

CaWaAGT.dvi


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Applied General Topology

c© Universidad Politécnica de Valencia

Volume 8, No. 2, 2007

pp. 187-205

The Alexandroff Duplicate and its subspaces

Agata Caserta
∗

and Stephen Watson

Abstract. We study some topological properties of the class of the

Alexandroff duplicates and their subspaces. We give a characterization

of metrizability and Lindelöf properties of subspaces of the Alexan-

droff duplicate. This characterization clarifies the potential for finding

Michael spaces among the subspaces of Alexandroff duplicates.

2000 AMS Classification: 54B05, 54B10.

Keywords: Resolution, Alexandroff duplicate, Lindelöf property, Michael-
type line.

In their famous 1922 memoir on compact spaces [1], Alexandroff and Urysohn
defined a topological space that has become known as the Alexandroff double
circle or the Alexandroff duplicate. In this paper we study several version of the
Alexandroff duplicate by viewing it as particular resolution by constant maps.

Alexandroff duplicates have been studied and used by many topologists.
In particular, Michael’s 1963 example of a Michael spaces is subspace of an
Alexandroff duplicate.

This paper is an organized study of the topological properties of the class of
the Alexandroff duplicates and of their subspaces. In particular, we characterize
when subspaces of Alexandroff duplicates have the Lindelöf property. This
suggests that the potential for finding Michael spaces among the subspaces of
Alexandroff duplicates is not high.

In this note, P and Q denote the set of the irrational and rational numbers,
respectively. Ordinal numbers are denoted by Greek letters; when viewed as
topological spaces, they are given the order topology. Products of topological
spaces are endowed with the standard product topology.

The symbol [A]λ denotes the family of subsets of A having size exactly λ.
The symbols [A]≤λ and [A]<λ have similar meaning.

∗Corresponding author.


188 A. Caserta and S. Watson

Let ≤∗ be the quasi-order on a countable product of ordered sets that is
associated to the coordinate-wise order on each set. Thus f ≤∗ g stands for
f(n) ≤ g(n) for all but finitely many n ∈ ω. A subset of ωω is unbounded if
it is unbounded in (ωω,≤∗). A dominating family is an unbounded set that is
cofinal in (ωω,≤∗). A subset of

ωω is said to be a scale if it is a dominating
family and is well-ordered by ≤∗.

Recall that P can be identified with ωω with the product topology. For
each ξ ∈ <ωω = {η | η : [0,n] → ω for some n}, a basic open neighborhood
of ξ in the product topology is {f ∈ ωω : ξ ⊆ f}. For every g ∈ ωω, the
sets {f ∈ ωω : f ≤ g} and {f ∈ ωω : f ≤∗ g} are respectively compact and
σ-compact (see [14]).

Let X and Y be topological spaces. A set A ⊆ X is Y-analytic if it is a
projection on X of a closed subset of X × Y . In particular, A ⊆ X is analytic
if it is P-analytic.

Given a function f : X → Y , the small image of A ⊆ X is defined by
f♯(A) = {y ∈ Y : f−1(y) ⊆ A}. Sometimes we abuse of terminology and say
that f♯ is open, with the meaning that for each open subset A of X, f♯(A) is
an open subset of Y .

In most cases we will employ the notation used in [6] and [9].

1. Basic definitions and preliminary results

We begin with the definition of the Alexandroff double circle as a resolution
by constant map.

Definition 1.1. Let (X,τ) be a topological space. For any U ∈ τ, denote

Û = U × 2. Define a base for a topology on Y = X × 2 by B = B0 ∪ B1, where
B0 is the family of all subsets Û \ (F × {1}) of Y , with U ∈ τ, and F ∈ [X]

<ω,
and B1 = {(x, 1) : x ∈ X}. This topological space is the resolution of X at
each point into the two point space by the constant zero function (see [15] and
[7]). However we use the notation Y = X ×ad 2 (the subscript ad stands for
Alexandroff duplicate).

