@
Applied General Topology

c© Universidad Politécnica de Valencia

Volume 13, no. 1, 2012

pp. 11-19

Some results and examples concerning

Whyburn spaces

Ofelia T. Alas, Maira Madriz-Mendoza and Richard G. Wilson

Abstract

We prove some cardinal inequalities valid in the classes of Whyburn
and hereditarily weakly Whyburn spaces and we construct examples of
non-Whyburn and non-weakly Whyburn spaces to illustrate that some
previously known results cannot be generalized.

2010 MSC: Primary 54D99; Secondary 54A25; 54D10; 54G99

Keywords: Whyburn space, weakly Whyburn space, submaximal space,
scattered space, semiregular, feebly compact

1. Introduction

A Hausdorff space X is said to be Whyburn if whenever A ⊆ X is not closed
and x ∈ cl(A) \ A, there is B ⊆ A such that cl(B) \ A = {x}. The space
is weakly Whyburn if whenever A ⊆ X is not closed, there is B ⊆ A such
that |cl(B) \ A| = 1. These classes of spaces have been studied previously in
[1], [4] and also earlier in [7] and elsewhere under the names AP-spaces, and
WAP-spaces. If A ⊆ X, then the Whyburn closure of A, denoted by wcl(A) is
defined as

A∪
⋃
{cl(C) : C ⊆ A, |cl(C) \A| = 1}.

It follows immediately that a space is weakly Whyburn if and only if every
Whyburn closed set is closed. Undefined terminology can be found in [2] or [5]
and all spaces are assumed to be (at least) Hausdorff.



12 O. T. Alas, M. Madriz-Mendoza and R. G. Wilson

2. Whyburn and weakly Whyburn spaces

In [4], a pseudocompact Whyburn space which is not Fréchet was constructed
and Proposition 2.1 of [7] states that a weakly Whyburn compact Hausdorff
space must have a non-trivial convergent sequence. It is easy to see that this
latter result generalizes to countably compact Hausdorff spaces and also to fee-
bly compact spaces with an infinite set of isolated points (recall that a space is
feebly compact if every locally finite family of non-empty open sets is finite).
The question then arises whether this result is true for all pseudocompact or
feebly compact spaces. To answer this question we need the following termi-
nology.

If Y is a non-empty scattered space, then we set

Y0 = {x : {x} is open} and for each ordinal α,
Yα = {x : {x} is open in Y \

⋃
{Yβ : β < α}}.

The dispersion order of Y is then the least ordinal for which Yα = ∅. For
the sequel, we note that for each n ∈ ω, the dispersion order of the countable
ordinal ωn + 1 is n + 1.

We also need two lemmas, the simple proof of the first of which we omit.

Lemma 2.1. For n ∈ ω, an infinite scattered subset A of a T1-space X has
dispersion order at most n if and only if it is the union of n discrete subspaces.

Lemma 2.2. If Y is a scattered metric space of finite dispersion order n + 1,
where n ≥ 1, and x ∈ Yn, then for any � > 0, there is an embedding h :
ωn + 1 → Y such that h(ωn) = x and diam(h[ωn + 1]) ≤ �.

Proof. The proof is by induction on the dispersion order of Y . If n = 1, then
each point x ∈ Y1 is the limit of a sequence S in Y0; S can be taken to have
arbitrarily small diameter and S ∪{x} is homeomorphic to ω + 1.

Suppose now that the result is true for each n < k and let Y be a scattered
space of dispersion order k + 1. Suppose that x ∈ Yk and � > 0; pick a
sequence 〈xm〉 = S ⊆ Yk−1 converging to x such that diam(S) < �/2. Since
Y is hereditarily collectionwise normal, we may find mutually disjoint open
sets Um such that xm ∈ Um; each set Um is scattered and has dispersion
order k. Applying the inductive hypothesis, for each m ∈ ω, we may find an
embedding hm : ω

k−1 + 1 → Um such that hm(ωk−1) = xm and such that
diam(Tm) < �/4

m, where Tm = h[ω
k−1 + 1]. Let T =

⋃
{Tm : m ∈ ω}∪{x};

it is straightforward to check that T is homeomorphic to ωk + 1 since each
neighbourhood of x contains all but finitely many of the sets Tm; furthermore,
diam(T) < �/2 + �/4 + �/16 < �. �

Example 2.3. There is a Whyburn H-closed (hence feebly compact) Hausdorff
space with no non-trivial convergent sequences.

