@
Applied General Topology

c© Universidad Politécnica de Valencia

Volume 4, No. 1, 2003

pp. 71–77

Dense Sδ-diagonals and linearly ordered
extensions

Masami Hosobuchi

Abstract. The notion of the Sδ-diagonal was introduced by H. R.
Bennett to study the quasi-developability of linearly ordered spaces. In
an earlier paper, we obtained a characterization of topological spaces
with an Sδ-diagonal and we showed that the Sδ-diagonal property is
stronger than the quasi-Gδ-diagonal property. In this paper, we define
a dense Sδ-diagonal of a space and show that two linearly ordered
extensions of a generalized ordered space X have dense Sδ-diagonals if
the sets of right and left looking points are countable.

Keywords: Sδ-diagonal, dense Sδ-diagonal, linearly ordered space (LOTS),
generalized ordered space (GO-space), linearly ordered extension.

2000 AMS Classification: 54F05.

1. Sδ-diagonals

We review in this section the definitions of Sδ-set and Sδ-diagonal, and state
our results obtained in [5].

The following definition is a generalization of a Gδ-set and was introduced
by H. R. Bennett [2] to study the quasi-developability of linearly ordered (topo-
logical) spaces.

Definition 1.1. Let X be a topological space. A subset A of X is called an
Sδ-set if there exists a countable collection {U(1),U(2), . . .} of open subsets of
X such that, for two points p ∈ A and q ∈ X \A, there exists an n such that
p ∈ U(n) and q /∈ U(n).

It is easy to see that a Gδ-set is an Sδ-set. Hence the notion of Sδ-set is
a generalization of Gδ-set. See [3] for a description of S-normal spaces whose
closed subsets are Sδ-sets.

Definition 1.2. Let X be a topological space. X has an Sδ-diagonal if the
diagonal subset ∆X of X × X is an Sδ-set of X × X, where ∆X denotes the



72 M. Hosobuchi

diagonal set {(x,x) : x ∈ X} in the Cartesian product X×X. The symbol ( , )
is used to stand for a point of X ×X.

It is useful to show the following lemma that relates to the property (∗)
given in [2]. N denotes the set of natural numbers.

Lemma 1.3. [5] Let X be a topological space. Let {G(n) : n ∈ N} be a
family of countable collections of open subsets of X. Suppose that, for any three
points x,y and z with y 6= z, there exists an m ∈ N such that x ∈

⋃
G(m)

and that no element of G(m) contains the set {y,z}, where
⋃
G(m) denotes⋃

{U : U ∈ G(m)}. Then, there exists a family {F(n) : n ∈ N} of countable
collections of open subsets of X such that, for such three points above, there
exists an m ∈ N such that x ∈

⋃
F(m) and any two distinct points of {x,y,z}

do not belong to the same member of F(m).

Theorem 1.4. [5] Let X be a topological space. X has an Sδ-diagonal if and
only if there exists a family {G(n) : n ∈ N} of countable collections of open
subsets of X such that, for three points x,y and z with y 6= z, there exists an
m ∈ N such that x ∈

⋃
G(m) and any two distinct points of {x,y,z} do not

belong to the same member of G(m).

2. Two linearly ordered extensions and notation

Recall that a generalized ordered space (GO-space) is a triple (X,τ,<),
where < is a linear ordering of the set X and τ a Hausdorff topology on X
having a base of order-convex sets. We will denote by λ the order topology
on (X,<). It is known that λ ⊂ τ. A space of the form (X,λ,<) is called
a linearly ordered topological space (LOTS). Every LOTS is a GO-space, but
not conversely. In fact it is known that the class of GO-spaces coincides with
the class of subspaces of LOTS. Given a GO-space X there are two well-known
linearly ordered extensions of X. One of these is X∗ and was defined by D. J.
Lutzer [7]. The other one is L(X) and was studied in [8]. We review here the
definitions of those linearly ordered extensions. The intervals in a GO-space
or a LOTS are written in the form [a,b], [a,b[, ]a,b] and ]a,b[. For example,
[a,b] = {x : a ≤ x ≤ b}, [a,b[ = {x : a ≤ x < b} and so on. For a GO-space
X, we set R = {x ∈ X : [x,→ [ ∈ τ −λ} and L = {x ∈ X : ]← ,x] ∈ τ −λ},
where λ denotes the order topology as mentioned above. R (resp. L) is called
the set of right (resp. left ) looking points. Then X∗ is defined as follows:

X∗ = (X ×{0}) ∪{(x,k) : x ∈ R,k < 0,k ∈ Z}∪{(x,k) : x ∈ L,k > 0,k ∈ Z}
⊂ X ×Z,

where Z denotes the set of integers. On the other hand, L(X) is defined as
follows:

L(X) = (X ×{0}) ∪{(x,−1) : x ∈ R}∪{(x, 1) : x ∈ L}⊂ X ×{−1, 0, 1}.

X∗ and L(X) are linearly ordered topological spaces equipped with the lex-
icographic order topologies. We, furthermore, need some technical notation



Dense Sδ-diagonals 73

for the proof of the theorems in Section 5. For a convex open subset U of a
GO-space X, we define a convex open subset Ũ of E(X), where E(X) denotes
either X∗ or L(X). Then eight cases can occur. In the following, the intervals
must be considered in E(X).

(1) If a is the minimum point of U, then we define Ũ1 = [(a, 0),→ [ ⊂ E(X).
(2) Let a = inf U and a ∈ X \ U. If E(X) = X∗, then Ũ1 = {(x,k) ∈

X∗ : a < x} = ](a, +∞),→ [ ⊂ X∗, where (a, +∞) ∈ X × (Z∪{+∞})
and the interval is taken in X∗. Likewise, if E(X) = L(X), then
Ũ1 = {(x,k) ∈ L(X) : a < x} = ](a, 1),→ [ ⊂ L(X). Note that (a, 1)
may not belong to L(X).

(3) If there is a gap u = (A,B) such that u is the left end-point of U, then
we define Ũ1 =](u, 0),→ [ ⊂ E(X).

(4) If none of Cases 1 − 3 occurs, then we define Ũ1 = E(X).
(5) If b is the maximum point of U, then we define Ũ2 = ] ← , (b, 0)] ⊂

E(X).
(6) Let b = sup U and b ∈ X \ U. If E(X) = X∗, then Ũ2 = {(x,k) ∈

X∗ : x < b} = ] ← , (b,−∞)[ ⊂ X∗ (cf. (2)). If E(X) = L(X), then
Ũ2 = {(x,k) ∈ L(X) : x < b} = ] ← , (b,−1)[ ⊂ L(X). Note that
(b,−1) may not belong to L(X).

(7) If there is a gap v = (A,B) such that v is the right end-point of U,
then we define Ũ2 = ]← , (v, 0)[ ⊂ E(X).

(8) If none of Cases 5 − 7 occurs, then we define Ũ2 = E(X).
We set Ũ = Ũ1 ∩ Ũ2. Ũ is called the convex open set associated with U. Let U
be an open set of a GO-space X. Then U is decomposed into a union of open
convex subsets {Uα : α ∈ A}. In this case, we define Ũ =

⋃
{Ũα : α ∈ A},

where Ũα is the open set associated with Uα. Then Ũ is an open subset of
E(X), and called the open set associated with U.

3. Sδ-diagonals in linearly ordered extensions

The following theorems are proved in our paper [5]. Let X be a GO-space.
It is easily seen that X∗ contains X as a closed subset and L(X) contains X
as a dense subset. See [7, 8] for further information about X∗ and L(X). In
both cases, X and X ×{0} are identified by the correspondence of x to (x, 0).

Theorem 3.1. [5] Let X be a generalized ordered space with an Sδ-diagonal.
If R∪L is countable, then X∗ has an Sδ-diagonal.