For each x ∈ X, we denote τ(x) = {U ∈ τ : x ∈ U} and B(x) = {(x, 1)} ∪

{Û \ {(x, 1)} : U ∈ τ(x)}. Further, let B
′

=
⋃

x∈X B(x).

Lemma 1.2. If X is a T1 space, then B
′

is a base, such that B
′

⊂ B, and B(x)
is a local base at each x ∈ X. Moreover, if Z = (A×{1})∪(B×{0}) ⊆ X ×ad 2

and U is a base in X at x ∈ B \ A, then {Û ∩ Z : U ∈ U} is a base at (x, 0) in
Z.

Proof. Let (x, 0) ∈ Û \ (F × {1}). Set F
′

= F \ {x} and V = U \ F
′

. Then

V̂ \ {(x, 1)} ⊂ Û \ (F × {1}) and V̂ \ {(x, 1)} ∈ B(x). �

In this paper, unless otherwise stated, topological spaces are considered

T1. Furthermore, if U = {Ui : i ∈ I}, let Û = {Ûi | Ui ∈ U} and U
∗ =

{Ûi \ (F × {1}) : F ∈ [X]
<ω,Ui ∈ U}.


The Alexandroff Duplicate and its subspaces 189

Remark 1.3. The space X is homeomorphic to the subset Z = (I×{1})∪((X\
I) × {0}) of its duplicate X ×ad 2. Indeed, the function φ : X → Z defined by

φ(x) = (idX (x),χI (x)) is clearly a bijection. Now, let U
∗ = (Û \(FU ×{1}))∩Z

be a basic open set in Z. Then φ−1(U∗) = U \ (FU ∩ I) which is open in X.
Moreover, if U ∈ τ, then φ(U) = (U ∩ I) × {1} ∪ (U ∩ (X \I)) × {0} is an open
set in Z.

Given the topological space X ×ad 2, we consider the following functions:

• r : X ×ad 2 → X × {0} such that for each x ∈ X, r(x, 1) = (x, 0), and
r ↾ X × {0} = idX×{0};

• π0 : X ×ad 2 → X such that π0(x,i) = x for each x ∈ X and i = 0, 1;
• ι : X → X × {0} ⊂ X ×ad 2 such that for each x ∈ X, ι(x) = (x, 0);

and its inverse map π0 ↾ (X × {0}) = ι̂.

Note that the projection map π0 is continuous.
In the following we show some properties of this functions.

Lemma 1.4. Let X be a topological space and X × {0} ⊂ X ×ad 2, then
X ∼= X × {0}. Further, ι and ι̂ are homeomorphisms.

Proof. Let (Û \ (F × {1})) ∩ (X × {0}) = U × {0} be an open set in the

subspace X × {0}. Hence ι−1(Û \ (F × {1})) = U, therefore ι is continuous.
The continuity of ι̂ follows from the continuity of π0. Since ι̂◦ ι = idX and ι◦ ι̂
= idX×{0}, both ι and ι̂ are homeomorphisms, i.e., ι

−1 = ι̂. �

Definition 1.5. Let Y ⊂ X and f : X → Y be a function. Then f is
called a retraction of X onto Y (Y is the retract of X) if it is continuous and
f ↾ Y = idY . If f is continuous only at the points of Y , then it is called a weak
retraction of X onto Y .

Lemma 1.6. The map r is a retraction of X ×ad 2 onto X × {0}. Moreover
r is a closed map.

Proof. From r = î◦π0, it follows that r is a continuous map. Since r ↾ X ×{0}
= idX × {0}, r is a retraction of X ×ad 2 onto X × {0}. We show that r

# is

open on B. Let Û \ (F ×{1}) ∈ B. Then r#(Û \ (F ×{1})) = U ×{0}\F ×{0},
which is an open set in X × {0}. Thus r is a closed map. �

Lemma 1.7. Let Z = (A × {1}) ∪ (B × {0}) ⊆ X ×ad 2 and suppose φ : Z →

B × {0} is a continuous map. Define φ
′

: Z → B × {0} by

φ
′

(x,i) =

{
φ(x, 0) if x ∈ A ∩ B ∧ i = 1
φ(x,i) otherwise.

