Mem. Fac. Sci. Eng. Shimane Univ.
Series B: Mathematical Science
38 (2005), pp. 31–38

UNIQUE MIDPOINT PROPERTYIN METRIZABLE SPACES: A SURVEY
YURI KITAI
Communicated by Yasunao Hattori(Received: March 1, 2005)
Abstract. We shall give a brief survey of metric spaces with the unique midpoint property (abbreviated UMP).

1. Introduction
A metric space (X, ρ) is said to have the unique midpoint property (abbreviated
UMP) if for every pair of distinct points x and y of X there is an exactly one
point p ∈ X such that ρ(x, p) = ρ(y, p). We call the point p the midpoint of x
and y. For example, the real line with the usual metric and its subspaces of therational numbers and the closed interval [0, 1] have UMP, while the metric subspaceof the irrationals, the usual Cantor set and the union [−1, 0] ∪ [1, 2] of exactly twoof disjoint closed intervals do not have UMP. Several authors studied the relation
between the unique midpoint property and the embeddability of spaces into thereal line R, cf., [1], [4], [6], [9] and [10]. In the present note, we shall present a brief
survey in this field.
In sections 2 and 3, we shall describe some results about connected and nonconnected metric spaces with UMP from [1] and [9] respectively. We shall present
a proof of Theorem 2.2, because the original proof from [1] contains a gap andwe correct it. In section 4, we shall consider UMP as a topological property, notmetric one: A metrizable space X is said to have UMP if there is a compatible
metric ρ on X such that (X, ρ) has UMP in original sense. As mentioned above,
although the space of irrational numbers, the Cantor set and [−1, 0] ∪ [1, 2] do nothave UMP as metric subspaces of R, it seems to be natural that we ask whether
if they have any compatible metrics with UMP. The main interest in the section is
the following question: Which subspaces of the real line R have UMP?
2000 Mathematics Subject Classification. 54E35.
Key words and phrases. Metric spaces, unique midpoint property (UMP), connected space,
separable metric space, real line, the space of the rationals, the space of the irrationals andCantor set.
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We refer to [2] and [5] for general terminology and basic facts on general topology
and to [3] for other results on metric spaces having special properties which relateswith UMP.
2. Connected metric spaces with UMP
We will use the following notation throughout this section: For a metric space(X, ρ), a point x ∈ X and ε > 0,
B(x, ε) = {q ∈ X : ρ(x, q) < ε},R+ = {x ∈ R : x > 0}.
A point c of a topological space X is said to be a cut point of X if and only ifthere exist disjoint open sets U and V of X such that X − {c} = U ∪ V .
It is well known that (cf. [5]) if X is a metric contimuun (=comact connencted
space) and X has just two non-cut points, then X is a homeomorphic to the closed
unit interval [0, 1] in the real line R. In [1], A. D. Berard considered non-comapact
metric space with UMP and obtained the following characterization of such spaces.
Theorem 2.1 ([1]). If X is a connected metric space with UMP, then X is homeomorphic to an interval in R. Furtheremore, we have the following:(i) If X has two distinct non-cut points, then X is a closed interval.
(ii) If X has an exactly one non-cut point, then X is a half open interval.
(iii) If X has no non-cut points, then X is an open interval.
We notice that every connected metric space with UMP has at most two distinct
non-cut points ([1, Theorem 3]).
To prove the theorem above, A. D. Berard introduced a partial order in a connected metric space with a non-cut point as follows: Let (X, ρ) be a connected metric space with a non-cut point z. Then for x, y ∈ X we say x < y if ρ(x, z) < ρ(y, z).
The following theorem is a key of the proof of Theorem 2.1.
Theorem 2.2 ([1]). Let (X, ρ) be a connected metric space with the UMP and z
a non-cut point of X. Then the order topology induced by the partial order defined
above coincides the original (metric) topology.
We shall describe the proof of the theorem, because the original proof due to A.