Proof. We consider the space X = [0, 1] with the usual metric topology µ. Let
τ be the topology on X generated by

µ∪{X \D : D ⊆ X is µ−discrete}.



Some results and examples concerning Whyburn spaces 13

Since {X \D : D ⊆ X is µ−discrete} is a filter of dense subsets of (X,µ) it
follows that (X,τ) is H-closed. Furthermore, it is clear that (X,τ) is Hausdorff
and has no convergent non-trivial sequences. Even more is true: It follows from
Lemma 2.1 that every scattered subspace of (X,µ) of finite dispersion order is
closed in the topology τ. We will show that (X,τ) is a Whyburn space. To
this end, suppose that A ⊆ X is not closed and let x ∈ clτ (A) \ A. Now in
(X,µ), A is the union of a scattered subset C ⊆ A and a dense-in-itself subset
B ⊆ A, hence either (i) x ∈ clτ (B) or (ii) x ∈ clτ (C). We consider the cases
separately.
(i) Since B is dense-in-itself, every non-empty open subset of B contains a
dense subset homeomorphic to the rationals, Q. Choose a nested local base at
x of µ-closed sets V = {Vn : n ∈ ω}; we may assume that Vn+1 ⊆ int(Vn) and
B ∩ (int(Vn) \ Vn+1) 6= ∅ for each n ∈ ω. Since Q is universal for countable
metric spaces, for each n ∈ ω, in the open subset B ∩ (int(Vn) \Vn+1) of B we
may find a subspace Dn homeomorphic to the compact ordinal ω

n + 1 which
has dispersion order n + 1; let D =

⋃
{Dn : n ∈ ω}. It is easy to see that D

is scattered and has dispersion order ω and since x ∈ clµ(D) a straightforward
argument shows that clτ (D) \D = {x}.
(ii) Choose a nested local base at x of µ-closed sets V = {Vn : n ∈ ω}. If
x ∈ clτ (C), since each scattered subspace of (X,µ) of finite dispersion order
is τ-closed, it follows that for each n ∈ ω, C ∩ Vn has (countably) infinite
dispersion order κ and since every countable limit ordinal has cofinality ω, we
may assume without loss of generality that κ = ω. Then, for each n ∈ ω, using
the previous lemma we may find embeddings hn : ω

n + 1 → Vn ∩ C and it
is not hard to see that the maps hn may be chosen so that if m 6= n, then
Tn ∩Tm = ∅, where Tk = hk[ωk + 1]. Each of the sets Tk is µ-compact and τ-
discrete but T =

⋃
{Tk : k ∈ ω} has infinite dispersion order and so x ∈ clµ(T).

Furthermore, since for each µ-neighbourhood V of x, the set T \V is τ-closed,
it follows that cl(T) \C = {x}. �

In the sequel d(X), L(X), t(X) and ψ(X) will denote respectively the den-
sity, tightness, Lindelöf number and pseudocharacter of a space (X,τ) and
ψ(x,X) will denote the pseudocharacter of x in X. If (X,τ) is a Hausdorff space
and x ∈ X, then let ψc(x,X) = min{|U| : {x} =

⋂
{cl(U) : x ∈ U ∈U ⊆ τ}}.

Theorem 2.4. A k-space is weakly Whyburn if and only if for each non-
closed set A ⊆ X, there is some compact set K ⊆ X and x 6∈ A such that
cl(K ∩A) = (K ∩A) ∪{x} = K.