Then φ
′

is continuous too. Moreover if φ is a retraction, then φ
′

is a retraction
too.

Proof. Let H = Z \((A∩B)×{1}) and K = ((A∩B)×{1})∪(B×{0}) be two
closed sets of Z. Let ψ = φ ↾ H and θ = r ↾ K two continuous maps, such that
ψ ↾ K∩H = θ ↾ K∩H. Then, by the pasting lemma, φ

′

= ψ∪θ is a continuous
map. Since φ ↾ B × {0} = idB×{0}, then φ

′

↾ B × {0} = idB×{0}. �


190 A. Caserta and S. Watson

Definition 1.8. Let X and Y be two topological spaces, and f : X → Y be a
continuous map. Then f is hereditarily closed if f ↾ Z : Z → f(Z) is closed, for
each subset Z of X. The function f is hereditarily perfect if it is hereditarily
closed and all fibers (f ↾ Z)−1(y) are compact subsets of Z respectively, for
each subset Z of X.

Remark 1.9. Observe that for an Hausdorff space X, the function f : X → Y
is hereditarily perfect if and only if f is hereditarily closed and all fibers are
finite. Indeed, assume that |(f ↾ Z)−1(y)| ≥ ℵ0. Let {zn}n be an infinite
subset of (f ↾ Z)−1(y). Then {zn}n has a cluster point z which is an element
of (f ↾ Z)−1(y), because it is a closed subset of Z. Hence, {f(zn)}n does not
have accumulation point in f(z) which is a contradiction.

Note that for each Z ⊆ X ×ad 2 the function r ↾ Z has finite fibers. The
following example shows that in general the retraction is not a hereditarily
closed map, hence not hereditarily perfect.

Example 1.10. Let r : [0, 1] ×ad 2 → [0, 1] × {0} be a retraction. Let Z =
[0, 1) × 2 ∪ {(1, 1)} ⊆ [0, 1] ×ad 2, and F = [0, 1) × 2. Then F is a closed subset
of Z such that r ↾ Z(F) = [0, 1)×{0}, which is not closed in r(Z) = [0, 1]×{0}.

Next we show a characterization for a retraction of a subspace of the Alexan-
droff duplicate. Before we need the following definitions.

Definition 1.11. Let A and B be subsets of a topological space X. We say
that A is closed in B if for each b ∈ B such that b ∈ A, then b ∈ A.

Definition 1.12. Let {Aα}α∈κ, B be subsets of a topological space X. The
family {Aα}α∈κ is locally finite in B if for each point b of B there exists a
neighborhood of b in X which intersects finitely many elements of {Aα}α∈κ.

Lemma 1.13. Let X and Y be topological spaces, B ⊂ X and {Aα}α∈κ a
family of sets closed in B which is locally finite in B. Let f : B → Y and for
each α ∈ κ, gα : Aα → Y are continuous maps which are compatible with f
and each other. Then h = f ∪ (

⋃
α∈κ gα) is a continuous map at each point of

B.

Proof. Since h is the extension of each gα and f, it remains to check the
continuity of h at points of B. Let (xσ)σ∈Σ a net, with b ∈ B such that
xσ → b. Then for each neighborhood of b, IX (b), there exists σ0 ∈ Σ such
that for each σ ≥ σ0, xσ ∈ IX (b). By hypothesis the family {Aα}α∈κ is locally
finite in B, hence for each σ ≥ σ0, IX (b) ∩ Aαi = ∅ for i /∈ {1, ...,n}. Set
Σ′ = {σ ∈ Σ : σ ≥ σ0}. Then (xσ)σ∈Σ′ ⊆ B ∪ (

