Prove that $d_X$ is a metric on $X$, where $X$ is the set of all closed subsets of the unit disk in $\mathbb{R}^2$.

Question

Consider the closed unit disk $$D = \big\{(x, y) \in\mathbb{R}^2\,\big| x^2+y^2\leq 1\big\} \subseteq \mathbb{R}^2$$ with the metric $d$ induced from the Euclidean metric on $\mathbb{R}^2$, and let $X$ be the set of all closed subsets of $D$. For $A\in X$ and $x \in D$, define $$\text{dist}(x, A) := \min_{a\in A}\, d(x, a)\,.$$ For $A, B \in X$, define $$d_X(A, B) := \max \left\{\max_{a\in A}\,\text{dist}(a, B),\max_{b\in B} \text{dist}(b, A)\right\}\,.$$ Show that $d_X$ defines a metric on $X$.

How can I prove that $d_X$ satisfies the triangle inequality?

Answer 1

Hint. For simplicity, I shall write $\delta$ for $\text{dist}$. Note that, for $A,B,C\in X$, we have $$\begin{align}d_X(A,B)&+d_X(B,C)\\&\geq \max\left\{{\small\max_{a\in A}\,\delta(a,B)+\max_{b\in B}\,\delta(b,C)}\,,\,\,{\small\max_{b\in B}\,\delta(A,b)+\max_{c\in C}\,\delta(B,c)}\right\}\,.\end{align}$$ Next, verify that $$\max_{a\in A}\,\delta(a,B)+\max_{b\in B}\,\delta(b,C)\geq \max_{a\in A}\,\delta(a,C)$$ and that $$\max_{b\in B}\,\delta(A,b)+\max_{c\in C}\,\delta(B,c)\geq \max_{c\in C}\,\delta(A,c)\,.$$ One probably important remark is that every set in $X$ is a compact subset of $\mathbb{R}^2$.