To prove a similar theorem concerning L(X), it is necessary to assume the
existence of sequences in X that witnesses first-countability for points of R∪L:

Theorem 3.2. [5] Let X be a GO-space with an Sδ-diagonal. Assume that,
for every point s ∈ L, there exists a decreasing sequence {x(s,n) : n ∈ N}
in X such that inf{x(s,n)} = s and, for every point s ∈ R, there exists an
increasing sequence {y(s,n) : n ∈ N} in X such that sup{y(s,n)} = s. If R∪L
is countable, then L(X) has an Sδ-diagonal.



74 M. Hosobuchi

4. Dense Sδ-diagonals

The following definition gives an analogy to the dense Gδ-diagonal in [1].

Definition 4.1. A Hausdorff space X has a dense Sδ-diagonal if there exists
a dense subset D of ∆X such that D is an Sδ-subset of X × X, where ∆X
denotes the diagonal subset of the Cartesian product X ×X.

We show the following theorem that is analogous to a result concerning
spaces that have a dense Gδ-diagonal [1].

Theorem 4.2. Let X be a Hausdorff space. Then X has a dense Sδ-diagonal
if and only if X has a dense subset Y such that Y is an Sδ-subset of X and Y
has an Sδ-diagonal.

Proof. If D ⊂ ∆X is a dense Sδ-set in X×X, then D∩∆X is a dense Sδ-set in
∆X. Now the map h : ∆X → X defined by h(x,x) = x is a homeomorphism,
and the homeomorphic image of a dense Sδ-set is a dense Sδ-set.

Conversely, suppose Y is a dense Sδ-subset of X. Then h−1(Y ) is a dense
Sδ-subset of ∆X. The rest is easily verified. �

5. Theorems concerning dense Sδ-diagonals of linearly ordered
extensions

Theorem 5.1. Let X = (X,τ) be a GO-space with a dense Sδ-diagonal. If
R∪L is countable, then X∗ has a dense Sδ-diagonal.

We first show the following lemma.

Lemma 5.2. Let X be a GO-space and X∗ the linearly ordered extension of
X. For a subspace Y of X, set Z = Y ∪ (X∗ \X). If Y is dense in X and an
Sδ-subset of X, then Z is a dense subspace of X∗ and an Sδ-subset of X∗.

Proof. To see that Z is dense in X∗, let x ∈ X∗ \ Z = X \ Y and V be a
neighborhood of x in X∗, where X is identified with X ×{0} as usual. Since
V ∩ X is a neighborhood of x in X, it follows that V ∩ X ∩ Y 6= ∅. Since
V ∩X ∩Y ⊂ V ∩Z, it follows that V ∩Z 6= ∅. Hence Z is a dense subspace
of X∗. To show the last part, let {U(n) : n ∈ N} be a countable collection
of open subsets of X such that, for y ∈ Y and x ∈ X \ Y, there exists an
m ∈ N such that y ∈ U(m) and x 6∈ U(m). For every n ∈ N, let Ũ(n) be the
open subset associated with U(n) as in Section 3. Set Ũ(0) = X∗ \ X. Then
it is obvious that Ũ(0) is an open subset of X∗. We show that the countable
collection {Ũ(n) : n ≥ 0} of open subsets of X∗ assures that Z is an Sδ-subset
of X∗. Let z ∈ Z and x ∈ X∗ \Z = X \Y.

Case 1. Let z ∈ X∗\X. Then it is easy to see that z ∈ Ũ(0) and x 6∈ Ũ(0).
Case 2. Let z ∈ Y. Since x ∈ X \ Y, there exists an m ∈ N such that

z ∈ U(m) and x 6∈ U(m). By the definition of Ũ(m), it follows that z ∈ Ũ(m)
and x 6∈ Ũ(m). This completes the proof. �



Dense Sδ-diagonals 75

Now we shall prove Theorem 5.1.