⋃n
i=1

Aαi ) and xσ → b. Since
every net has a subnet which is an ultranet , we have that if A is a finite cover
of X, there exists A ∈ A and subnet (xσλ )λ∈Λ such that (xσλ )λ∈Λ ⊂ A. Then
assume, without loss of generality, that (xσλ )λ∈Λ ∈ B or (xσλ )λ∈Λ ∈ Aαi for
some i ∈ {1, ...,n}. First assume that (xσλ )λ∈Λ ∈ B. Since xσλ → b and
(h(xσλ ))λ∈Λ = (f(xσλ ))λ∈Λ, by continuity of f, follow that h(xσλ ) → h(b). Let
(xσλ )λ∈Λ ∈ Aαi for some i ∈ {1, ...,n}. From Aαi closed in B and b ∈ Aαi ∩B,


The Alexandroff Duplicate and its subspaces 191

it follows that b ∈ Aαi . By continuity of gαi we have gαi (xσλ ) → gαi (x). Thus
h(xσλ ) → h(b). �

Note that in the previous Lemma, if the family of Aα’s is finite, we only
need that all sets Aα are closed in B.

Definition 1.14. Let A, B be subsets of a topological space X with A ⊆ B.
We say that A is discrete in B if for each b ∈ B there exists a neighborhood of
b in X which intersects A at most in one point.

Lemma 1.15. Let X, Y be topological spaces and B ⊂ X such that the points
of X \ B ∩ B can be separated by a disjoint family of open sets in X. Then

there exists {Aα : α ∈ κ}, with κ = |X \ B ∩ B|, a family of sets closed in B
such that:

(i) X = (
⋃

α∈κ Aα) ∪ B,
(ii) |Aα ∩ B| = 1 for each α ∈ κ,
(iii) Aα ∩ Aβ = ∅ for each α and β distinct.

Moreover, under the additional assumption that X \ B ∩ B is discrete in B, it
follows that the family {Aα : α ∈ κ} is locally finite in B.

Proof. Let {xα : α ∈ κ} = X \ B ∩ B and let {Uα : α ∈ κ} be a disjoint family
of open sets, such that xα ∈ Uα for each α ∈ κ. Let {Aα : α ∈ κ} be defined
by: Aα = (Uα \B)∪{xα} for each α ∈ κ and A0 = (X \ (

⋃
α∈κ Uα ∪B))∪{xκ}.

Each Aα with α ∈ κ is closed in B, because the only accumulation points of
X \B in B are xα’s and {xα} = Aα ∩ B. Moreover, since xα’s are distinct and
Uα’s are disjoint, {Aα : α ∈ κ} is disjoint.

Assume that {xα : α ∈ κ} is discrete in B, we prove that{Aα : α ∈ κ} is
locally finite in B. Note that, for all α ∈ κ, Aα ∩ (B \ {xα : α ∈ κ}) = ∅. If

b ∈ B \ {xα : α ∈ κ}, then b /∈ X \ B. Thus there exists a neighborhood IX (b)

of b such that IX (b) ∩ (X \ B) = ∅. Since X \ B ∩ B is discrete in B, there
exists a neighborhood I′X (b) of b such that |I

′
X (b) ∩ {xα : α ∈ κ}| ≤ 1. Let

I′X (b) ∩ {xα : α ∈ κ} = xα with α ∈ κ. Assume that (IX (b) ∩ I
′
X (b)) ∩ Aβ 6= ∅

for β ∈ κ and β 6= α. Let x ∈ (IX (b) ∩ I
′
X (b)) ∩ Aβ , then either x ∈ X \ B

hence x /∈ B, or x /∈ B\{xα : α ∈ κ}, hence x = xβ , a contradiction. Therefore
IX (b) ∩ I

′
X (b) intersects {Aα : α ∈ κ} in xα. Now, if b ∈ {xα : α ∈ κ}, then

b = xα for α ∈ κ. Since, by construction, Uα ∩Uβ = ∅ for α 6= β and Aβ ⊂ Uβ,
it follows that Uα ∩Aβ = ∅ for β 6= α except for β = 0. Therefore Uα intersects
finitely many elements of {Aα : α ∈ κ}. �

Lemma 1.16. Let X, Y be topological spaces and B ⊂ X such that X \ B ∩B
is discrete in B and its points can be separated by a disjoint family of open sets
in X. Then any f : B → Y continuous map can be extended to X, so that it
remains continuous at points of B.

