@
Appl. Gen. Topol. 21, no. 2 (2020), 326-329

doi:10.4995/agt.2020.13836

c© AGT, UPV, 2020

A disjointly tight irresolvable space

Angelo Bella a and Michael Hrušák b

a Department of Mathematics and computer science, University of Catania, Città universitaria,

viale A. Doria 6, 95125 Catania, Italy (bella@dmi.unict.it)
b Centro de Ciencias Matemáticas, Universidad Nacional Autónoma de México, Campus Morelia,

Morelia, Michoacán, 58089, México. (michael@matmor.unam.mx)

Communicated by S. Garćıa-Ferreira

Abstract

In this short note we prove the existence (in ZFC) of a completely
regular countable disjointly tight irresolvable space by showing that
every sub-maximal countable dense subset of 2c is disjointly tight.

2010 MSC: 54G15.

Keywords: irresolvable; disjointly tight; empty interior tightness.

In [2] the first author and V. Malykhin constructed using the Continuum
Hypothesis a completely regular countable disjointly tight irresolvable space,
and asked if it exists in ZFC. In this note we show how to combine their ideas
with those of Juhász, Soukup and Szentmiklóssy [6] to obtain such a space.

Our notation is standard and follows [7, 4]. All topological spaces considered
are completely regular.

Recall that a topological space X is irresolvable if it has no isolated points
and no disjoint dense sets. Following [2] we say that a space X is disjointly
tight (or it has disjoint tightness) if whenever x ∈ X is an accumulation point
of a set A there are disjoint A0, A1 ⊆ A each containing x in its closure. A
space X is sub-maximal if it has no isolated points and every dense subset of
X is open, or equivalently (see [1]), if every open subspace of X is irresolvable
and every nowhere dense subset of X is closed. These notions were introduced
by Hewitt [5] and have bee extensively studied in the past fifty years.

Received 11 June 2020 – Accepted 25 August 2020

http://dx.doi.org/10.4995/agt.2020.13836


A disjointly tight irresolvable space

Answering a question in [1], Juhász, Soukup and Szentmiklóssy [6] showed
that

Theorem 1.1 ([6]). 2c contains a countable dense subspace which is sub-
maximal.

The content of the paper is essentially reduced to proving the following:

Lemma 1.2. Let X be a countable dense subspace of 2κ, for some infinite
cardinal κ. If U ⊆ X is open, and x ∈ X is an accumulation point of U, then
there are disjoint open U0, U1 ⊆ U such that x is an accumulation point of both
U0 and U1.

Proof. As 2κ is c.c.c., we can write U as a countable union of disjoint basic
clopen sets

U =
!

n∈ω
Wn ∩ X.

For each n ∈ ω there is a finite Fn ⊆ κ and σn : Fn → 2 such that
Wn = {y ∈ 2κ : σn ⊆ y}.

Let J ⊇
"

n∈ω Fn be a countable subset of κ such that for every y ∕= y
′ ∈ X

there is an α ∈ J such that y(α) ∕= y′(α). Then the projection function
H : 2κ → 2J defined by H(y) = y ↾ J is one-to-one on X. Let Z = H[X].
Then Z is a dense subset of 2J, and the sets H[Wn] are disjoint clopen subsets
of 2J having H(x) = x ↾ J as an accumulation point of their union. As J is
countable, 2J is a compact metrizable space, hence there are disjoint infinite
sets A0, A1 ⊆ ω such that every neighborhood of H(x) in 2J contains all but
finitely many of the sets H[Wn], n ∈ A0 ∪ A1.

Let, for i = 0, 1

Ui =
!

n∈Ai

Wn ∩ X.

Both U0 and U1 are then open in X and we claim that x is in their closure.
To see this let W ⊆ 2κ be a basic open neighborhood of x, i.e. there is a

finite set F ⊆ κ such that
W = {y ∈ 2κ : y ↾ F = x ↾ F}.

