# Distance between a cone and a disjoint hyperplane

I seek to prove the following, which I guess is true:

Define $$A:=\{x \ge 0\} \subset \mathbb{R}^m$$ and assume that $$U\subset \mathbb{R}^m$$ is an affine subspace with $$A \cap U=\emptyset$$. Show that

$$\delta:=\inf_{\substack{x\ge 0\\x'\in U}}\lVert x-x'\rVert>0.$$

I can easily handle the cases where $$\mbox{dim}(U)=0$$ or $$1$$.

Not sure how to solve this, but I can help you with one more case. If U is a hyperplane then there is some vector y and constant c such that $$U=\{ x: x \cdot y =c \}$$. Disjointness implies that $$y \geq 0, c<0$$. This implies the distance is at least $$- c/ \| y \|$$.

• Thank you for the answer, but I do not see why it should blow up? U could be parallel to one of the faces of A and even if not, I do not see a simple argument. – crankk Apr 15 at 14:21
• @crankk you're absolutely right. I edited my answer to solve one more special case. – Amichai Lampert Apr 16 at 6:22
• Assume that $dim(U)<m$ and that $U\cap A=\emptyset$. Does it follow that there is a hyperplane $U'$ with $U\subset U'$ and $U'\cap A = \emptyset$? That would solve the problem... – crankk Apr 16 at 11:03
• I think I am one step away from the solution: Is it true that, given to disjoint convex closed sets $U$ and $V$ there is a $c\in\mathbb{R}$ and a non zero $v$ such that either $(x,v)\ge c > (y,v)$ or $(y,v) \ge c >(x,v)$ for all $x\in U$ and $y\in V$? – crankk Apr 17 at 7:49
• @crankk: I don't think that's true for general convex closed sets. E.g. take $U$ to be the closed convex hull of points of the form $(0,y,0)$ and $(1,y,e^y)$ with $y\leq 0,$ and take $V$ to be the closed convex hull of points of the form $(0,y,-e^y)$ and $(1,y,0)$ with $y\leq 0.$ The only separating hyperplane is $z=0,$ which has points from both. – Dap Apr 17 at 19:42

I'll assume $$U$$ is non-empty. $$U$$ is of the form $$u+V$$ for some point $$u$$ and some linear subspace $$V.$$ Consider the projection map $$\phi:\mathbb R^m\to \mathbb R^m/V.$$ It takes $$U$$ to a single point which I will denote $$[u].$$ The distance between $$A$$ and $$U$$ is bounded below by (in fact equal to) the distance between $$\phi(A)$$ and $$[u].$$ So it remains to show that the distance between $$\phi(A)$$ and $$[u]$$ is positive.

Each standard basis vector $$e_i$$ is taken to a point other than $$[u]$$ because $$e_i\in A$$ which is disjoint from $$U.$$ Since $$A$$ is a convex cone generated by the rays $$e_1,\dots,e_m,$$ the set $$\phi(A)$$ is a convex cone generated by the rays $$\phi(e_1),\dots,\phi(e_m).$$ A finitely generated convex cone is polyhedral (the "Farkas-Minkowski-Weyl theorem", see for example Schrijver's "Theory of Linear and Integer Porgramming") and therefore closed. The distance between a point $$[u]$$ and the closed set $$\phi(A)$$ disjoint from $$[u]$$ is positive, which is what we needed to prove.

• Thank you! I hoped for a more elemtary way, which does not need the "Farkas-Minkowski-Weyl Theorem"... – crankk 2 days ago