Proof. Let κ be a cardinal with |X \ B ∩ B| = κ. By Lemma 1.15, there exists
{Aα : α ≤ κ} a disjoint family of closed sets in B such that X = (

⋃
α∈κ Aα)∪B

and Aα ∩ B = {xα} for some xα. Let gα : Aα → Y given by gα(x) = f(xα).


192 A. Caserta and S. Watson

Then gα’s are continuous maps. By Lemma 1.13, there exists h = f ∪
⋃

α∈κ gα,
which extends f to X and is continuous at B. �

Note that in the previous Lemma, if |X \ B ∩ B| < ℵ0, we only need the
space X to be Hausdorff.

Corollary 1.17. Let B ⊆ X. If X \ B ∩ B is locally discrete in B and its
points can be separated by a disjoint family of open sets in X, then there is a
weak retraction of X onto B.

Proof. Let idB be the identity map on B. Apply Lemma 1.16 with Y = B. �

With reference to Definition 1.5, the following proposition gives a character-
ization for a retraction of a subspace Z of X ×ad 2.

Proposition 1.18. Let Z = (A × {1}) ∪ (B × {0}) ⊆ X ×ad 2. Then Z can be
retracted onto B × {0} if and only if A ∪ B can be weakly retracted onto B.

Proof. Let φ be a retraction of Z onto B × {0}, and assume, without loss of
generality, by Lemma 1.7, that for each x ∈ A ∩ B, φ(x, 1) = φ(x, 0) = (x, 0).
For each x ∈ A ∪ B choose ix ∈ {0, 1} such that (x,ix) ∈ Z. Since φ(x, 1) =
φ(x, 0) for each x ∈ A∪B the choice of ix does not matter. Define f : A∪B → B
by f(x) = ι−1 ◦ φ(x,ix). For each x ∈ B, f(x) = ι

−1 ◦ φ(x, 0) = x. It remains
to prove the continuity of f at the points of B. Let (xσ)σ∈Σ a net in X, x 6= xσ
for each σ ∈ Σ and x ∈ B, such that xσ → x. Choose iσ ∈ {0, 1} where
(xσ, iσ) ∈ Z. Then (xσ, iσ) → (x, 0). Since ι

−1 ◦ φ is a continuous function,
ι−1 ◦ φ(xσ, iσ) → ι ◦ φ(x, 0). Then f(xσ) → f(x).

Vice versa, let f : A ∪ B → B be a weak retraction and (r ↾ Z) : Z →
(A ∪ B) × {0} the retraction of Z onto B × {0}. Define φ : Z → B × {0} such
that φ(x,i) = ι ◦ f ◦ ι−1 ◦ (r ↾ Z)(x,i). Then φ is a continuous map at the
points of B × {0}. Since A× {1} is a discrete subset of X ×ad 2, it follows that
φ is continuous. Moreover φ ↾ B × {0} = idB×{0}. �

Lemma 1.19. Let X be a complete metrizable space, Y ⊆ X such that Y is
a countable dense set with no isolated points. Then there is no weak retraction
of X onto Y .