Proof. The sufficiency is clear, since K∩A is not closed in K. For the necessity,
suppose that (X,τ) is a Hausdorff weakly Whyburn k-space and that A ⊆ X
is not closed in X. Then there is some compact set C ⊆ X such that C ∩ A
is not closed in C. Since C is a closed subset of X, C is weakly Whyburn and
hence there is some x ∈ C \A and a set B ⊆ C ∩A such that cl(B)\A = {x}.
Clearly cl(B) is the required compact subset of X. �

Corollary 2.5. A weakly Whyburn k-space is pseudoradial.



14 O. T. Alas, M. Madriz-Mendoza and R. G. Wilson

Proof. This is an immediate consequence of the previous lemma and the fact
that a compact weakly Whyburn space is pseudoradial (see [7]). �

The next result extends Theorem 3 of [1] to the class of Hausdorff spaces.

Theorem 2.6. If X is a weakly Whyburn Lindelöf P-space and for each x ∈ X,
ψ(x,X) < ℵω, then X is pseudoradial.

Proof. For any Hausdorff space ψc(x,X) ≤ L(X)ψ(x,X) (see 2.8(c) of [3]) and
hence ψc(x,X) < ℵω for each x ∈ X. Let A ⊆ X be a non-closed set and B ⊆ A
such that cl(B)\A = {x} for some x ∈ X. Let U = {Uα : α < κ} be a family of
minimal cardinality κ of open sets in cl(B) such that

⋂
{cl(Uα) : α < κ} = {x}.

Since X is a P-space and x is not isolated, κ is a regular uncountable cardinal.
Since κ is minimal, for each α ∈ κ we may choose xα ∈

⋂
{cl(Uβ) : β <

α}\{x}⊆ A. Since cl(B) is Lindelöf, the set so constructed {xα : α < κ} must
have a complete accumulation point z ∈ cl(B). Since

⋂
{cl(Uα) : α < κ} = {x}

and the well-ordered net S = 〈xα〉α∈κ is finally in each set cl(Uβ) it follows that
z = x and x is the unique complete accumulation point of S. Furthermore,
S = 〈xα〉α∈κ must converge to x, for otherwise there would exist a subset of S
of size κ with no complete accumulation point. �

Theorem 2.7. The product of two Whyburn spaces, one of which is a k-space
and the other is locally compact is weakly Whyburn.

Proof. Suppose that X is a Whyburn k-space and Y is a Whyburn locally
compact space. It is known (see [7]) that a compact Whyburn Hausdorff space
is Fréchet-Urysohn and it is easy to see that the same is true of a Whyburn
Hausdorff k-space. It then follows from 3.3.J of [2] that X × Y is sequential
and hence weakly Whyburn. �

Question 2.8. Is the product of two Whyburn k-spaces, weakly Whyburn?

Theorem 2.9. If X is weakly Whyburn, then |X| ≤ d(X)t(X).

Proof. If X is finite, the result is trivial; thus we assume that X is infinite.
Suppose that d(X) = δ, t(X) = κ and D ⊆ X is a dense (proper) subset
of cardinality δ. Let D = D0 and define recursively an ascending chain of
subspaces {Dα : α ≤ κ+} as follows:

Since X is weakly Whyburn, there is some x ∈ X\D and Bx ⊆ D such that
cl(Bx) \D = {x}; clearly, we have |cl(Bx)| ≤ δ ≤ δκ and we may assume that
|Bx| ≤ κ.

We then define

D1 =
⋃
{cl(B) : B ⊆ D0, |B| ≤ κ, |cl(B) \D0| = 1}.

Clearly D1 ! D0 and since there are at most δκ such sets B it follows that
|D1| ≤ δκ.

Suppose now that for each β < α ≤ κ+ we have defined dense sets Dβ such
that |Dβ| ≤ δκ and Dγ ⊆ Dλ whenever γ < λ < α. If α is a limit ordinal, then
define Dα =

⋃
{Dβ : β < α} and then |Dα| ≤ |α|.δκ ≤ κ+.δκ = δκ. If on the



Some results and examples concerning Whyburn spaces 15

other hand α = β + 1, and Dβ  X, then since X is weakly Whyburn there is
some x ∈ X \Dβ and Bx ⊆ Dβ such that cl(Bx) \Dβ = {x}. Again we have
that |cl(Bx)| ≤ δκ and we may assume that |Bx| ≤ κ. Now we may define

Dα =
⋃
{cl(B) : B ⊆ Dβ, |B| ≤ κ, |cl(B) \Dβ| = 1}.

