@
Appl. Gen. Topol. 23, no. 1 (2022), 225-234

doi:10.4995/agt.2022.15586

© AGT, UPV, 2022

Alexandroff duplicate and βκ

Andrzej Szymanski

Department of Mathematics and Statistics, Slippery Rock University of Pennsylvania, U.S.A.

(andrzej.szymanski@sru.edu)

Communicated by S. Garćıa-Ferreira

Abstract

We discuss spaces and the Alexandroff duplicates of those spaces that
admit a Č-S embedding into the Čech-Stone compactification of a dis-
crete space.

2020 MSC: 54G05; 03E75; 54B99.

Keywords: extremally disconnected; ultrafilter space.

1. Introduction

The Alexandroff duplicate of a topological space X, denoted by A (X), is
the topological space defined as follows.

The underlying set consists of two disjoint copies of the set X, say X ×{0}
and X ×{1}; for the sake of some technical simplicity, elements and subsets of
the first copy are going to be denoted the same as it was for the original set,
whereas the elements and subsets of the second copy are going to be denoted
by priming the corresponding symbols, i.e., x′, Y ′, etc. If Y ⊆ X, D (Y ) stands
for the duplicate of Y , i.e., for the set D (Y ) = Y ∪Y ′.

The topology of the space A (X) is generated by the subsets of the form
D (U) − F ′ = U ∪ (U −F)′, where U is open in X and F ⊆ X is finite, and
by {x′} where x ∈ X. Thus each point of X′ is an isolated point of the space
A (X). The original space X is contained in A (X) as its closed subset and also
as its exactly two-to-one retract.

The concept itself originated with P.S. Alexandroff and P.S. Urysohn in
1929 [1] for X = the unit circle; the generalization for arbitrary X, as defined
above, is due to R. Engelking [12] - it has been extensively utilized and studied

Received 08 May 2021 – Accepted 10 December 2021

http://dx.doi.org/10.4995/agt.2022.15586
https://orcid.org/0000-0002-6992-8953


A. Szymanski

since (cf. [16], [7], [6], [14], [4], and [5]). Of the other possible generalizations,
the most general generalization of Engelking’s one was accomplished by R.P.
Chandler et al. [8], but it didn’t get as much traction as Engelking’s.

If X is a crowded (i.e., without isolated points) space, then X′ is a dense
subset of isolated points and X is the remainder of A (X). In this regard, the
structure of A (X) is akin to that of βκ, so the following question arises:

When is A (X) embeddable into the Čech-Stone compactification of the dis-
crete space X′?

Specifically, we say that A (X) is Č-S embeddable if there exists an embed-
ding h : A (X) → β (X′) such that h (x′) = x′ for each x ∈ X.

Suppose that A (X) is Č-S embeddable. There are some obvious necessary
conditions A (X) has to satisfy. To list a few.

(0) X has to be a completely regular space;
(1) A (X) has to be extremally disconnected (since X′ ⊆ A (X) ⊆ β (X′);
(2) Any bounded real function on X′ has to have a continuous extension to

A (X) (since it has a continuous extension to β (X′));
(3) Any two disjoint subsets of X cannot have a common accumulation point.
Finally, considering A (X) as a subspace of β (X′), one can also observe the

following phenomenon: Let p ∈ X be a non-isolated point of X. Considering p
as a free ultrafilter on X′, if A ⊆ X′ is such that A ∈ p, then there is an open
neighborhood U of p in X and a finite subset F of X such that U −F ⊆ A. It
means that

(4) For each non-isolated point p ∈ X, the family

{U −F : U is an open neighborhood of p and F is a finite subset of X}

is a base for the ultrafilter p.
The main goal of this note is to show that any of the properties (1) − (4)

stated above constitutes also a sufficient (thus, equivalent) condition for any
(completely) regular crowded space and its Alexandroff duplicate to be Č-
S embedabble. Additional mutual relationships between those properties are
discussed in Section 1 and Section 2.

Spaces satisfying condition (4), above, are defined and studied in Section 1
under the name ultrafilter spaces. They play a crucial role in establishing the
aforementioned equivalences.

Spaces satisfying condition (3), above, are called perfectly disconnected.
They were first defined and studied by E. van Douwen in [9].

The equivalence between perfectly disconnected space and extremal discon-
nectedness of Alexandroff duplicate, i.e., the equivalence of the conditions (1)
and (3), above, was first established by P. Bhattacharjee, M. Knox, and W.
McGovern, [3].

