# Why does the Hausdorff metric need to be defined on bounded subsets only?

Claim: Suppose $X$ is a non-empty set and $d$ is a metric on $X$. Let $S(X)$ denote the collection of all non-empty closed bounded subsets of $X$. For each $A$ and $B$ in $S(X)$, define

$$h(A,B) = \max~ \big ( ~\sup ~\{~ dist~(b,A)~|~b \in B ~\},~ \sup ~\{~dist~(a,B)~|~a \in A~ \}~ \big )$$ Then $h$ is a metric on $S(X)$. It is called the Hausdorff metric.

Proof is as follows :

However, I haven't been able to find an exact use for taking $S(X)$ as the set of all bounded subsets of $X$ which are also non empty and closed.

Could someone please tell me the use of taking bounded subsets here?

Thank you very much for your help in this regard.

If you allow unbounded sets there is no reason for $h(A,B)$ to be finite. For instance, if $X=A=\mathbb{R}$ and $B=\{0\}$, then $\sup \operatorname{dist}(a,B)$ is clearly infinite.