We start with studying the space .
Fix a point and set . We are interested, above all, in unbounded metric spaces, i.e., in the situation when , but all the results obtained from now on hold also in the case of . Set
and define the function in the following way:
(the latter occurs only if ).
We will need the following properties of this function: for every , the function is non-decreasing; for every , the function is non-increasing.
Show that is non-decreasing. Let . If , then from it follows that . In the case when , (this situation appears only for ), one gets
Finally, if , , then .
Let us illustrate the definition of the function D on some concrete metric spaces.
Example 1 For any linear normed space X with any choice of , the function is given by , .
If , , then no matter what point is chosen as , for any , , the value of is equal to the number of integers belonging to the interval .
One gets the same values for if , , and . If , the value coincides with the number of integers in the set when or , and is equal to the number of integers in the set when .
Next, consider , , . Then , , and for , the function is defined as follows:
if , then
if , then
For arbitrary closed and , we determine now . Next, we show that this distance is finite and also give the properties of the function ; these properties will be used in the sequel.
Lemma 1 Let be given. Then for every and the following statements hold true:
the function is non-decreasing;
for every , the inequality
and the relation
denoted , the inequality holds for each ;
denoted , the inequality holds for each .
Proof First of all, note that for any sets , one has
To show this inequality, we estimate the deviation . Since
for every and
for every , we get the inequality . Similarly, it can be checked that . So, (8) is proved.
From inequality (8), it follows that for every . Indeed, this is obvious if ; in the case when , one has , and hence,
Using (8) one gets relation (6)
Next, for , , hold. So, as a consequence of (8), the following holds:
Thus, the function is non-decreasing.
Now, we prove equality (7). If , then , for , so (7) is true.
Let . First, suppose that . Since the function is non-decreasing and bounded, there exists . Then for each r, the inequality takes place, and it follows that for any , every , and , one has and . Thus, taking into account , one gets , i.e., . As above, for each point , holds. Hence, for any , which means that . The inequality is easily obtained by passing to the limit in (6).
Now, let . If (7) is not true, then there exists . So, arguing as before, one gets the inequality , which contradicts the initial assumption. Thus, equality (7) is proved.
In order to prove statement (3), note that for every , holds and hence . If , , then, according to the inclusions
Similarly, it follows that . So, . For every , , holds and therefore,
Property (4) is a consequence of the well-known estimate (see, e.g., )
Indeed, assuming , and taking into account that for every , holds, one gets
Theorem 1 Equality (5) defines a metric in the space . If X is a complete metric space, then is also complete.
Proof Show that satisfies all the axioms of a metric, i.e., that for any , the following hold:
, and if and only if ;
It is obvious that and for any . According to Lemma 1, one has
Therefore, . Next,
The symmetry of is straightforward; property (3) follows from the inequalities
Thus, is a metric space.
Now, let X be complete; show that is also a complete metric space.
Consider a fundamental sequence . For any , there is a number such that for all , the following inequality takes place:
From (10), it follows that the number sequence is fundamental, so it is convergent; denote its limit by ; obviously, .
Take an arbitrary radius and consider the sequence . It follows from (10) that for any and all , the inequality takes place. Show that for the sequence and all , the similar inequality holds:
Without loss of generality, assume . Pick an arbitrary . If , then since . In the case when , there exists a point satisfying the estimate . Held , one gets , therefore . So,
Similarly, and relation (11) is proved.
The sequence is fundamental with respect to the metric dist and since the space is complete (see ), this sequence converges to some ‘limit’ set . According to inequality (8), the following holds:
and therefore . Moreover, (this equality follows from statement (4) of Lemma 1).
For every , the set is not empty for . Prove that for any , holds. Let . Then there exists a sequence , , convergent to x. This means that ; moreover, . Conversely, let , then , , and hence there is a sequence , , . Starting with some index I, all the members of this sequence satisfy the condition , and therefore . Since converges to , it follows that . So, .
From what has been proved, one may conclude that as r increases, the sets ‘expand’ in the following way: if , then (i.e., is a subset of containing elements x such that .) Define now the set and show it is closed. Note that for every r, holds and the set is closed. Consider a sequence convergent to y. This sequence is bounded, so starting with some index i, the inclusions , , where , hold. Thus, since is closed, it follows that ; on the other hand, . So, and F is closed.
Finally, show that the sequence converges to the set F with respect to the metric . For an arbitrary , put . Since the sequence converges (in the Hausdorff metric) to the set , for every i, starting with some I, the following inequalities hold:
According to Lemma 1, the function is non-decreasing. Hence, from (12), it follows that for every . Next, for , one has . Thus, for all , and therefore , i.e., . □
Remark 1 The function , given by (4), also defines a metric in the space , but for a complete metric space X, the space , unlike , may not be complete. For example, the sequence of the sets (each consisting of a natural number) in the space is fundamental :
yet there is no nonempty set such that
Let us show this. If such a set does exist, then taking any of its points , one gets the following estimates: for every and each ,
Thus, , where is a solution of the equation . So, the inequality
holds, which contradicts relation (13).
The Hausdorff distance between sets has a simple geometric interpretation:
Regarding the metric , the analogue of the ε-neighborhood of a set A is given by
(i.e., by the ε-neighborhood of the set ). This means that the ‘main part’ of the distance can be defined by the equality
Let us now give a criterion for a sequence of closed sets to be convergent with respect to the metric .
Lemma 2 Given , , the convergence implies and for any .
Conversely, let . If there exist and such that for every , the sequence converges in the metric dist to some set , then for any r, such that , holds, and in the space the sequence converges (with respect to ) to the set ; moreover, .
We omit the proof of this statement since it repeats that of the corresponding result for the finite dimensional case .
We conclude this section by proving a theorem on the connection between convergence in the metric and convergence in the metric dist. Note that if a sequence converges in the Hausdorff metric, then by statement (2) of Lemma 1, it converges (to the same limit) in the metric ; moreover, as it will be shown later, in the metric . The converse is not true. For example, in ℝ the sequence of sets , , converges in the metric to the set . On the other hand, for any i, j, one has , i.e., the sequence is divergent in the Hausdorff metric.
Theorem 2 Let , . Then the following statements hold:
if , then ;
if and there is a such that , , then and .
Proof (1) Suppose , . Then, for an arbitrary , there is a number such that for every . So, according to inequality (9), one gets , and from (6) it follows that the inequality holds for every . Thus, by Lemma 2, .
(2) Now, let . Then from Lemma 2, it follows that for every . Next, since for every , holds, then . Indeed, if this inclusion fails, then there exists an such that , so one gets the estimate
that contradicts the relation . Hence .
Let . Since , for every i one has
Next, for every , , we have
and therefore , i.e., . Thus, and
Similarly, . So, it is proved that
hence . □