We refer to R. Engelking’s book [13] for all undefined topological notions.

2. Topological Ultrafilter Spaces

All considered topological spaces are T1 and let X be a topological space.

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Alexandroff duplicate and βκ

For A ⊆ X, ∂ (A) denotes the set of all accumulation points of the set A,
i.e., ∂ (A) = {x ∈ X : |U ∩A| ≥ ω for each open neighborhood U of x}

An ultrafilter on a non-empty set X is a family ξ of non-empty subsets of
X closed under finite intersections and maximal with respect to that property,
i.e., if X = A∪B, then A ∈ ξ or B ∈ ξ.

We say that the space X is an ultrafilter space at p if the family
ξp = {A ⊆ X : U −F ⊆ A, where p ∈ U is open and F ⊆ X is finite}
is an ultrafilter on the set X.

Proposition 2.1. A space X is an ultrafilter space at p if and only if the
following condition holds true:
(∂) If Y and Z are disjoint subsets of X, then p /∈ ∂ (Y ) ∩∂ (Z).
Proof. (⇒) Let X be an ultrafilter space at p and let Y and Z be disjoint subsets
of X. We may assume that they cover X. There exists an open neighborhood
U of p and a finite subset F of X such that U −F ⊆ Y or U −F ⊆ Z; say the
former holds true. Hence U ∩Z ⊆ F, which means that p /∈ θ (Z).

(⇐) Suppose to the contrary that X is not an ultrafilter space at p. Thus
X = A∪B, where A∩B = ∅, and for each open neighborhood U of p and a
finite subset of X, U −F " A and U −F " B. Thus both sets (U −F) ∩A
and (U −F) ∩B are infinite, i.e., p ∈ ∂ (A) ∩∂ (B); a contradiction. �

We say that the space X is an ultrafilter space if X is an ultrafilter space at
each point p ∈ X. Thus.
Proposition 2.2. A space X is an ultrafilter space if and only if the following
condition holds true
(∂∂) If Y and Z are disjoint subsets of X, then ∂ (Y ) ∩∂ (Z) = ∅.

Condition (∂∂) implies:

Corollary 2.3. Let X be an ultrafilter space. Then
(a) X is hereditarily extremally disconnected and nodec (= each nowhere dense
subset of X is closed and, consequently, discrete).
(b) (see also van Douwen [9]) ∂ (E) = cl (intE) for arbitrary E ⊆ X.
Proof. Part (a) follows immediately from (∂∂). Part (b) needs some additional
argument.

Inclusion ∂ (E) ⊇ cl (intE) is obvious. To show the converse inclusion, let
x /∈ cl (intE) and let U be an open neighborhood of x disjoint from intE. It
means that U ∩E is boundary which, in turn, implies that U ∩E is nowhere
dense. Hence, by (a), x is not an accumulation point of E. �

Topological spaces satisfying condition (∂∂) were introduced by E. van Douwen
under the name perfectly disconnected space. He also gave the following charac-
terization of such spaces. For the sake of completeness, we provide (a different)
proof.

Theorem 2.4 (van Douwen [9]). X is a perfectly disconnected space if and
only if X is extremally disconnected and each dense subset of X is open, i.e.,
X is submaximal.

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A. Szymanski

Proof. If X is perfectly disconnected, then, X is (hereditarily) extremally dis-
connected. It is submaximal as well for if D is dense in X, then X −D cannot
have any accumulation point. Thus X −D is closed and discrete and so D is
open.

Let X be extremally disconnected and submaximal and assume, to the con-
trary, that X does not satisfy the condition (∂∂). So let Y , Z be disjoint
subsets of X such that there is a p ∈ ∂ (Y ) ∩ ∂ (Z). Hence p ∈ clY ∩ clZ.
Set U = int (cl (Y )) and V = int (cl (Z)). It follows from submaximality that
U ∩V = ∅ for no non-empty open subset of X can contain two disjoint dense
subsets. The space X, being submaximal, is also nodec. Hence Y − U and
Z −V are two disjoint closed discrete subsets of X. Hence
cl (Y ) ∩ cl (Z) = [clU ∩ (Z −V )] ∪ [clV ∩ cl (Y −U)]. So if p ∈ clY ∩ clZ,

then p ∈ clU ∩ (Z −V ) or p ∈ clV ∩ cl (Y −U); assume the former holds
(similar argument applies when the latter holds true). There exists an open
neighborhood W of p such that W ∩ (Z −V ) = {p} and W ⊆ clU. Hence
W ∩ clZ = {p}, thus p /∈ ∂ (Z); a contradiction. �

It is well known that the two aforementioned properties, i.e., that of submax-
imality and extremal disconnectedness, characterize maximal crowded topolog-
ical spaces (cf. E. Hewitt [15] and M. Katĕtov [17]). By the Kuratowski-Zorn
Lemma, maximal crowded topology majorizes any given T1 or T2 crowded
topology. However examples of maximal crowded topologies which are T3 are
unknown with the exception of van Douwen’s example from 1993 which hap-
pens to be countable (see [9] ). Uncountable examples can be obtained by
taking a disjoint union of the van Douwen’s example.