Proof of Theorem 5.1. By Theorem 4.2, there exists a dense subspace Y
of X such that Y is an Sδ-subset of X and that Y has an Sδ-diagonal. Let
{G(n) : n ∈ N} be a family of countable collections of open subsets of Y such
that, for three points x,y and z of Y with y 6= z, there exists an m ∈ N such
that x ∈

⋃
G(m) and no element of G(m) contains {y,z}. The existence of the

above family is guaranteed by Theorem 1.4. For an open subset V of Y, there
exists an open set VX of X such that VX ∩Y = V. Let ṼX be the open subset
of X∗ associated with VX as explained in Section 2. Set VZ = ṼX ∩Z, where
Z is as mentioned in Lemma 5.2.

It is clear that VZ is open in Z and that VZ ∩Y = V . For every n ∈ N, set
G̃(n) = {VZ : V ∈ G(n)} and G̃(0) = {{x} : x ∈ X∗ \ X}. Let S = R ∪ L =
{si : i ∈ N} be an enumeration of the countable set S. Let (si,k) ∈ X∗\X. Set
G̃+(si,k) = {](si,k),→ [ ∩Z} and G̃−(si,k) = {]← , (si,k)[∩Z}, where these
intervals are considered in X∗. By virtue of Lemma 5.2 and Theorem 4.2, it is
sufficient to show that a family of those countable collections of open subsets
of Z witnesses the Sδ-diagonal of Z. To see this, let x,y and z be three points
of Z with y 6= z. We may assume without loss of generality that y < z.
Case 1. If {x,y,z} ⊂ Y, then there exists an m ∈ N such that x ∈

⋃
G(m)

and {y,z} 6⊂ V for any V ∈G(m). Hence it follows that x ∈
⋃
G̃(m) and that

{y,z} 6⊂ VZ for any VZ ∈ G̃(m).
Case 2. Let x ∈ Y and y or z belong to Z \Y.

(i) We assume that y ∈ Z \ Y . Then we can write y = (si,k), where
k 6= 0. If x < y, then x ∈ ]← ,y[∩Z and {y,z} 6⊂ ]← ,y[. Hence, by
the definition, it follows that x ∈

⋃
G̃−(si,k) and that {y,z} 6⊂ V for

V ∈ G̃−(si,k). If y < x, then x ∈ ]y,→ [∩Z and {y,z} 6⊂ ]y,→ [. Hence
it follows that x ∈

⋃
G̃+(si,k) and that {y,z} 6⊂ V for V ∈ G̃+(si,k).

(ii) Let z ∈ Z \Y. Then the proof is analogous to (i).
Case 3. Let x ∈ Z \ Y. Then it follows that x ∈

⋃
G̃(0) and {y,z} 6⊂ V for

any V ∈ G̃(0). Therefore, by virtue of Lemma 1.3 and Theorem 1.4, X∗ has a
dense Sδ-diagonal. This completes the proof. �

Theorem 5.3. Let X = (X,τ) be a GO-space with a dense Sδ-diagonal. If
R∪L is countable, then L(X) has a dense Sδ-diagonal.

Proof. By Theorem 4.2, there exists a dense subspace Y of X such that Y is
an Sδ-subset of X and Y has an Sδ-diagonal. Since X is dense in L(X), it
follows that Y is a dense subspace of L(X). To prove that L(X) has a dense
Sδ-diagonal, it is sufficient to show, by Theorem 4.2, that Y is an Sδ-subset of
L(X). Let {U(n) : n ∈ N} be a countable collection of open subsets of X such
that, for y ∈ Y and x ∈ X \Y, there exists an m ∈ N such that y ∈ U(m) and
x 6∈ U(m). For every n ∈ N, let Ũ(n) be the open subset of L(X) associated
with U(n). For si ∈ S = R ∪ L and ε ∈ {−1, 1}, set Ũ+(si,ε) = ](si,ε),→ [
and Ũ−(si,ε) = ]← , (si,ε)[, where the intervals are considered in L(X). The



76 M. Hosobuchi

countable collection {Ũ(n), Ũ+(si,ε), Ũ−(si,ε) : n ∈ N, i ∈ N, ε = ±1} of
open subsets of L(X) guarantees that Y is an Sδ-subset of L(X). To see this,
let y ∈ Y and z ∈ L(X) \Y.