D. Berard contains a gap (see Remark 1 below).
We need three lemmas ([1, Lemma 4], [1, Lemma 7] and [1, Lemma 8]) to prove
the theorem.
Lemma 2.1 ([1]). Let (X, ρ) be a connected metric space with UMP, and z1 and z2
two distinct non-cut points of X. Then X ⊆ [B(z1 , ε) ∩ B(z2 , ε)] ∪ {z1 , z2 }, whereε = ρ(z1 , z2 )Lemma 2.2 ([1]). Let (X, ρ) be a connected metric space with UMP and z a noncut point of X. Then for each a, b ∈ X with a < b, I(a, b) is a connected metricspace with UMP and exactly two non-cut points a and b, where I(a, b) = {x ∈ X :a ≤ x ≤ b}.

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Lemma 2.3 ([1]). Let (X, ρ) be a connected metric space with UMP and z a
non-cut point of X. Then for each cut point m ∈ X and each ε > 0 there exist
x, y ∈ B(m, ε) such that x < m < y.
Now, we are ready for proving Theorem 2.2.
Proof of Theorem 2.2. It is not difficult to show that the order topology inducedby the partial order is weaker than the original (metric) topology. Hence, it suffices
to show that the order topology induced by the partial order is stronger than theoriginal (metric) topology. To show it, it suffices to show that for any m ∈ X andany ε > 0, there exist x and y such that m ∈ (x, y) ⊆ B(m, ε), where x ∈ X or x
is the symbol −∞ and y ∈ X or y is the symbol +∞.
Case 1. Let m be a cut point of X. Then by Lemma 2.3, it follows that there
exist x, y ∈ B(m, ε/4) with x < m < y. Then it follows from Lemma 2.2 that
I(x, y) ⊆ B(m, ε).
Case 2. If m = z, the result is obvious with x = −∞ and y being the uniquepoint which is a distance ε from z.
Case 3. Let m be a non-cut point of X with m 6= z. We may assume thatε < ρ(z, m). Let x be the unique point with ρ(m, x) = ε. By Lemma 2.1, it follows
that X ⊆ B(z, ρ(z, m)) ∪ {m}, hence we have ρ(z, x) < ρ(z, m). Hence x < m. By
Lemma 2.2, I(x, m) is a connected metric space with UMP and has two distinct
non-cut points x and m. Therefore, I(x, m) ⊆ [B(x, ε) ∩ B(m, ε)] ∪ {x, m} byLemma 2.2, and I(x, m) ⊆ B(m, ε) ∪ {x}, i.e., (x, +∞) ⊆ B(m, ε). This complets
the proof.
¤
Remark 2.1. In the third case of the original proof of [1], A. D. Berard took theunique point x ∈ X with ρ(z, x) = ρ(z, m) − ε/4 and y = +∞. But, the next
example shows that they do not satisfy (x, +∞) ⊆ B(m, ε). Consider
X = {(x, y) : (x − 1)2 + y 2 = 1, 0 ≤ y ≤ 1} ⊆ R2 .
Then, (X, d) is a connected metric space with UMP and has two distinct non-cut
points z = (0, 0) and m = (2, 0), where d is the metric on X induced by the
Euclidean metric on R2 . Now, for each ε with 0 < ε ≤ 1/4, let x be the uniquepoint with d(x, z) = d(z, m) − ε/4 = 2 − ε/4. Then,r³
ε ´2 1 pd(x, m) = 22 − 2 −=ε(16 − ε).
44This implies that
1pd(x, m) − ε =ε(16 − ε) − ε
4
√
1√ √
=
ε( 16 − ε − 4 ε)
4
√
√
1√ √
ε( 16 − ε − 4 ε)
>
4
√
1√
=
ε(4 − 5 ε) > 0.
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Hence, (x, +∞) * B(m, ε).
