We continue our work on endpoints and startpoints in

During his investigations on the hyperconvex (or injective) hull of a metric space Isbell [

In [

Improving on a result from [

We also specialize some of our earlier results in [

In this section we first recall some of the basic definitions from asymmetric topology needed to read this paper. Then we recall some fundamental facts of the theory of the

Let

We will say that

Let

Given

A map

Furthermore a map

Given two real numbers

Given a

For further basic concepts used from the theory of asymmetric topology we refer the reader to [

Many facts about hyperconvexity in metric spaces can be found in [

We next recall some facts mainly from [

Let

Let

Then

We will call a function pair

Furthermore

Obviously the second component

Given any real-valued function

One readily checks that

Furthermore, of course,

Hence

It is known (see [

Moreover

For each

Then

Note also that

The following important result (see [

Our first example shows that in that formula the step of taking the supremum with

Let

Furthermore

The following definition was given in [

A finite sequence

An element

An element

Let us finally recall that a quasipseudometric space

The next result says intuitively that the points in the remainder of the

Let

As stated above, for

Fix

Then

It follows that

Let

Assume that

Let

According to the proofs of [

We next prove the result mentioned in Section

Let

In [

Here we will show that each endpoint of

Suppose that

By join compactness of

Taking limits, therefore

We conclude that

The assertion about startpoints is proved analogously.

Let

Given a partially ordered set

We will call

Observe that, if

As we noted in the Preliminaries section, with the help of the specialization order we can equip each

Fix

Let

Hence

Note that obviously

Let us consider two specific examples of this construction.

(a) Let

(b) Let

Let

In the special case of

Let

Let

Of course

We are next going to illustrate the concepts of a startpoint (resp., endpoint) in the case of

Let

Suppose that

Assume that

Let

We next illustrate Lemma

Let

Let

The following definition can essentially be found in [

(a) Each completely

(b) Let

(a) Suppose first that

Suppose now that

(b) For the convenience of the reader we include a proof, which follows [

If

(a) Each completely

(b) Let

Recall that an element

Of course, in a complete lattice the completely

Let

In a partially ordered set

Our characterizations (Lemma

Hasse diagram of

As usual, an element

Let

We recall that a subset

It is known (see [

Let

(a) Suppose that

Since

(b) We continue the proof of part (a) dealing with startpoints. By the additional assumption of meet density of

In this section we will discuss various examples in the light of the results in Section

Let

Indeed we have the following result.

Let

Since

Let

Let

Suppose that

In order to prove the converse suppose that

In

In the set

The closed unit interval of the set of rational numbers equipped with its usual linear order and the natural

Let

Suppose that

Each atom in a partially ordered set

For a set

In fact, for each

Let

Let us give a proof of this statement just using the basic definitions. Suppose that

On the other hand fix

Assume that

Given a nonempty subset

Let

It is well known that each filter on

Furthermore

Let

Thus we are done, since no

Some of the results mentioned in the previous sections may have reminded the reader of the theory of the Dedekind-MacNeille completion (cf. also [

For the following discussion we need some basic facts from the theory of the Dedekind-MacNeille completion of a partially ordered set (see, e.g., [

Let

Let

For the proof we consider

We continue the discussion of Example

Considering

Hasse diagram of

We will say that a partially ordered set

Let

Let

Furthermore by our assumption, obviously we have that

Then

As an illustration it may be useful to include here the following simple example (see [

Let

For the following result compare the discussion preceding [

Let

Suppose that

Let

Consider now arbitrary

Suppose now that

We conclude that there is

Consequently

Recall that

Let

Indeed, more generally, our next result shows explicitly how, given a partially ordered set

Let

Consider an arbitrary pair

Since the functions

Similarly one verifies that

Since

Let

Then the map

By Lemma

(Of course this means exactly that

We conclude that the set

Given a partially ordered set

Let

Given a

In this paper we established that these two sets are equal if

Given a partially ordered set

Let

We note however that, for any

Let

Furthermore

Suppose that

For the converse suppose that

Let us consider the set

We showed that for any join compact

The authors would like to thank the National Research Foundation of South Africa for partial financial support. This research was also supported by a Marie Curie International Research Staff Exchange Scheme Fellowship within the 7th European Community Framework Programme. Finally the authors would like to thank the referee for his or her suggestions.