Let A, B and C bounded sets of $\mathbb{R}$...

Question

Let $A$, $B$ and $C$ bounded sets of $\mathbb{R}$. Define $D_H(A,B)$ as

$$D_H(A,B)=\max\Bigg\{\underset{x\in A, y \in B}{\inf}|x-y|, \underset{x\in A, y \in B}{\sup}|x-y|\Bigg\}$$

Prove that: $$D_H(A,B)\leq D_H(A,C)+D_H(B,C)$$

This is and old exercise from the begin of the course I'm studying, but until today I am not able to prove this. I only need a hint. I'm stucked for a few hours in this exercise. In the moment this exercise was presented we were only allowed to use the definition of $\sup$ and $\inf$ of a set, the completeness of $\mathbb{R}$, stuff like that, nothing more elaborated than that, like sequences, for example.

I tryed to use the triangle inequality, but I felt like I had to separate in many cases and it didn't look right.

Answer 1

Claim #1: $D_{H}(A,B)=\sup_{a\in A,b\in B}|a-b|$

Proof of Claim #1: Pick any $x\in A$ and $y \in B$. Then $$\inf_{a\in A,b\in B}|a-b|\leq |x-y| \leq \sup_{a\in A,b\in B}|a-b|$$ This implies $$\inf_{a\in A,b\in B}|a-b| \leq \sup_{a\in A,b\in B}|a-b|$$ and so $$D_{H}(A,B)=\sup_{a\in A,b\in B}|a-b|$$

Claim #2: For any fixed $c\in C$ we have $$\sup_{a\in A,b\in B}|a-b| \leq \sup_{a\in A}|a-c| + \sup_{b\in B}|b-c|$$

Proof of Claim #2: Choose any $x\in A$ and $y\in B$. Then $$|x-y|\leq |x-c|+|y-c|\leq \sup_{a\in A,b\in B}\Big[ |a-c|+|b-c|\Big] $$This shows $\sup_{a\in A,b\in B} \Big[ |a-c|+|b-c|\Big]$ is an upper bound of the set $$\Big\{|x-y|:x\in A,y\in B\Big\}$$ so by the definition of $\sup$ we get $$\sup_{a\in A,b\in B}|a-b| \leq \sup_{a\in A,b\in B}\Big[ |a-c|+|b-c|\Big]$$ and using this we have $$ \sup_{a\in A,b\in B}\Big[ |a-c|+|b-c|\Big]=\sup_{a\in A}|a-c| + \sup_{b\in B}|b-c|$$ So finally, $$\sup_{a\in A,b\in B}|a-b| \leq \sup_{a\in A}|a-c| + \sup_{b\in B}|b-c|$$

Claim #3: For any fixed $c\in C$ we have $$\sup_{a\in A}|a-c|\leq \sup_{a\in A,z\in C}|a-z|$$ $$\sup_{b\in B}|b-c| \leq \sup_{b\in B,z\in C}|b-z|$$

Proof of Claim #3: Choose $x\in A$ arbitrarily. Then $$|x-c|\leq \sup_{a\in A,z\in C}|a-z|$$ This shows $\sup_{a\in A,z\in C}|a-z|$ is an upper bound of the set $$\Big\{|x-c|:x\in A\Big\}$$ so using definition of $\sup$ we have $$\sup_{a\in A}|a-c|\leq \sup_{a\in A,z\in C}|a-z|$$ An identical argument also proves $$\sup_{b\in B}|b-c| \leq \sup_{b\in B,z\in C}|b-z|$$

Claim #4: $D_{H}(A,B)\leq D_H(A,C)+D_H(B,C)$

Proof of Claim #4: Combine all three previous claims to get the result.