We describe the examples of metric spaces with UMP which have several properties. A metric space (X, ρ) is said to be convex if for each pair of distinct points
x and y there is exactly one point p ∈ X such that ρ(x, y) = ρ(x, p) + ρ(y, p). The
convex metric seems to have any relation with UMP. The following example is acomplete connected metric space with UMP for which the metric is not convex.
Example 2.1 ([1]). There is a complete connected metric space (X, ρ) with UMP
for which the metric ρ is not convex. Indeed, for each x, y ∈ [−1, 1] = X letρ(x, y) = |x − y|/(1 + |x − y|). Then (X, ρ) is the desiered metric space.
The following examples show the connectedness of a metric space is not a necessary condition for which has UMP.
Example 2.2 ([9]). Let
X = {(x, x2 ) ∈ R2 : x ≥ 0} ∪ {(−1, 1)}.
Then, (X, d) is a locally compact separable metric space with UMP which is homeomorphic to a subspace of the real line, where d is the metric induced by the Euclidean metric on R2 . Furthermore, it is neither connected nor totally-disconnected.
Example 2.3 ([1]). There is a totally disconnected complete metric space (X, ρ)such that (X, ρ) has UMP.
Indeed, consider()∞
X
Y = x : x ∈ [0, 1], x =ai 4−i (ai = 0 or 3) .
i=1

Notice that Y is homeomorphic to the Cantor set. Let h and k be linear homeomorphisms defined by
h : Y → [0, 1/4] × {0},k : Y → [1, 5/4] × {0}.
Let q = (5/8, 1) ∈ R2 and put X = {q} ∪ h(Y ) ∪ k(Y ). Let ρ : X × X → R+ be
defined byρ(q, q) = 0,
ρ(x, q) = σ(q, x) = 1
for x ∈ h(Y ) ∪ k(Y ),ρ(x, y) = σ(y, x) = |x − y| for x, y ∈ h(Y ) ∪ k(Y ).
Then it is easy to see that (X, ρ) is a totally disconnected complete metric spacewith UMP.
3. Non-connected metric spaces with UMP
In this section, we shall consider certain non-connected metric spaces with UMP.
As is shown at the end of the previous section (cf. Examples 2.2 and 2.3), we do
not need the connectedness for being homeomorphic to subspaces of the real line
R. Follows from Theorem 2.1, one may expect that non-connected metric spaces

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with UMP are homeomorphic to subspaces of the real line R. In this direction, S.
D. Nadler, Jr. proved the following theorem.
Theorem 3.1 ([9]). If (X, d) is a locally compact separable metric space with UMP,
then X is homeomorphic to a subspace of the real line R.
A result from [7] is used in the proof of the theorem.
The next example shows that the separability is essential in Theorem 3.1.
Example 3.1 ([9]). Let (X, σ) be any uncountable metric space with UMP (e.g.,
the real line R with the usual metric). We define the new metric ρ as follows:(ρ(x, y) = 1 + σ(x, y) if x 6= y,ρ(x, y) = 0
if x = y.
Then, (X, ρ) is a discrete metric space and ρ has UMP. It is obvious that (X, ρ) is
not separable and hence, (X, ρ) can not be topologically embedded in R.
There are examples of totally disconnected separable metric spaces with UMP,
e.g., the space of rationals. We also notice that the space of irrationals and theCantor set have compatible metirc with UMP (see the next section). Although
there are totally disconnected separable metric spaces of arbitrary dimension (cf.
[8, Theorem 3.9.3]), we don’t know any totally disconnected separable metric spacewith UMP which have the positive dimension. Hence we can ask the following.
Question 3.1 ([9]). Is every totally disconnected separable metric space with UMP
0-dimensional?
We don’t know the local compactness can be dropped in Theorem 3.1, i.e., thefollowing question seems to remain open.
Question 3.2. Is separable metric space with UMP homeomorphic to a subspace ofthe real line?