The next characterization of ultrafilter spaces entails a new setting.

Lemma 2.5. A set W is an open set in the space A (X) if and only if W =
[D (U) −F ′] ∪ E′, where U = W ∩ X, F ⊆ U is discrete in U, and E is a
subset of X −U.

Proof. Let W ⊆ A (X) be open. By setting U = X ∩ W we have: W =
(W ∩D (U)) ∪ [W −D (U)]. Set E = W − D (U). Since W ∩ D (U) =
U ∪ [W ∩U′], we need to show that U′ − W = F ′ for some discrete in U
subset F of U. Indeed, if x ∈ U = W ∩ X, then there exists an open neigh-
borhood V of x and a finite subset S of X such that D (V ) −S′ ⊆ W . Hence
V ∩F ⊆ S, which shows that F is discrete in U.

The converse is obvious. �

Theorem 2.6. A space X is an ultrafilter space viz. perfectly disconnected
space if and only if A (X) is extremally disconnected.

Proof. Necessity is obvious. To prove sufficiency, pick any open subset W of
A (X) and let us show that ClW is open in A (X). By Lemma 2.5, W =
[D (U) −F ′]∪E′, where U is an open subset of X, F ⊆ U is discrete in U, and

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Alexandroff duplicate and βκ

E is a subset of X −U. Thus ClW = Cl [D (U) −F ′] ∪ClE′. One can easily
verify that Cl [D (U) −F ′] = cl (U) ∪ U′ − F ′ and that ClE′ = E′ ∪ ∂ (E).
The perfect disconnectedness of X yields that clE ∩ clU = ∅ and that clU
is clopen. Subsequently, Cl [D (U) −F ′] = D (clU) − (clU −U)′ − F ′ is an
open subset of A (X). Using part (b) of Corollary 2.3, we get: ClE′ = E′ ∪
cl (intE) =

[
D (cl (intE)) − (cl (intE) − intE)′

]
∪ (E − cl (intE))′ is an open

subset of A (X) too. �

The characterization of the extremal disconnectedness of the Alexandroff
duplicate in terms of perfect disconnectedness (= Theorem 2.6) was first es-
tablished by P. Bhattacharjee, M. L. Knox, and W. McGovern, [3], in 2020.
The aforementioned van Douwen’s example of a regular countable perfectly
disconnected space provides an affirmative answer to a problem posed by K.
Almontashery and L. Kalantan, [2]

The following corollary recaps the main results of this section,

Corollary 2.7. For arbitrary T1 crowded space X, the following conditions are
pairwise equivalent.
(j) X is an ultrafilter space;
(jj) X is a perfectly disconnected space;
(jjj) X is a maximal crowded space;
(Ij) A (X) is an extremally disconnected space.

Remark 2.8. The ultrafilters ξp induced by the topology on an ultrafilters space
X may not be uniform, i.e., that any member of ξp has to be of cardinality
|X|. However there exists an open subset G of X such that if G is considered
as an ultrafilter space, then the ultrafilters ξp are going to be uniform. To see
this, let’s recall some (known) definitions.
If Z is an arbitrary topological space, Y ⊆ Z, and p ∈ Z is its non-isolated
point, then:
∆ (p,Z) = min{|U| : U is an open neighborhood of p}− the dispersion char-
acter of Z at p; and ∆ (Z) = min{∆ (p,Z) : p ∈ Z and p is non-isolated}−
the dispersion character of Z.
Thus any open subset G of X such that |G| = ∆ (X) will yield an ultrafilter
space with all the ultrafilters ξp to be uniform.

There may be a variety of maximal filters of open sets, The question arises
whether all types of maximal filters of open sets on an ultrafilter space X are
ultrafilters on the set X. It turns out, it depends on the separation axioms of
X.

A space X is said to be a strong ultrafilter space if any maximal filter of
open sets on the space X generates an ultrafilter on the set X.

