• Abstract
• Introduction
• Proof of the theorems
• NEXT: An example and remarks
• References

Proof of the theorems

To prove Theorems 1 and 2, we use the following theorem due to Purisch [8] and some lemmas. A space is called suborderable if it is homeomorphic to a subspace of a linearly ordered space.

In [8, Theorem 3.3], the set U in the condition (iv) is not assumed to be closed; however, we can add ^^ closed' by the remark after the proof of [8, Theorem 3.3]. Nadler [7] essentially proved that a connected subspace, consisting of more than one point, of a metric space having the property (a) has UMP, and hence it is homeomorphic to an interval by Berard's theorem stated in the introduction. Hence, the following lemma is a direct consequence of [7, Lemma 2.3]; the statement is included for reader's convenience. Recall that a point p of a connected set C is a cut point of C if is disconnected.

Recall from [2] that a quasi-component of a point p of a space X is the intersection of all clopen sets of X containing p.

Proof: Let Q be a quasi-component of a point of X. It suffices to show that every pair of distinct points in Q is included in a connected subset of Q. To show this, let and be distinct points in Q. Put and . Then by the property (a). If , then is clopen in X and contains but not . This contradicts the fact that Q is a quasi-component. Thus, we put . Suppose that b is not in Q and G is a clopen set of X such that . Then is clopen in X and contains but not , which is a contradiction. Hence, . Put and for i = 1, 2. It is clear that and for i = 1, 2. It remains to show that each is connected. First, we show that for each positive number and each i = 1, 2. If , then , because by the property (b). If and , choose a clopen set H in X such that and . This can be done because . Then is clopen in X and contains but not , which is a contradiction. Hence, for each i = 1, 2. Since , this implies that for each i = 1, 2. Hence, by the property (b), . To complete the proof, suppose that is disconnected. Then there is a proper clopen subset V in with . Pick a point p of and let . Since , is clopen in X and contains b but not p. This contradiction completes the proof.

Proof of Theorem 1: By [6, Corollary 5.6], a separable suborderable metric space X can be embedded in a separable linearly ordered metric space , and can be embedded in by [2, 6.3.2(c) p.373]. Hence, it suffices to show that a separable metric space X having the properties (a) and (b) is suborderable. We show that X satisfies the conditions (i)-(iv) in Theorem 3. To prove (i)-(iii), let C be a component of X. Since C is a connected metric space having the property (a), C has UMP. Thus, C is a singleton or homeomorphic to an interval by Berard's theorem stated in the introduction. Hence, (i) is satisfied. The number of non-cut points of C is at most two, and by Lemma 1, each cut point of C is in the interior of C. Hence, (ii) is satisfied. Moreover, it follows from Lemma 2 that C is an intersection of clopen sets in X. In case |C| = 1, say , it is easily checked that C has a clopen neighborhood base, because for each by the property (b). In case C has no boundary point, C itself is clopen in X. Thus, it remains to settle the case where C is homeomorphic to an interval and has a boundary point. By Lemma 1, the boundary of C consists of at most two points. We put . Then Y has the properties (a) and (b) and is a component in Y for each boundary point y of C. Hence, similarly to the case where |C| = 1, we can find a clopen neighborhood base of y in Y. If the boundary of C consists of only one point y, put . If the boundary of C has two points and , put . In each case, is a clopen neighborhood base of C in X, thus proving (iii). Finally, to prove (iv), let U be a hereditarily disconnected, closed subspace of X. Since U has the properties (a) and (b), by the same way as above, we can find a clopen neighborhood base of x in U for each . Hence, . Since U is a separable metric space, it follows from [2, Theorem 7.1.11] that . This completes the proof.

As stated in the preceding proof, a separable suborderable metric space is homeomorphic to a subspace of . Hence, Theorem 2 follows from the next theorem.

Proof: We show that X satisfies the conditions (i)-(iv) in Theorem 3. Similarly to the proof of Theorem 1, (i) and (ii) are proved. Since X is locally compact, a component C of X with |C| = 1 has a clopen neighborhood base by the proof of [2, Theorem 6.2.9]. Hence, by the same argument as in the proof of Theorem 1, (iii) can be proved. Finally, (iv) follows from [2, Theorem 7.1.12]. The proof is complete.

Remarks 1. The example in the next section shows that local compactness cannot be removed from Theorems 2 and 4. By [7, Example 3.4], separability cannot be removed from Theorems 1 and 2. A space is called peripherally compact if each point has a neighborhood base consisting of open sets with compact boundary. All locally compact spaces and all metric spaces having the property (b) are peripherally compact. It is open whether a peripherally compact and separable metric space (X, d) having the property (a) is homeomorphic to a subspace of .