# Proof: the Hausdorff metric is finite under the boundedness assumption

Let $(E,\rho)$ be a metric space and $\mathcal{F}$ the family of non-empty, closed and bounded subsets of $E$. For any $X,Y\in\mathcal{F}$, define $d(X, Y)=\max\{\sup\{\rho(x,Y):x\in X\},\sup\{\rho(y,X):y\in Y\}\},$

where $\rho(x,Y)$ denotes the distance of point $x\in X$ from set $Y$ and $\rho(y,X)$ is the distance of point $y\in Y$ from set $X$, that is $\rho(x,Y)=\inf\{\rho(x,y):y\in Y\},~\rho(y,X)=\inf\{\rho(y,x):x\in X\}.$

It can be shown that $(\mathcal{F},d)$ is a metric space, and the metric $d$ is known as the Hausdorff metric. In order to prove that $d:\mathcal{F}\times\mathcal{F}\rightarrow \mathbb{R}$ is a metric, the following three properties of (the definition of) metric need to be shown:

1. $d(X,Y)=0\Leftrightarrow X=Y,$
2. $d(X,Y)=d(Y,X),$
3. $d(X,Y)\le d(X,Z)+d(Z,Y).$

Proofs of the validity of these three properties for the Hausdorff metric are readily available in several books of topology. To ensure that the Hausdorff metric is well-defined, its finiteness will be shown.

Let $x\in X$ and $y\in Y$. Then, since $\rho(X,Y)=\inf\{\rho(x,y):(x,y)\in X\times Y\}$,

it follows that for every $n\in\mathbb{N}$ there exist $z\in X$ and $w\in Y$ such that $\rho(z,w)\le\rho(X,Y)+\frac{1}{n}.\ \ \ \ (1)$

Indeed, assume that (1) is not true. Then there exists $n_0\in\mathbb{N}$ such that for all $z\in X$ and $w\in Y$ it holds that $\rho(z,w)>\rho(X,Y)+\frac{1}{n_0},$

whence $\inf\{\rho(z,w):(z,w)\in X\times Y\}=\rho(X,Y)\ge \rho(X,Y)+\frac{1}{n_0}\Rightarrow\frac{1}{n_0}\le 0,$

so a contradiction has been reached, which means that (1) actually holds.

According to the triangle inequality of metric $\rho$, $\rho(x,y)\le \rho(x,z)+\rho(z,w)+\rho(w,y).\ \ \ \ (2)$

It thus follows from (1) and (2) that for every $n\in\mathbb{N}$ $\rho(x,y)\le \delta(X)+\rho(X,Y)+\delta(Y)+\frac{1}{n}.\ \ \ \ (3)$

Note that (3) holds because the sets $X,~Y$ are bounded, which means that their respective diameters $\delta(X),~\delta(Y)$ satisfy $\rho(x,z)\le\delta(X)=\sup\{\rho(q,q):q\in X\}<\infty,$ $\rho(w,y)\le\delta(Y)=\sup\{\rho(q,q):q\in Y\}<\infty.$

Since (3) holds for all $n\in\mathbb{N}$, it follows that $\rho(x,y)\le \delta(X)+\rho(X,Y)+\delta(Y).\ \ \ \ (4)$

Indeed, assume that (4) is not true. Then it holds that $\delta(X)+\rho(X,Y)+\delta(Y)<\rho(x,y),$

and, in combination with (3), it is deduced that for any $n\in\mathbb{N}$ $0<\rho(x,y)-(\delta(X)+\rho(X,Y)+\delta(Y))\le\frac{1}{n},$

which is a contradiction because it follows from the Archimedean property that there exists $n_0\in\mathbb{N}$ such that $\frac{1}{n_0}<\rho(x,y)-(\delta(X)+\rho(X,Y)+\delta(Y)),$

hence (4) is true.

Recalling that $\rho(x,Y)=\inf\{\rho(x,y):y\in Y\}\le\rho(x,y),$

it follows from (4) that for any $x\in X$ $\rho(x,Y)\le \delta(X)+\rho(X,Y)+\delta(Y).\ \ \ \ (5)$

Since (5) holds for any $x\in X$, it follows that the set $\{\rho(x,Y):y\in Y\}$ has an upper bound, therefore according to the completeness axiom for real numbers, $\{\rho(x,Y):y\in Y\}$ has a supremum. In particular, $\sup\{\rho(x,Y):x\in X\}\le \delta(X)+\rho(X,Y)+\delta(Y).\ \ \ \ (6)$

By means of symmetry of (6), this also means that $\sup\{\rho(y,X):y\in Y\}\le \delta(X)+\rho(X,Y)+\delta(Y),\ \ \ \ (7)$

which completes the proof.