# Hausdorff metric satisfies triangle inequality

Let $A$ be a nonempty compact subset of $\mathbb{R}^n$. Let $\epsilon$ be some positive real number. Define $$U_\epsilon (A) = \{ x\in \mathbb{R}^n \mid \inf_{a\in A} d(x,a) <\epsilon \}$$ Also, for two nonempty compact subset $A$, $B$, define $$d_H (A,B) = \inf \{ \epsilon >0 \mid A\subset U_\epsilon (B) \text{ and } B\subset U_\epsilon(A)\}$$ , where $d(x,y)$ is the Euclidean distance.

Show that for nonempty compact subset $A,B,C$, we have $d_H (A,C) < d_H (A,B) + d_H (B,C)$

How should I show this? Should I use $d(x,z)<d(x,y)+d(y,z)$ somewhere along the way? I tried to unpack the definitions (which worked when I tried to show that $d(A,B)=0 \implies A=B$) but it seems to complicated.

Let $\epsilon,\epsilon'>0$ be such that $$A\subseteq U_\epsilon (B), \quad B\subseteq U_\epsilon(A);$$ $$B\subseteq U_{\epsilon'} (C), \quad C\subseteq U_{\epsilon'}(B).$$ Note that if $x\in U_\epsilon (B)$, then by definition there exists $b\in B$ such that $d(x,b)<\epsilon$. As $b\in B\subseteq U_{\epsilon'} (C)$, there exists $c\in C$ such that $d(b,c)<\epsilon'$. It follows that $$d(x,c)\leq d(x,b)+d(b,c)<\epsilon+\epsilon'\quad\Rightarrow\quad A\subseteq U_{\epsilon+\epsilon'}(C).$$ Similarly we can show that $C\subseteq U_{\epsilon+\epsilon'}(A)$. Therefore we have \begin{aligned}&\{ \epsilon >0 \mid A\subset U_\epsilon (B) \text{ and } B\subset U_\epsilon(A)\}+\{ \epsilon' >0 \mid B\subset U_{\epsilon'} (C) \text{ and } C\subset U_{\epsilon'}(B)\}\\ \subseteq&\{ \delta >0 \mid A\subset U_\delta (C) \text{ and } C\subset U_\delta(A)\}.\end{aligned} Take infimum on both sides yields $$d_H(A,B)+d_H(B,C)\geq d_H(A,C).$$