Prove that $\delta$ is a metric in $\mathcal{K}(X)$

Question

Let $(X,d)$ be a complete metric space. We define

$\mathcal{K}(X)=\{K \subset X : K \text{ is compact and non empty}\}$

Define $d'(A,B)=sup_{a \in A}\{d(a,B)\}$

Show that $\delta$ defined as $\delta(A,B)=max\{d'(A,B),d'(B,A)\}$ is a metric in $\mathcal{K}(X)$

Ok, so I was able to show that $\delta (A,B) \geq 0$ and it's $0$ iff $A=B$. I was also able to prove that $\delta$ is symmetric. However, I am not able to show the triangle inequality. I don't know how to use the hypothesis that $X$ is complete. (The exercise consists of two more other items I was able to prove without using completeness of $X$.) Any hint?

Answer 1

We need to show that:

$$d'(A,C)\le d'(A,B)+d'(B,C)\le\delta(A,B)+\delta(B,C).$$

Reversing the roles of $A$ and $C$ shows the other inequality in order to show the full triangle inequality for $\delta$. Now let's unravel all the suprema and infima.

• Goal: $d'(A,C)\le d'(A,B)+d'(B,C)$
• Let $a\in A$, goal: $d(a,C)\le d'(A,B)+d'(B,C)$
• $d(a,B)\le d'(A,B)$, thus goal: $d(a,C)\le d(a,B)+d'(B,C)$
• Let $b\in B$, goal: $d(a,C)\le d(a,b)+d'(B,C)$
• $d(b,C)\le d'(B,C)$, thus goal: $d(a,C)\le d(a,b)+d(b,C)$
• Let $c\in C$, goal: $d(a,C)\le d(a,b)+d(b,c)$
• $d(a,C)\le d(a,c)\le d(a,b)+d(b,c)$ and we're done.

So actually there is no completeness necessary for verifying the triangle inequality - just follow your nose on the definitions and it proves itself. The above is a "backwards proof" along the lines that one would normally approach the solution; it can be rewritten in forward proving style as follows:

Let $a\in A,b\in B,c\in C$. Then $d(a,C)\le d(a,c)$, and $d(a,c)\le d(a,b)+d(b,c)$, so by $d(b,C)=\inf_{c\in C}d(b,c)$ we have $d(a,C)\le d(a,b)+d(b,C)\le d(a,b)+d'(B,C)$. Since this is true for every $b$ we have $d(a,C)\le d(a,B)+d'(B,C)\le d'(A,B)+d'(B,C)$, and since this is true for every $a$ we have $d'(A,C)\le d'(A,B)+d'(B,C)$ as desired.

Answer 2

You have $\delta(A,B)$ = X or Y ; $\delta(A,C)$ = Z or W; $\delta(C,B)$ = U or V thus you have to verify (in (X,d)!) the six resulting possibilities.