Triangle inequality of a nasty metric $\rho(A,B)=\max\left\{\sup_{x\in A}\inf_{y\in B}|x-y|,\sup_{y\in B}\inf_{x\in A}|x-y|\right\}$

This was a qual problem in Winter 2012 at my university. I feel comfortable verifying everything but the triangle inequality.

Let $\Omega$ denote the set of all nonempty closed subsets of $[0,1]$ and let $\rho\colon\Omega\times\Omega\to[0,1]$ be defined by $$\rho(A,B)=\max\left\{\sup_{x\in A}\inf_{y\in B}|x-y|,\sup_{y\in B}\inf_{x\in A}|x-y|\right\}$$ Show that $(\Omega,\rho)$ is a metric space.

Intuitively, I think of $\sup_{x\in A}\inf_{y\in B}|x-y|$ as the closest distance from each $x\in A$ to the set $B$, and then taking the largest of those.

Is there a clever way to see $\rho$ respects the triangle inequality?

• What topological space are you refering to when you claim that $\Omega$ is a closed subset of a compact space? Aug 30 '13 at 0:23
• @OlivierBégassat Sorry, I wasn't thinking. I meant to say the closed sets in $\Omega$ are compact in $[0,1]$. Aug 30 '13 at 0:25
• If we're picky, it's not a metric space since $\rho(A,\varnothing) = \infty$. Make it "nonempty closed sets" or make a special provision for the empty set. Aug 30 '13 at 0:29
• @DanielFischer Right, I'll change that. Aug 30 '13 at 1:10
• In textbooks this is called the "Hausdorff metric", in case you wanted to look-up a proof. Aug 30 '13 at 1:22