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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 $d(a,b) < d(A,B) + \epsilon$...2012-01-29
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    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 $$d(a,c) thus $$\inf_{z\in C} d(x,z) 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.2012-01-29
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    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.)2012-01-29
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    But what if A,B,C sets aren't bounded? Is triangle inequality hold in this case? Can anyone help?2012-02-27

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