Proposition 2.9. If X is normal ultrafilter space, then X is a strong ultrafilter
space.

Proof. Let X be an ultrafilter space with an underlying set being a cardinal
number κ. Let ξ be a maximal filter of open subsets of X, and let us show that
ξ is an ultrafilter on κ.

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A. Szymanski

Suppose to the contrary that κ = A∪B, where A,B are disjoint and U∩A 6=
∅ 6= U ∩B for each U ∈ ξ. By Proposition 2.2, int (clA) ∩ int (clB) = ∅. Let
us assume int (clA) ∈ ξ. Since B ∩ int (clA) is a nowhere dense subset of
X, it is closed and nowhere dense by Corollary 2.3. Hence int (clA) − B ∈ ξ
and (int (clA) −B) ∩ B = ∅, which contradicts the initial assumption. By
the similar argument, int (clB) /∈ ξ. By maximality of ξ, there exists V ∈ ξ
such that V ∩ int (clA) = ∅ = V ∩ int (clB). Hence U ∩ (A− int (clA)) 6=
∅ 6= U ∩ (B − int (clB)) for each U ∈ ξ. The sets E = A − int (clA) and
F = B − int (clB) are nowhere dense, thus closed, and disjoint. By normality
of X, there are disjoint open subsets W1, W2 of X such that E ⊆ W1 and
F ⊆ W2. But then W1, W2 ∈ ξ; a contradiction. �

3. Ultrafilter Spaces vs. Spaces of Ultrafilters

Let κ be an infinite cardinal number. βκ stands, as usual, for the set of all
ultrafilters (free or principal) on the set κ endowed with the topology generated

by the sets  = {ξ ∈ βκ : A ∈ ξ}, where ∅ 6= A ⊆ κ.
In what follows, all considered ultrafilters spaces are assumed to be crowded.
Let X be an ultrafilter space with the underlying set κ, where κ is a cardinal

number. For each α ∈ κ, let
ξα = {U −F : U is an open neighborhood of α and F is a finite subset of κ}.
Thus ξα ∈ βκ−κ for each α ∈ κ. We can define a function ϕ : X → βκ−κ

by setting: ϕ (α) = ξα for each α ∈ κ = X. Thus ϕ (X) can be thought off as a
pointless copy of X. Let E (X) = κ∪ϕ (κ) ⊆ βκ. In what follows, both ϕ (X)
and E (X) are considered to be subspaces of the space βκ.

Proposition 3.1.
(a) The function ϕ : X → βκ−κ is continuous;
(b) ϕ is one-to-one if and only if X is T2;
(c) ϕ is an embedding if and only if X is T3.

Proof. (a) Let V be an open set in βκ − κ and let α ∈ κ = X be such that
ξα ∈ V . There exists A ⊆ κ such that ξα ∈ {ξ ∈ βκ−κ : A ∈ ξ} ⊆ V . Since
A ∈ ξα, there exists an open set U of X such that α ∈ U ⊆ A. Thus α ∈ U ⊆
ϕ−1 (V ).

(b) Assume that ϕ is one-to-one and let α 6= β ∈ κ = X. Since ξα 6= ξβ,
there exist disjoint sets A and B such that ξα ∈ {ξ ∈ βκ−κ : A ∈ ξ} and
ξβ ∈ {ξ ∈ βκ−κ : B ∈ ξ}. Hence there exist open sets U, V in X such that
α ∈ U ⊆ A and β ∈ V ⊆ B. Thus α ∈ U and β ∈ V and U ∩ V = ∅. The
converse implication is obvious.

(c) Assume that ϕ is an embedding and let α ∈ U ⊆ X, where U is open.
Thus ξα ∈ ϕ (U) and since ϕ (U) is open, there exists A ⊆ κ such that ξα ∈
{ξ ∈ βκ−κ : A ∈ ξ} ⊆ ϕ (U). Since the set {ξ ∈ βκ−κ : A ∈ ξ} is clopen
and since A ∈ ξα, there exists an open set V of X such that α ∈ V ⊆ A.
Thus α ∈ V ⊆ clV ⊆ ϕ−1 (U) = U. Conversely, let U be open in X and let
ξα ∈ ϕ (U). There exists an open set V in X such that α ∈ V ⊆ clV ⊆ U.

