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this is my first post and wasn't able to find this question anywhere. I am trying to show, that $$\text{dist}(A,B):=\text{inf}\left \{ d(a,b): a\in A, b\in B\right \}$$ is a metric on $\mathcal{P}(X) \backslash \emptyset$ at whereat $(X,d)$ is a metric room, $\mathcal{P}(X)$ is the power set of $X$ and $\emptyset \neq A,B \in X$. The two attributes "identity of indiscernibles" and "symmetry" are done. These just come from the metric $d$. Now I am trying to show the triangle inequality and I am not sure either my proof is okay: $$\text{dist}(A,C)\leq d(a,c)\leq d(a,b) +d(b,c)$$ Then we have

$$ \inf \left \{ \text{dist}(A,B) \right \}\leq \inf\left \{ d(a,b) +d(b,c)\right \}\leq \inf\left \{ d(a,b) \right \}+\inf\left \{ d(b,c) \right \}$$ The last inequality comes from non-negativity of $d$.

Stefan Hamcke
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Niklas Hebe
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2 Answers2

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Hint: Consider what happens when $A$ intersects $B$ and $B$ intersects $C$, but $A$ and $C$ are a positive distance apart.

The reason why the triangle inequality does not hold, is that the $b$ in the term $\inf\{d(a,b):a∈A,b∈B\}$ can be different from the $b$ in $\inf\{d(b,c):b∈B,c∈C\}$. That's why your last inequality isn't valid.

On the other hand, we could have argued that $A$ and $B$ need not be equal, even though $d(A,B)=0$.

If you would still like to have some metric on a subset of the $\mathcal P(X)$, there is indeed a way to achieve this.

Stefan Hamcke
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Definition of (usual) metric requires that $d(A,B)=0$ if and only if $A=B$. But here if $A$ and $B$ intersect the distance is zero without sets being same. So the identity of indiscernibles is on hold.

Consider $A=(0,1)$, $B=(1,2)$ and $C=(2,3)$. Then $d(A,B)=d(B,C)=0$ while $d(A,C)=1$ so triangle inequality is on hold too.

What if you used $sup$ instead of $inf$ in the definition?

Maesumi
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