# Does one-sided Hausdorff distance satisfy the triangle inequality for non-closed bounded subset?

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?

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Hint: given $\epsilon > 0$, for each $a \in A$ there is $b \in B$ with $d(a,b) < d(A,B) + \epsilon$... –  Robert Israel Jan 29 '12 at 8:28
Therefore given $\epsilon_1, \epsilon_2 >0$ we have 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 $d(A,B)+d(B,C)+\epsilon_1+\epsilon_2$ is an upper bound of $\inf_{z\in C} d(x,z)$ thus $$d(A,C)\leq d(A,B) + d(B,C) + \epsilon_1 + \epsilon_2$$, for arbitrary $\epsilon_1,\epsilon_2 >0$. is it OK? Thanks Robert. –  Ajat Adriansyah Jan 29 '12 at 9:07
Your work seems correct. You could post this as an answer so that others can check and upvote it. (In fact, you can also accept your own answer, after waiting for a day or so.) –  Srivatsan Jan 29 '12 at 10:49
But what if A,B,C sets aren't bounded? Is triangle inequality hold in this case? Can anyone help? –  user25873 Feb 27 '12 at 23:26

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.

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