Concerning Question 3.2, we have the following.
Theorem 3.2 ([4]). The following hold.
(1) A separable metric space (X, ρ) is homeomorphic to a subspace of the realline if (X, ρ) satisfies the following two conditions.
(a) The cardinality of any midset is at most 1, i.e., |{z ∈ X : ρ(x, z) =ρ(y, z)}| ≤ 1 for each distinct points x, y ∈ X.
(b) The cardinality of any subset consisting of points which are equidistantfrom a point is at most two, i.e., for each x ∈ X and each positive
number ε > 0 |{y ∈ X : ρ(x, y) = ε}| ≤ 2.
(2) If X is locally compact and satisfies (a), then X is homeomorphic to a
subspace of the real line R.
We don’t know whether if the second part of the theorem holds for every rimcompact space (= every point has a neighborhood base consisting of the open setswith compact boundaries) (see [4, Remark 1]).

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4. Topological aspects of the unique midpoint property
In this section, we shall consider the unique midpoint property as a topologicalproperty. As is mentioned in Introduction, the real line R with the usual metricand its metric subspace of rational numbers Q have UMP, while the usual Cantor
set and the space of irrational numbers P do not have UMP as the metric subspaceof R. Then the following question arises: Does there exist a compatible metricwith UMP on P or the Cantor set? Now, we say that a metrizable space X has
the unique midpoint property (abbreviated UMP) if there is a compatible metric d
on X such that (X, d) has UMP. Connecting the question above, M. Ito, H. Ohta
and J. Ono [6] proved the following.
Theorem 4.1 ([6]). The discrete space D with cardinality n has UMP if and only
if n 6= 2, 4 and n ≤ c, where c denotes the cardinality of the continuum.
Theorem 4.2 ([6]). Let D be a discrete space with cardinality not greater than c.
Then the countable power DN of D has UMP. In particular, the Cantor set 2N andthe space NN of irrational numbers have UMP.
We notice that not all infinite subspaces of (Dc )N have UMP. Let xn → x be aconvergent sequence in (Dc )N such that xm 6= xn for m 6= n. Y. Hattori and H.
Ohta showed that the subspace S = {x} ∪ {xn : n ∈ N} does not have UMP ([4,Remark 2]). On the other hand, since the countable power S N is homeomorphic to
the Cantor set, it follows from Theorem 4.2 that S N has UMP.
We alo notice that the metric of the space of irrationals (or the Cantor set) withUMP described in the proof of Theorem 4.2 is not intutive and it may be difficult
to how to metrize the spaces. Hence the following question seems to be interested.
Question 4.1 ([6]). Is there a subspace X of the Euclidean soace Rn or the Hilbert
space R∞ such that X is homeomorphic to 2N or NN and the metric on X inducedby the usual metric of Rn or R∞ has UMP?
Remark 4.1 ([6]). We have an intuitive way of showing that the topological sum
Q ⊕ P of the space Q of rational numbers and the space P of irrational numbers
has UMP. Indeed, let h : R → (−1, 1) be the homeomorphism defined by h(x) =x/(1 + |x|) for x ∈ R. Define√√
X = {(x, 1 − x2 ) : x ∈ h(Q)} ∪ {(x, − 1 − x2 ) : x ∈ h(P)} ⊆ R2 .
Then X is homeomorphic to Q ⊕ P and the metric on X induced by the Euclidean
metric on R2 has UMP.
Theorems 4.1 and 4.2 are proved by an approach from the graph theory. We
shall make a short description of this approach, because we may be intersted initself.
The edge connecting vertices x and y is denoted by xy. By a colouring of agraph G we mean a map defined on the set of edges E(G) of G. For a colouring
ϕ : E(G) → A, we call ϕ(e) the colour of e for e ∈ E(G). A colouring ϕ of G is
said to have the the unique midpoint property (abbreviated UMP) if for every pairof distinct vertices x, y of G there is an exactly one vertex p of G such that xp

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and yp are edges of G and ϕ(xp) = ϕ(yp). Theorem 4.1 is a direct consequenceof the following theorem which connects with the unique midpoint property of thediscrete space and of the complete graph.
