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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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    But what if A,B,C sets are$n$'t bounded? Is triangle inequality hold in this case? Can anyone help?2012-02-27

1 Answers 1

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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)

thus

$\inf_{z\in C} d(x,z)

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.