Proof. By contradiction, let f : X → Y be a weak retraction and A ⊆ X the set
of all points in which f is continuous. Then Y ⊆ A and A is a Gδ set in X, hence
A is a complete metrizable space. Observe that A =

⋃
y∈Y (f ↾ A)

−1({y}), i.e.,
A is an Fσ in X. By Baire Category Theorem, there exists y ∈ Y such that
(f ↾ A)−1({y}) is not nowhere dense, hence (f ↾ A)−1({y}) contains U, an
open set in A. Since U ∩ Y 6= {y}, we have a contradiction. �

Corollary 1.20. Let Z = (P × {1}) ∪ (Q × {0}) ⊂ [0, 1] ×ad 2. Then Z cannot
be retracted onto Q × {0}.

Proof. By Lemma 1.19, there is not a weakly retraction of R onto Q. Proposi-
tion 1.18 ends the proof. �


The Alexandroff Duplicate and its subspaces 193

2. Properties preserved by the Alexandroff Duplicate and its
subspaces

As a special case of the the Fundamental Theorem of Resolutions, we have
that if X is a compact space, then X ×ad 2 is also compact.

Next we show that many properties of X are preserved by its duplicate.

Lemma 2.1. X is Tychonoff space if and only if X ×ad 2 is Tychonoff.

Proof. First we show that T1 is preserved. Any isolated point (x, 1) ∈ X ×ad 2
is clopen. Moreover, since X is T1, and X ∼= X × {0} is a closed subspace
of X ×ad 2, it follows that every point is closed in X ×ad 2. Observe that for
the points of X × {1}, since they are isolated points, there exists always a
continuous map f : X ×ad 2 → I such that f((x, 1)) = 0 and f((x,i)) = 1 for

any (x,i) 6= (x, 1). Let (x0, 0) ∈ X×ad 2 and U
∗ = Û \{(x0, 1)} a neighborhood

of (x0, 0). Since X is Tychonoff, x0 ∈ X and U ∈ τX , there exists g : X → I
continuous map such that g(x0) = 0 and g(U

c) = 1. We define f : X ×ad 2 → I
such that f((x,i)) = 1 for (x,i) = (x0, 1) and f((x,i) = g(x) otherwise. Then
f is continuous map such that f((x0, 0)) = 0 and f((x,i)) = 1 for any (x,i) ∈
(X ×ad 2) \ U

∗.
Since X ∼= X × {0}, the vice versa holds as well. �

Lemma 2.2. If X is normal, then also X ×ad 2 is normal.

Proof. Let K and C be two closed disjoint subsets of X×ad2. Let K∩X×{0} =
K1 and C∩X ×{0} = C1 closed disjoint subsets in X ×{0}, by normality of X,
there exists U1 and V1 open sets in X ×{0} such that K1 ⊆ U1and C1 ⊆ V1 and

U1∩V1 = ∅. Then V = (V̂1 \K)∪(C\X×{0}) and U = (Û1\C)∪(K\X×{0})
are open disjoint subsets in X ×ad 2 containing C and K respectively. Then
X ×ad 2 is normal. �

Let A and B subsets of X and Z = (A × {1}) ∩ (B × {0}) ⊆ X ×ad 2.
Henceforth, unless we state otherwise, we denote a subset of X ×ad 2 simply
with Z.

In general a subspace Z ⊆ X ×ad 2 need not to be normal even if X is
normal. Indeed, if X is not hereditarily normal, there exists A ⊆ X that is not
normal. Take Z = A × {0}. Then Z is not normal as a subset of that normal
space X ×ad 2.

Lemma 2.3. Let X be a normal space and B a closed subset of X. Then
Z ⊆ X ×ad 2 is normal.