Theorem 4.3 ([6]). Let n be a cardinal and Kn the complete graph (i.e., eachvertex of Kn is adjacent to every other vertices) with n vertices. Then we have thefollowing.
(1) The discrete space D with cardinality n has UMP if and only if n ≤ c andthe complete graph Kn has a colouring with UMP.
(2) The complete graph Kn has a colouring with UMP if and only if n 6= 2, 4.
We turn our attention to the graphs which have a colouring with UMP. Let G bea finite graph having a colouring with UMP. Then we denote the smallest numberof colours we need to colour G with UMP by ump(G):ump(G) = min{|ϕ(E(G))| : ϕ is a colouring of G with UMP}.
Then M. Ito, H. Ohta and J. Ono [6] proved the following.
Proposition 4.1 ([6]). For each k ≥ 0, ump(K2k+1 ) = k.
Proposition 4.2 ([6]). For each k ≥ 3, k ≤ ump(K2k ) ≤ 2k − 1.
We notice that the following equalities are already known ([6]):ump(K6 ) = 4, ump(K8 ) = 5 and ump(K10 ) = 5.
Moreover, we also have the following inequalities ([6]):ump(K12 ) ≤ 8 and ump(K14 ) ≤ 10.
The following question still remains open.
Question 4.2 ([6]). Determine the values of ump(K2k ) for each k ≥ 6.
Now, we shall consider subspaces of the real line R which have UMP. For n ∈ N,
let In be the union of n-many disjoint closed intervals in R. H. Ohta and J. Ono[10] proved the following theorems.
Theorem 4.4 ([10]). Let X be a compact disconnected metrizable space. If either
exactly one of the components is or exactly two of the components are nondegenerate, then X does not have UMP. In particular, I2 does not have UMP. On theother hand, I2n−1 has UMP for each n ∈ N.
Theorem 4.5 ([10]). Let X be a disconnected metrizable space and Y a subspaceof X such that Y is homeomorphic to either I1 or I2 . If |X − Y | < c, then X does
not have UMP.
Theorem 4.6 ([10]). If X is a subspace of the real line R satisfying the following
conditions, then X has UMP:(i) Each nondegenerate component of X is an open set of X.
(ii) The union of one-point components of X is an open set of X.
(iii) At least one component is not compact.

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Y. KITAIThe following is a direct consequence of Theorems 4.4, 4.5 and 4.6.
Corollary 4.1 ([10]).
(1) Let I and J be disjoint intervals in R. Then I ∪ J has UMP if and only ifeither I or J is not compact.
(2) The union of an odd number of disjoint closed intervals in R has UMP.
(3) The subspaces [0, 1] ∪ Z and [0, 1] ∪ Q of R do not have UMP, where Z
denotes the set of integers.
(4) Let X be a space which is the union of at most countably many subspaces{Xn : n ∈ A} of R. Assume that at least one of the spaces Xn is a
non-compact interval and others are either intervals or totally disconnected.
Then, X has UMP.
(5) Let X be a space with exactly two components. Then, X has UMP if andonly if X is not compact and each component is either homeomorphic to aninterval or a singleton.
J. Ono showed that I4 does not have UMP, but I6 , I10 and Iω have UMP. This
arises the following question.
Question 4.3 ([10]). For n = 8 or an even number n ≥ 12, does In have UMP?
Connecting Corollary 4.1, we can also ask:
Question 4.4 ([10]). Does every subspace of the real line R containing a noncompact interval as a clopen set have UMP?
Acknowledgments. The author would like to express her thanks to Professor Y.
Hattori for his advice.
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Department of Mathematics, Shimane University, Matsue, 690-8504, Japan