Clearly Dα ! Dβ and since there are at most (δκ)κ such sets B it follows that
|Dα| ≤ δκ.

To complete the proof it suffices to show that for some α ≤ κ+, we have
that Dα = X. Suppose to the contrary that ∆ =

⋃
{Dα : α < κ+} 6= X;

|∆| ≤ κ+.δκ = δκ. Then, since X is weakly Whyburn and has tightness κ,
there is some z ∈ X \ ∆ and some set B ⊆ ∆ of cardinality at most κ, such
that cl(B) \ ∆ = {z}. Since the sets {Dα : α < κ+} form an ascending chain
and cf(κ+) > κ, it follows that for some γ < κ+, B ⊆

⋃
{Dα : α < γ} and

hence z ∈ Dγ+1, a contradiction. �

Lemma 2.10. If X is hereditarily weakly Whyburn, then |X| ≤ 2d(X).

Proof. Suppose to the contrary that |X| > 2d(X). Let ∆ be a dense subset
of X of minimal cardinality, A = {A ⊆ ∆ : |clX(A)| ≤ 2d(X)} and Y =⋃
{clX(A) : A ∈ A}. Since |P(∆)| = 2d(X), it follows that |Y | ≤ 2d(X) and

hence if we put Z = ∆ ∪ (X \ Y ), then |Z| > 2d(X). Now if B ∈ P(∆) \A,
then |clX(B) ∩ Z| > 2d(X) thus showing that ∆ is Whyburn closed in Z but
not closed. Thus Z is not weakly Whyburn and hence X is not hereditarily
weakly Whyburn. �

3. The Whyburn property in scattered and submaximal spaces

We recall our convention that all spaces are Hausdorff. A space is said
to be submaximal if every dense subset is open. A standard procedure for
constructing submaximal topologies is as follows. Suppose that (X,τ) is a
(Hausdorff) space and D is a maximal filter in the family of dense subsets of
X. Then the topology σ generated by the subbase τ ∪D is submaximal and
is called a submaximalization of τ. Note that σ is semiregular if and only if τ
is semiregular and submaximal (then σ = τ). Obviously, a scattered space is
submaximal if and only if it has dispersion order 2.

As we mentioned earlier, every regular scattered space is weakly Whyburn
and the Katětov extension of ω (see 4.8(n) of [5]) shows that this is not true
in the class of Urysohn spaces. Thus it is natural to ask the following two
questions.

(1) Must a dense-in-itself submaximal Whyburn space be regular?, and
(2) Is every scattered semiregular space Whyburn?

We give a partial answer to the first question by showing that a submaxi-
malization of a resolvable space is never Whyburn and answer the second by



16 O. T. Alas, M. Madriz-Mendoza and R. G. Wilson

constructing a semiregular scattered space of dispersion order 2 which is not
weakly Whyburn.

Recall that a space is resolvable if it possesses two mutually disjoint dense
subsets.

Theorem 3.1. A submaximalization of a resolvable Hausdorff space is not
weakly Whyburn.

Proof. Suppose that (X,τ) is a resolvable T2-space and F is a maximal filter
of dense sets in X. We first show that there is F ∈ F such that X \ F is
somewhere dense in X. To this end, suppose to the contrary that no such F
exists, then for each F ∈ F, UF = X \ F is nowhere dense. Now let D and
D′ be complementary dense subsets of X; clearly D,D′ 6∈ F. For each F ∈F,
since int(F) = X \ cl(X \ F), it follows that int(F) is dense in X and so too
are D ∩ int(F) ⊆ D ∩F and D′ ∩ int(F)) ⊆ D′ ∩F. Since F is maximal, any
dense set which meets each element of F in a dense set is an element of F and
so it follows that D ∈F and D′ ∈F contradicting the fact that F is a filter.

