An example and remarks

The following example shows that a separable metric space having the property (a) alone need not be homeomorphic to a subspace of .

Example. There exists a separable metric space (X, d) having the property (a) which is not suborderable.

Proof: Let be an upper bound of and put . Define . Let X be the subspace of the Euclidean plane defined by , where for each and {(0,0)}. Since the component has no clopen neighborhood base, it follows from Theorem 3 that X is not suborderable. We define a compatible metric d on X for which (X, d) has the property (a). To do this, put for each and define a sequence for each by the following rules: In case , 0 for each . In case , and

Then

and

For every pair of points and , with , in X, define

Note that n may be and by our convention. We must show the triangle inequality for , and r = (z, . Now, we prove it only in the cases where m ;SPMlt; k ;SPMlt; n and m ;SPMlt; n ;SPMlt; k ; the proof for other cases is left to the reader.

Case 1. m ;SPMlt; k ;SPMlt; n. Since ,

which implies that

Since and , it follows that

Case 2. m ;SPMlt; n ;SPMlt; k. Since

and

we have

and

respectively. Hence,

Thus, d is a metric on X and it is easily checked that d is compatible with the topology of X. To show that (X, d) has the property (a), let and be distinct points in X. Put . All we have to show is that . If m = n, then | B| = 1 by the definition of d. Thus, we suppose that m ;SPMlt; n. Then for each by the definition of d. Hence, it suffices to show that for at most one k. For this end, define a sequence by the following rules: For ,

for ,

If for k ;SPMlt; m, then , where . Since , this implies that . Conversely, if the last inequalities hold, then , where . Thus, we have:

(1) For k ;SPMlt; m, if and only if .

Similarly, putting for each k, we have the
following:

(2) if and only if
;

(3) for m ;SPMlt; k ;SPMlt; n, if and only if ;

(4) if and only if ;

(5) for , if and only ;

(6) if and only if .

By the definition of the 's, (1)-(5) above imply the following (1')-(5'), respectively:

(1') for k ;SPMlt; m, 0 if ;

(2') if ;

(3') for m ;SPMlt; k ;SPMlt; n, if ;

(4') if ;

(5') for , if .

Hence, if the following (7)-(9) were proved, then it would follow from (1')-(5') and (6) that for at most one k.

(7) If , then .

(8) If , then .

(9) If , then .

It remains to prove (7)-(9).

Proof of (7). If , then

Since

and

we have

and

Proof of (8). If , then

Since

and

we have

On the other hand, the last term of (10) is:

Hence, .

Proof of (9). If , then

Since

and

we have

and

Hence, the proof is complete.

Remarks 2. The authors do not know whether the example can be strengthened by replacing the property (a) by UMP or UTP. We now show that a metric space need not have UMP even if it is homeomorphic to a subspace of . The simplest example is a metric space consisting of two points. A less trivial one is a convergent sequence. To see this, let be a convergent sequence, where , and suppose that has a compatible metric d for which has UMP. Then for every pair of distinct points m and n in , there is a unique such that d(m, p) = d(n, p). Choose such that for each . Then, p(1, n) ;SPMlt; k for each . For, if for some , then , which is a contradiction. Hence, there exist and an infinite such that for each . Similarly, using instead of 1, we can find and an infinite such that for each . Choose distinct m and n in . Then, for each i = 1, 2, which contradicts UMP. It is interesting to know what kind of subspace X of has a compatible metric d for which (X, d) has UMP. For examples of such a subspace, see [1] and [7].

The authors would like to thank N. Murase and K. Yamada for helpful discussions on several points of the paper. They also acknowledge L. D. Loveland, who read the first version of the paper and kindly informed them the existence of Nadler's paper [7].