted to the first quadrant), these sets are separable and a separating hyperplane of the form $a^Tx \le b$ could be $a = (0,1)^T$ and $b=0$. This cannot hold with strict inequality since set $D$ approaches arbitrarily close to the boundary of set $C$ in the limit – Casey May 13 '18 at 17:09
  • Related – Rodrigo de Azevedo Apr 07 '21 at 20:35

1 Answers1

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Depends on what "separating" means in the context. The line $x_2=0$ provides non-strict separation: the set $D$ is contained in the open half-space while the set $C$ is contained in the closed half-space.

One may want strict separation, under which both sets are contained in disjoint open half-spaces. Such separation is impossible for $C$ and $D$. Indeed, a line that stays strictly above $C$ must be of the form $x_2=c$ for some $c>0$; and any such line intersects $D$.