#### Abstract

We construct a class of continuous quasi-distances in a product of metric spaces and show that, generally, when the parameter (as shown in the paper) is positive, is a distance and when , is only a continuous quasi-distance, but not a distance. It is remarkable that the same result in relation to the sign of was found for two other classes of continuous quasi-distances (see Peppo (2010a, 2010b) and Peppo (2011)). This conclusion is due to the fact that is a product space. For the purposes of our main result, a notion of density in metric spaces is introduced.

#### 1. Introduction

In this paper a quasi-distance on a set is defined as a function with the usual properties of a metric and a weaker version of the triangle inequality: where .

This function is not always continuous with respect to the -topology generated by itself in the same manner as by a distance. This is because the “open” balls , which form a base for a complete system of neighbourhoods of , are not always open sets in the -topology. Examples of “open” balls that are not open sets can be found in [1].

It is known that for every quasi-distance , there exists another quasi-distance , whose topology and uniformity are the same as those of and the open -balls are open sets [2].

But if any condition or relation is satisfied with respect to a quasi-distance , we do not know if, in general, the same condition or relation will be satisfied with respect to the “new” quasi-distance , even if is equivalent to .

For this reason, the continuity of a quasi-distance cannot be omitted without “danger.”

In fixed-point theory, some authors require the continuity of the used quasi-distance as complementary condition (see, among others, [3]).

The most interesting fact about the quasi-metric spaces is that, in many applications, they constitute a more general setting than the metric spaces without losing the good properties of these last spaces (see, among others, [4]).

In [5–7] we proved that, generally, the functions defined in a product space for two points, and , respectively, by
with for all , are distances when but* only* continuous quasi-distances when .

In this paper we revisit this problem for another family of functions in a product space defined by with and we find the same result.

This makes us think that the conclusions reached are not coincidental.

We have reason to believe that a similar result holds for other classes of quasi-distances in product spaces.

Additionally, we notice that the iterated quasi-triangle constants of our quasi-distances,
which we are calling* generalized improved quasi-triangle inequality*, improve the general one:
(the two last coefficients are ; it is not a mistake).

In Section 2 we introduce the notion of a density condition in metric spaces for the purposes of our main result.

In Section 4 we give a counterexample proving the importance of the -density condition in part III of our main result.

#### 2. Density in Metric Spaces

In the third part of our main result (Section 3) we will need metric spaces satisfying an additional property of density that we are calling -*density condition*.

*Definition 1. *For an and a positive number , the pair of metric spaces and satisfies the -density condition if the space contains at least three distinct points, , satisfying the relations
and for every there exist at least three distinct points , satisfying the relations

In terms of segments, Definition 1 may be expressed as follows.

For an and a positive number , the pair of metric spaces and satisfies the -*density condition* if contains at least a segment, , containing a point dividing it at the ratio and, for every , contains at least a segment , time “longer” in than in and containing a point dividing it at the ratio .

By segment we mean the set defined by the following:

*Definition 2. *For two points in a metric space , the segment is the set of points satisfying

It is obvious that the extremities and belong to the segment , but it may only be reduced at its two extremities.

We note also that, despite appearances, if , , and , this does not mean, in general, that , as in the example below (of course, in the particular case when , we have ).

*Example 3. *Let be two strictly positive numbers satisfying , , a number satisfying , and a set containing whatever four points , , , and of . Define the function by

The two first properties of a distance are obviously satisfied.

So, to prove that is a distance, it suffices to prove the triangle inequality in all possible “triangles” formed with the points of , that is, , , , and (Figure 1).

In the “triangles” and , the “sides” are and the greatest one is , so the “triangle” inequality is satisfied.

In the “triangles” and , the “sides” are, respectively, and and the greatest one is, by our choice, ; or , and , so the “triangle” inequality is satisfied.

The points and belong to , because and . On the other hand , but, generally, .

We note that the -*density condition*, for an and a positive number , as it is required in the third part of our main result (Section 3), is not a very restrictive condition. For example, if is the usual distance in , , and , in order that the pair (; ) satisfies the -*density condition*, it suffices that contains the set , while contains the set as in Figure 2.