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Alexandroff duplicate and βκ

Hence clV is clopen and such that ξα ∈{ξ ∈ βκ−κ : clV ∈ ξ}⊆ ϕ (U). Thus
ξα ∈ intϕ (U). �

Theorem 3.2. Let X be an ultrafilter space with an underlying set being a
cardinal number κ. If X is regular, then E (X) is homeomorphic to A (X).

Proof. Let h : A (X) → E (X) be defined as follows: h (α′) = α for each α′ ∈ X′
and h (α) = ξα for each α ∈ X. Let’s show that h is a homeomorphism.

Clearly, h is a bijection. To see that h is continuous, let A ⊆ κ and α ∈ X
satisfy ξα ∈ Â (i.e., h (α) = ξα ∈ Â ). There exists an open neighborhood
U of α and a finite set F ⊆ X such that U − F ⊆ A. It is obvious that
h (D (U) −F ′) ⊆ Â∩E (X).

To see that h−1 is continuous, let W ⊆ A (X) be open and let ξα ∈ h (W).
Since h and ϕ coincide on X, h|X is a homeomorphism (cf. Proposition 3.1
(c) ). There exists A ⊆ κ such that ξα ∈ Â∩h (X) ⊆ h (W). By Lemma 2.5,
W = [D (U) −H′]∪G′, where H is a discrete and closed (in U) subset of U and
G ⊆ X−U is arbitrary. There exists an open set V ⊆ X and a finite set F ⊆ X
such that α ∈ V , V ∩H = ∅, and U −F ⊆ A. Hence ξα ∈ Û ⊆ h (W). �

Corollary 3.3. X is a regular ultrafilter space if and only if A (X) is Č-S
embeddable.

Let us consider a function ψ : E (X) → ϕ (κ) ⊆ βκ−κ, given by: ψ (α) =
ϕ (α) = ξα for each α ∈ κ, and ψ (ξα) = ξα for each α ∈ κ. Clearly, ψ is
continuous (see also Proposition 3.1 (a)): ψ−1 (U∗) = Û ∩E (X) for each open
subset U of X. Thus ψ is an exactly two-to-one retraction onto ϕ (κ).

There exists a continuous extension ψ̂ of ψ to βκ into clϕ (κ) ⊆ βκ − κ.
Hence ψ̂ is a retraction from the space βκ onto its subspace clϕ (κ). Let us
note the following.

Lemma 3.4. Let X be a regular ultrafilter space and A ⊆ κ, Â∩ clϕ (κ) = ∅
if and only if A is a discrete and closed (in the topology on X). Consequently,

if ξ ∈ A∗, then ψ̂ (ξ) ∈ clϕ (A) −ϕ (κ).

Proof. Since ϕ (κ) consists only of those ultrafilters that contain an open subset

of X, Â∩clϕ (κ) = ∅ means that A (considered as a subspace of X) is boundary
and so it is an infinite closed and nowhere dense subset of X.
Since ξ ∈ Â = clA (when considered in βκ), ψ̂ (ξ) ∈ clψ̂ (A) = clϕ (A). Since
A is closed in X, ϕ (A) is closed in ϕ (κ) (by Proposition 3.1). Hence ψ̂ (ξ) ∈
clϕ (A) −ϕ (κ). �

Theorem 3.5. Let X be a regular ultrafilter space., The retraction ψ̂ is one-
to-one on the subspace βκ− clϕ (κ) if and only if X is a normal space.

Proof. Assume that X is normal and let ξ 6= ζ ∈ βκ−clϕ (κ). The three cases
when at least one of the two points ξ, ζ belongs to κ are obvious. So assume
that ξ, ζ ∈ βκ−κ. There exist disjoint sets A,B ⊆ κ such that A ∈ ξ, B ∈ ζ,
and  ∩ clϕ (κ) = ∅ = B̂ ∩ clϕ (κ). By Lemma 3.4, ψ̂ (ξ) ∈ clϕ (A) − ϕ (κ)

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A. Szymanski

and ψ̂ (ς) ∈ clϕ (B) −ϕ (κ). Since X is normal, clϕ (A) ∩ clϕ (B) = ∅. Hence
ψ̂ (ξ) 6= ψ̂ (ζ).

Assume that the retraction ψ is a bijection outside of ϕ (κ). To show that
then X is a normal space (given that X is perfectly disconnected space), it
suffices to show that any two disjoint discrete subsets of X can be separated.

For suppose that A,B ⊆ X are disjoint and discrete but they cannot be
separated. It follows that clϕ (A) ∩ clϕ (B) 6= ∅; let p ∈ clϕ (A) ∩ clϕ (B).
By Lemma 3.4, ψ restricted to  is a homeomorphism between  and clϕ (A).

