TITLE: Does one-sided Hausdorff distance satisfy the triangle inequality for non-closed bounded subset?
QUESTION [5 upvotes]: Given $(X,d)$ is a metric space. Suppose that $A,B,$ and $C$ are subsets of $X$ which are bounded but non-closed. 
One side Hausdorff distance is defined by $$d(A,B)= \sup_{x\in A} \inf_{y \in B} d(x,y).$$ Does triangle inequality
$$d(A,B)+d(B,C) \geq d(A,C)$$
hold?

REPLY [4 votes]: From Robert's Hint: $d(a,b) < d(A,B) + \epsilon$
Therefore given $\epsilon_1,\epsilon_2>0$ for each $a\in A$ there exist $b\in B$ and $c\in C$ such that
$$d(a,c)<d(a,b)+d(b,c)<d(A,B)+d(B,C)+\epsilon_1+\epsilon_2$$
thus
$$\inf_{z\in C} d(x,z)<d(a,c)<d(A,B)+d(B,C)+\epsilon_1+\epsilon_2$$
so that $d(A,B)+d(B,C)+\epsilon_1+\epsilon_2$ is an upper bound of  $\inf_{z\in C}d(x,z)$ and we have
$$d(A,C)\leq d(A,B) + d(B,C) + \epsilon_1 + \epsilon_2$$
for arbitrary $\epsilon_1,\epsilon_2>0$, and hence the result.