Tag Archives: Hausdorff

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



\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


and, in combination with (3), it is deduced that for any n\in\mathbb{N}


which is a contradiction because it follows from the Archimedean property that there exists n_0\in\mathbb{N} such that


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.