Clearly, every pair of normed vector spaces over satisfies the *-density condition* for each and for each positive number .

#### 3. Main Result

Theorem 4. *Let for be metric spaces, and real numbers, and positive numbers satisfying , and and two arbitrary points of . Then*(i)*the function defined by
is a continuous quasi-distance on satisfying the generalized improved triangle inequality for every points;*(ii)

*if or are all nonnegative or all nonpositive, is a distance;*(iii)

*if and there exist two indexes with , and an such that the spaces and satisfy the -*

*density condition*, the continuous quasi-distance is not a distance.*Proof. *(i) First, we will prove that is a quasi-distance and then that is a continuous function of its two variables with respect to the -topology.

It is clear that , that is symmetric, and that .

Let , , and .

Clearly, for whatever points and , we have

As is a distance, we can write, for whatever points , , ,

This means that is a quasi-distance with constant , and the same constant holds for whatever points. Let us show now that is continuous with respect to the -topology.

Inequalities (11) show that and are topologically equivalent. Hence to prove that is continuous, we will show that for every two sequences, and , converging, respectively, to and for , converges to .

For the points and , in the following, we will denote , , and .

If , an integer exists such that for , we have () (because the sequence converges to ).

For , , we have () and, consequently, converges to and ().

If , from we deduce
That is, .

It follows that and converge to the same limit . We can conclude that , formed by the terms of the two sequences above, also converges to . Thus, is continuous and the “open” balls really are, in fact, open sets for the -topology.

(ii) If s are all nonnegative, for every and ; if s are all nonpositive, for every and (recall that if , ). As and are distances, is a distance too.

If , for every , so for every and ; that is, is a distance.

Suppose now that ; for two arbitrary points and , denoting again we have

So, for it follows that

This means that if , , and since the maximum of two distances is a distance, * is a distance*.

Notice that in this part of the theorem we did not need any supplementary density condition on the spaces .

(iii) Suppose now that and that there exist an and two indexes such that , , and the pair satisfies the -*density condition*.

We choose a satisfying the relation Such a exists because, passing into limit in (16) as , we obtain that is clearly true.

As satisfies the -*density condition*, there exist three points , satisfying the relations and , and three points , satisfying the relations

With the points and we construct the points , , and of (except the th and the th coordinates, all the other coordinates are equal in , and ) and will prove that

For the points and we have from (18) that so .

For the points and , because and , so .

For the points and , because and , so .

We are ready to prove now the inequality , transforming it equivalently to From and , we have , so is positive and we can equivalently square (because and ), (we square) Dividing by we have The last inequality is true from (16). The proof is over. So, is not a distance.

*Example 5. *For and , the function defined by
is a continuous quasi-distance, but not a distance. Here, , , , , , and , while the function defined by
is a distance.

Here, , , , , , , and .

#### 4. A Counterexample

Showing the importance of the -*density condition* in part III of the theorem.

Let , be two metric spaces with distances, respectively, , , , , real numbers, and positive numbers satisfying, for , , , , , and .

For whatever , it is obvious that the pair does not satisfy the -*density condition*.

We will show that the function defined for every two points by is a distance.

For every three points of , we will show that the greatest “distance” (we put quotes because we do not yet know if is really a distance) between these points is inferior or equal to the sum of the two others.

##### 4.1. First Case

The six “distances” are

As , we have , so

The greatest of the six “distances” is .

In each of the four possible “triangles,” , , , and , there is a side equal to , another is equal to , and the third is .

Therefore we have to prove only one “triangle inequality,” valid for the four triangles: or equivalently

But from we have and from we have , so and the triangle inequality is satisfied.

##### 4.2. Second Case

The six “distances” are

As , we have , so , and .

The greatest of the six “distances” is . In each of the four possible “triangles,” , , , and , there is a side equal to , another is equal to , and the third is . Therefore we have to prove only one “triangle inequality,” valid for the four triangles: that we square equivalently

But from we have and from we have , so and the triangle inequality is satisfied. So, is a distance.

We note that if are normed vector spaces, , very slight modifications transform our main result in its normed version and that, in this case, no density condition is needed.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

The author would like to thank Redon A. Cabej for his valuable suggestions which improved the final version of this paper.