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Let $M$ be a metric space and consider $Y(M)$ the set of all closed and bounded subsets of $M$. Consider the function $ p:y\left( M \right)^2 \to R $ defined by:

$$ p\left( {X,Y} \right) = \max \left\{ {\mathop {\sup }\limits_{x \in X}\, d\left( {x,Y} \right),\mathop {\sup }\limits_{y \in Y}\, d\left( {y,X} \right)} \right\} $$ where the distance of a point $x$ and a set $A$ is defined by:

$ d\left( {x,A} \right) = \inf \left\{ {d\left( {x,a} \right);a \in A} \right\} $ Prove that this function is a distance.

I proved all the properties except the triangular inequality. How can I prove it?

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    $\begingroup$ It seems you're missing the condition that the sets need to be non-empty? Otherwise in $p(\emptyset,\emptyset)$ each of the suprema is the supremum of an empty set, which is $-\infty$, so the maximum would also be $-\infty$, and not $0$ as required. $\endgroup$ – joriki Mar 28 '12 at 17:04
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We have to assume that the sets in $Y(M)$ are non-empty. Take $(x,y,z)\in X\times Y\times Z$. Then $d(x,z)\leq d(x,y)+d(y,z)$ so take the infimum for $z\in Z$ we get $$d(x,Z)\leq d(x,y)+d(y,Z)\leq d(x,y)+p(Y,Z)$$ and now we take the infimum over $y\in Y$ to get $d(x,Z)\leq d(x,Y)+p(Y,Z)$ so $d(x,Z)\leq p(X,Y)+p(Y,Z)$ and $\sup_{x\in X}d(x,Z)\leq p(X,Y)+p(Y,Z)$. Now we go back to $d(x,z)\leq d(x,y)+d(y,z)$ and take the infimum over $x\in X$ to get $$d(z,X)\leq d(y,X)+d(y,z)\leq p(X,Y)+d(y,z)$$ then the infimum over $y$ we have $$d(z,X)\leq p(X,Y)+d(z,Y)\leq p(X,Y)+p(Y,Z)$$ which gives the result.

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