Similarly, ψ restricted to B̂ is a homeomorphism between B̂ and clϕ (B). There

exist ξ ∈  and ζ ∈ B̂ such that ψ̂ (ξ) = p = ψ̂ (ζ). Since Â∩ B̂ = ∅
and ξ,ζ /∈ clϕ (κ), the retraction ψ is not a bijection outside of clϕ (κ); a
contradiction. �

Remark 3.6. The retraction ψ̂ in Theorem 3.5 is an instance of a ≤ two-to-one
continuous maps on βκ onto a compact space of density κ. An extensive and
deep study of such maps was done by E. van Douwen in [9]. A. Dow, among
others, has published several papers on related subjects (cf. [10]).

The retraction ψ̂ is also an instance of a one-to-one retraction. A. Dow did a
thorough and profound study of one-to-one retractions on βω in [11].

Let f : X → Y be a function. The duplicate of f, D (f), is a function
D (f) : A (X) → Y defined as follows: D (f) (x) = f (x) and D (f) (x′) = f (x).
It is easy to see that the duplicate of a continuous function on X yields a
continuous function on A (X), regardless of the separation axioms or other
properties of its domain. In particular, X is C/C∗− embedded into A (X).
Can X′ be C∗− embedded into A (X)? In this regard, we have the following.

Corollary 3.7. If X′ is C∗− embedded into A (X), then X is an ultrafilter
space. Conversely, if X is an ultrafilter space, then X′ is C∗− embedded into
A (X) provided that X is a regular space.

Proof. Let A and B disjoint subsets of X. Without loss of generality, we may
also assume that A ∪ B = X. Define f : X′ → R setting f (x′) = 0 if x ∈ A
and f (x′) = 1 if x ∈ B. If f̃ is a continuous extension of f to A (X), then
f̃ (x) = 0 if x ∈ ∂ (A) and f̃ (x) = 1 if x ∈ ∂ (B). Hence ∂ (A) ∩ ∂ (B) = ∅,
which means, X is perfectly disconnected, viz. X is an ultrafilter space.

The converse part follows immediately from Theorem 3.2. �

Proposition 3.8. Let X be a regular ultrafilter space with the underlying set κ.
Then βX is homeomorphic to cl (ϕ (κ)). In particular, cl (ϕ (κ)) is extremally
disconnected.

Proof. Let f : ϕ (κ) → R be a continuous bounded function. Take the duplicate
D (f) of f, i.e., D (f) (ϕ (α)) = f (ϕ (α)) and D (f) (α) = f (ϕ (α)) for each
α ∈ κ. Since D (f) is continuos on E (X), it has a continuous extension onto
βκ since E (X) = κ∪ϕ (κ) ⊆ βκ. Since D (f) |ϕ (κ) = f, f has a continuous
extension to cl (ϕ (x)). Thus cl (ϕ (κ)) = β (ϕ (κ)) = βX since ϕ (κ) and X are

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Alexandroff duplicate and βκ

homeomorphic (see Proposition 3.1(c)). Since the Čech-Stone compactification
of an extremally disconnected space is extremally disconnected, the proof is
finished. �

Corollary 3.9. Let X be a regular ultrafilter space.
(a) If H ⊆ βX −X is a discrete set of remote points such that X ∩clH = ∅,
then Y = X ∪H is an ultrafilter space.
(b) If X is normal and H ⊆ βX − X is a countable discrete set of remote
points such that X ∩ clH = ∅, then Y = X ∪H is a strong ultrafilter space.

Proof. (a) Y is a perfectly disconnected space. Indeed, let A and B be dis-
joint subsets of Y , and suppose that p ∈ ∂ (A) ∩ ∂ (B). If p ∈ H, then
p ∈ ∂ (A∩X)∩∂ (B ∩X). Since p is remote, p /∈ cl (A∩X − int (A∩X)) and
p /∈ cl (B ∩X − int (X ∩X)). Hence p ∈ cl (int (A∩X))∩ cl (int (B ∩X)),
which is impossible since Y is extremally disconnected.

Suppose that p ∈ X. Then p /∈ clH and therefore p ∈ ∂ (A∩X)∩∂ (B ∩X).
This is impossible since X is perfectly normal.

(b) It follows momentarily from part (a) and Proposition 2.9 since H is
normal. �

Acknowledgements. The author thanks the referee for the recommended
suggestions and corrections.

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