Now let σ be the topology generated by τ ∪F and F ∈ F be such that
X \F is somewhere dense; thus intσ(clσ(X \F)) = U 6= ∅. Let V = U ∩F ,
x ∈ clσ(V ) \ F and note that V is infinite. Then if B ⊆ V is such that
x ∈ clσ(B), it follows that W = intσ(B) 6= ∅. But then, clσ(W) ∩ (X \F) =
clτ (W) ∩ (X \F) = clτ (clτ (W) ∩ (X \F)) ∩ (X \F) which is infinite. �

An example of a scattered submaximal Whyburn (even first countable) space
which is not regular (nor even semiregular) is easy to construct. Let Q denote
the rational numbers and X = Q×{0, 1} with the following topology:

Each point of Q×{0} is isolated and a basic open neighbourhood of (q, 1) is
of the form {(q, 1)}∪ [(Uq \{q})×{0}] where Uq is a Euclidean neighbourhood
of q ∈ Q.

A space X is said to be ω-resolvable if X possesses infinitely many mutu-
ally disjoint dense subsets. The construction of the next example depends on
the existence of a countable ω-resolvable Hausdorff space which is not weakly
Whyburn. Before constructing such a space, the following lemma is needed

Lemma 3.2. A space X, ω ⊆ X ⊆ βω is hereditarily weakly Whyburn if and
only if X is scattered.

Proof. The sufficiency is clear since a subspace of a scattered space is scattered
and it was proved in [4] that a regular scattered space is weakly Whyburn.
Furthermore, it is easy to see that if the dispersion order of X is 2, then it is
Whyburn also.

For the inverse implication, suppose that D ⊆ X\ω is dense in itself and let
Y = ω ∪D; if Y were weakly Whyburn, then we could find p ∈ D and B ⊆ ω
such that clY (B) \ ω = {p}, in other words, clY (B) = B ∪ {p}. However,
clY (B) = clβω(B) ∩ Y and so clY (B) ∩ D is an open subset of D to which p
belongs; since p is not isolated, this set must be infinite. �



Some results and examples concerning Whyburn spaces 17

By way of contrast to the last result we note that under CH the subspace
of P-points of βω\ω has character ω1 and it then follows from Proposition 2.7
of [4] that this space is Whyburn.

Consider a countable dense-in-itself subset D ⊆ clβQ(N)\N ⊆ βQ\Q (where
once again, Q denotes the set of rational numbers with the Euclidean topology).
Let X = Q∪D; X is a countable Tychonoff space which, is clearly ω-resolvable.
That N is Whyburn closed in X follows from the previous lemma and the fact
that clβQ(N) is homeomorphic to βω.

Example 3.3. There is a semiregular scattered space (of dispersion order 2)
which is not weakly Whyburn.

Proof. Let (Z,σ) be an ω-resolvable (dense-in-itself) countable Tychonoff space
which is not weakly Whyburn and let F be an infinite family of mutually
disjoint dense subsets of (Z,σ) and φ : Z →F a bijection. Let X = Z ×{0, 1}
and for each z ∈ Z, let Vz be an open neighbourhood base at z. We define a
topology τ on X = Z ×{0, 1} as follows:

Each point of Z ×{0} is isolated and an open neighbourhood of (z, 1) is of
the form

WV,z = {(z, 1)}∪ (V ×{0}) \ ({(z, 0)}∪φ(z)), where V ∈Vz.

The space (X,τ) is a scattered space of dispersion order 2 and we proceed
to show that it is neither regular nor weakly Whyburn.

It is easy to see that X is not regular since the open neighbourhood WV,z
of (z, 1) contains no closed neighbourhood of that point. To prove that X
is semiregular, it suffices to show that each of the sets WV,z is regular open.
To see this, suppose that (t, 1) ∈ clX(WV,z) where t 6= z; then since φ(z) is
dense in Z, each neighbourhood of (t, 1) meets the set φ(z)×{0} showing that
(t, 1) 6∈ intX(cl(WV,z)).