Then there is a k ∈ ω such that H[W ]∩H[Wn] ∕= ∅, for every n ∈ (A0 ∪A1)\k.
This implies that W ∩ Wn is a non-empty clopen subset of 2κ for every n ∈
(A0 ∪ A1) \ k. By density of X, W ∩ Wn ∩ X ∕= ∅ for n ∈ (A0 ∪ A1) \ k
and Wn ∩ X ⊆ Ui, for n ∈ Ai \ k and i = 0, 1. Hence, W ∩ U0 ∕= ∅ and
W ∩ U1 ∕= ∅. □

Proposition 1.3. Every countable dense sub-maximal subset of 2κ is disjointly
tight.

Proof. Note (or recall) that a subset with empty interior of a sub-maximal
space is closed discrete. Consequently, if x is an accumulation point of a subset
A of a submaximal space, then x is an accumulation point of int(A). Now, if

c© AGT, UPV, 2020 Appl. Gen. Topol. 21, no. 2 327



A. Bella and M. Hrušák

X is a countable dense sub-maximal subset of 2κ, then by Lemma 1.2 there are
disjoint open U0, U1 ⊆ int(A) such that x is an accumulation point of both U0
and U1. □

The main result of the note now easily follows.

Corollary 1.4. There is a countable irresolvable space which is disjointly tight.

Proof. Follows immediately from Proposition 1.3, Theorem 1.1 and the trivial
fact that every sub-maximal space is irresolvable. □

Another corollary of Proposition 1.3 is the result of [1] that no countable
dense subset of 2κ is maximal (i.e. every strictly stronger topology has an
isolated point).

Proposition 1.3 actually shows that the countable sub-maximal spaces which
are densely embedable in some Cantor cube are quite far from being maximal
(see [3], Theorems 2.1b and 2.2c).

However, every countable maximal space is homeomorphic to a subspace of
2c and, consistently, there are maximal spaces which are embedable into 2κ for
some κ < c.

Problem 1.5. Is it consistent that there is a countable irresolvable disjointly
tight space of weight strictly less than c?

Another variation on tightness was also considered in [2].

Definition 1.6. A space X has empty interior tightness if for any set A ⊆ X
and any point x ∈ A there exists a set B ⊆ A, with empty interior in X, such
that x ∈ B.

This notion is relevant here because it fully characterizes resolvability of a
countable space.

Theorem 1.7 ([2]). A countable space X is resolvable if and only if it has
empty interior tightness.

Theorem 1.7 and Corollary 1.4 provide a regular space with disjoint tightness
but not empty interior tightness. The next example shows that these two
notions are actually mutually independent.

Example 1.8. A countable regular space with empty interior tightness but
not disjoint tightness.

Proof. Let X = (Q × ω) ∪ {p}, where Q is the set of rationals and p a free
ultrafilter on ω. We topologize X by taking the collection

{U×{n} : U open in the Euclidean topology on Q, n ∈ ω}∪{{p}∪(Q×A) : A ∈ p}
as a basis. It is easily seen that X is a countable regular resolvable space.
Hence, Theorem 1.7 shows that X has empty interior tightness. On the other
hand, the set A = {0} × ω and p ∈ A witness that X does not have disjoint
tightness. □

c© AGT, UPV, 2020 Appl. Gen. Topol. 21, no. 2 328



A disjointly tight irresolvable space

Acknowledgements. The research of the first author was supported by the
grand PIACERI 2020/22 (linea 2) from the University of Catania. The research
of the second author was supported by PAPIIT grant IN104220, and by a
CONACyT grant A1-S-16164.

References

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Topol. Appl. 107 (2000), 259–273.

[2] A. Bella and V. I. Malykhin, Tightness and resolvability, Comment. Math. Univ. Car-
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[3] E. K. van Douwen, Applications of maximal topologies, Topol. Appl. 51 (1993), 125–139.
[4] R. Engelking, General topology, Heldermann Verlag, Berlin, 1989.
[5] E. Hewitt, A problem of set-theoretic topology, Duke Math. J. 10 (1943), 309–333.
[6] I. Juhász, L. Soukup and Z. Szentmiklóssy, D-forced spaces: A new approach to resolv-

ability, Topol. Appl. 153, no. 11 (2006), 1800–1824.
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dam, 1980.

c© AGT, UPV, 2020 Appl. Gen. Topol. 21, no. 2 329