Case 1. Let z ∈ X \Y. Then there exists an m ∈ N such that y ∈ U(m) and
z 6∈ U(m). Hence it follows that y ∈ Ũ(m) and z 6∈ Ũ(m).

Case 2. Let z ∈ L(X) \ X. We can write z = (si,ε), where ε ∈ {−1, 1}. If
y < z, then it follows that y ∈ Ũ−(si,ε) and z 6∈ Ũ−(si,ε). If z < y, then it
follows that y ∈ Ũ+(si,ε) and z 6∈ Ũ+(si,ε). Hence Y is an Sδ-subset of L(X).
This completes the proof of Theorem 5.3. �

6. Examples

Example 6.1. Theorems 3.1 and 3.2 do not hold without the assumption of
the countability of the set R∪L. Let us consider the Sorgenfrey line X = (R,S).
In this case, the right looking points R = R is uncountable. Since X has a Gδ-
diagonal, X has an Sδ-diagonal. However, X∗ does not have an Sδ-diagonal.
To prove this, it is sufficient to see that X∗ does not have a quasi-Gδ-diagonal
[5]. We easily see that there does not exist a family of countable collections
of open subsets of X∗ that separates two points of the form (x, 0) and (x, 1),
where x ∈ X.

Example 6.2. Theorem 3.2 does not hold without the existence of the se-
quences for points of R ∪ L. To show a counterexample, let Y be the set of
countable ordinals [0,ω1[ with the discrete topology. The right looking points
of Y comprise the set of limit ordinals. Let Y ∗ be the linear extension of Y
defined in Section 2. Let X = Y ∗∪{(ω1, 0)}, where X is ordered as (ω1, 0) > α
for all α ∈ Y ∗, and given the discrete topology. Then R = {(ω1, 0)} is a single-
ton and L(X) = Y ∗ ∪{(ω1,−1)}∪{(ω1, 0)}, where α < (ω1,−1) < (ω1, 0) for
all α ∈ Y ∗. There does not exist an increasing sequence in Y ∗ that converges to
(ω1,−1). Furthermore, L(X) does not have a quasi-Gδ-diagonal, because the
points of X and the point {(ω1,−1)} are not separated by a family of countable
collections of open subsets of L(X). Hence L(X) does not have an Sδ-diagonal.

Example 6.3. A generalized ordered space does not necessarily have a dense
Sδ-diagonal. To show this, consider the linearly ordered space Z that was
constructed by H. R. Bennett and D. J. Lutzer [4]. They proved that Z is not
first-countable at any point. Z is defined as follows:

Z = {(α1,α2, . . . ,αn,ω1,ω1, . . .) : αi < ω1, 1 ≤ i ≤ n, αi = ω1, i > n, n ≥ 1},

with the lexicographic order. Since Z is densely-ordered, a dense subset Y of
Z is a LOTS. If Y has a quasi-Gδ-diagonal, Y is quasi-developable. Since a
quasi-developable space is first-countable, Y does not have a quasi-Gδ-diagonal.
Therefore, Z does not have a dense Sδ-diagonal.



Dense Sδ-diagonals 77

Acknowledgements. The author is grateful to the referees for their helpful
comments.

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(1993), 133–140.

Received January 2002 Revised December 2002

Masami Hosobuchi

Tokyo Kasei Gakuin University,
Department of Housing and Planning,
2600 Aihara, Machida,
Tokyo 194-0292,
Japan.

E-mail address : mhsbc@kasei-gakuin.ac.jp


	Dense S-diagonals and linearly ordered extensions. By M. Hosobuchi