Finally, to show that (X,τ) is not Whyburn, it suffices to prove that there is
some A ⊆ Z such that A×{0} is Whyburn closed but not closed in X. However,
Z is not weakly Whyburn and hence there is some A ⊆ Z which is Whyburn
closed but not closed in Z and so if B ⊆ A is such that clZ(B)\A is nonempty,
we must have clZ(B)\A has no isolated points (and hence is infinite). We claim
that if B ⊆ A is such that clZ(B)\A is nonempty then clX(B×{0})\(A×{0})
is infinite. To prove our claim, suppose that s ∈ clZ(B) \ A; then either
(s, 1) ∈ clX(B ×{0}) \ (A×{0}) or not. If (s, 1) 6∈ clX(B ×{0}) \ (A×{0})
then there is some open neighbourhood U of s in Z such that clZ(U)∩B ⊆ φ(s)
and U contains infinitely many points of clZ(B)\A. If s 6= t ∈ U∩(clZ(B)\A),
then since B ∩U * φ(t), it follows that t ∈ clX(B ×{0}) \ (A×{0}), showing
that clX(B ×{0}) \ (A×{0}) is infinite. �



18 O. T. Alas, M. Madriz-Mendoza and R. G. Wilson

4. Some open questions

The space constructed in Example 2.3 is not regular, thus we are led to ask:

Question 4.1. Does every (weakly) Whyburn pseudocompact Tychonoff space
have a convergent sequence?

A number of dense pseudocompact subspaces of {0, 1}c and Ic have been
constructed which do not possess a non-trivial convergent sequence (for exam-
ple see [6]); however, the question of whether such constructions can produce
a weakly Whyburn space has apparently not been studied.

Question 4.2. Is the bound ψ(x,X) < ℵω necessary in Theorem 2.6?

Question 4.3. Suppose that |X| > 2d(X); can X be weakly Whyburn?

Question 4.4. Does there exist in ZFC a dense Whyburn subspace of βω\ω?

Acknowledgements. This research was supported by the network Algebra,
Topoloǵıa y Análisis del PROMEP, Project 12611243 (México) and Fundação
de Amparo a Pesquisa do Estado de São Paulo (Brasil). The third author
wishes to thank the Departament de Matemàtiques de la Universitat Jaume I
for support from Pla 2009 de Promoció de la Investigació, Fundació Bancaixa,
Castelló, while working on an early draft of this article.

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[2] R. Engelking, General Topology, Heldermann Verlag, Berlin, 1989.
[3] I. Juhász, Cardinal Functions in Topology - Ten years later, Mathematisch Centrum,

Amsterdam, 1980.
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spaces need not be Fréchet, Proc. Amer. Math. Soc. 131, no. 10 (2003), 3257–3265.

[5] J. R. Porter and R. G. Woods, Extensions and Absolutes, Springer Verlag, New York,
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[6] E. A. Reznichenko, A pseudocompact space in which only sets of complete cardinality

are not closed and not discrete, Moscow Univ. Math. Bull. 6 (1989), 69–70 (in Russian).
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(Received November 2009 – Accepted November 2011)



Some results and examples concerning Whyburn spaces 19

Ofelia T. Alas (alas@ime.usp.br)
Instituto de Matemática e Estat́ıstica, Universidade de São Paulo, Caixa Postal
66281, 05311-970 São Paulo, Brasil

Maira Madriz-Mendoza (seber@xanum.uam.mx)
Departamento de Matemáticas, Universidad Autónoma Metropolitana, Unidad
Iztapalapa, Avenida San Rafael Atlixco, #186, Apartado Postal 55-532, 09340,
México, D.F., México

Richard G. Wilson (rgw@xanum.uam.mx)
Departamento de Matemáticas, Universidad Autónoma Metropolitana, Unidad
Iztapalapa, Avenida San Rafael Atlixco, #186, Apartado Postal 55-532, 09340,
México, D.F., México


	Some results and examples concerning[6pt] Whyburn spaces. By O. T. Alas, M. Madriz-Mendoza and R. G. Wilson