Proof. Let K = K
′

∩Z and C = C
′

∩Z be two closed disjoint subsets of Z, with
K

′

and C
′

closed subsets of X×ad2. Let K∩B×{0} = K1 and C∩B×{0} = C1
be closed disjoint subsets in B × {0}. Since B is a closed subset of X, and
B × {0} is normal, there exists U1 = U ∩ B and V1 = V ∩ B, with U ,V ∈ τ,
which are disjoint open sets in B × {0} such that K1 ⊆ U1 and C1 ⊆ V1. Take

V ∗ = ((V̂1 ∩Z)\K)∪(C\B×{0}) and U
∗ = ((Û1∩Z)\C)∪(K\B×{0}). Then

V ∗ and U∗ are open disjoint subsets in Z containing C and K respectively. �


194 A. Caserta and S. Watson

Lemma 2.4. For any topological space X, the following hold:

(i) l(X) = l(X ×ad 2),
(ii) If |X| ≥ ℵ0, c(X ×ad 2) = |X|,
(iii) χ(X) = χ(X ×ad 2).

Proof. (i). Observe that l(X) = l(X × {0}) ≤ l(X ×ad 2), because X × {0}
is closed. It is sufficient to show that l(X ×ad 2) ≤ l(X). Let U be an open
cover for X ×ad 2, and assume, without loss of generality, that U ⊆ B. Thus,
U∩B0 = {Ûi\(FUi ×{1}) : i ∈ I∧FUi ∈ [X]

<ω}. Since l(X) = κ, there exists an
open refinement of {Ui}i∈I , {Vi}i∈J which cover X, such that |J| ≤ κ. For each
Vi ∈ {Vi}i∈J choose Ui ∈ {Ui}i∈I , such that Vi ⊆ Ui, and define FUi = FVi .

So {V̂i \ (FVi × {1}) : i ∈ J ∧ FVi ∈ [X]
<ω}, leaves uncovered

⋃
i∈J FVi × {1}

which can be covered by |J| many open sets. Thus l(X ×ad 2) ≤ κ.
(ii). Let |X| = κ ≥ ℵ0. The set X ×{1} ⊂ X ×ad 2 is a set of isolated points

of size κ , so c(X × {1}) = κ. Since X × {1} is an open subset of X ×ad 2,
c(X × {1}) ≤ c(X ×ad 2). Thus c(X ×ad 2) ≥ κ. From c(X × {0}) ≤ |X ×ad 2|,
if follows that c(X ×ad 2) ≤ κ.

(iii). For each a = (x, 1) ∈ X×{1}, {a} is a local base at a. Thus χ(a,X×ad
2) =1. Next, let a = (x, 0) ∈ X × {0} and χ(X) = k. Let U(x) be a local

base at each x ∈ X. Then {Û \ {(x, 1)} : U ∈ U(x)} is a local base at
a = (x, 0) ∈ X × {0}, which has the same size as U(x) . Then χ(X × {0}) =
χ(X ×ad 2) = χ(X). �

Lemma 2.5. If U is a locally finite family in X, then Û is a locally finite
family in X ×ad 2.

Proof. Let z = (x,i) ∈ X ×ad 2, observe that if V is a neighborhood of x in X,

then V̂ is a neighborhood of z in X ×ad 2. Now, since U is a locally finite family
of X, there exists a neighborhood Vx of x which intersects only finitely many
elements of U. Thus V̂x is a neighborhood of z in X ×ad 2, which intersect only
finitely many elements of Û. �

Remark 2.6. Furthermore, if U is a locally finite family of open subsets of X,
for any map f : U → [X × 2]<ω such that U → FU , the set {Û \ FU : U ∈ U}
is locally finite family of open subsets of X ×ad 2.

Proposition 2.7. If X is a paracompact space, then also X ×ad 2 is a para-
compact space.

Proof. Let U be an open cover for X ×ad 2, and assume, without loss of gener-
ality, U ⊂ B. So U ∩ B0 = {Û \ (FU × {1}) : U ∈ U

′

∧ FU ∈ [X]
<ω} where U

′

is a cover of X. By paracompactness of X, there exist an open refinement V
′

of U
′

which is locally finite. For each V ∈ V
′

, choose U ∈ U
′

such that V ⊆ U.
Define FV = FU . Then V0 = {V̂ \ FV × {1} : FV ∈ [X]

<ω,V ∈ V
′

} is also
a locally finite family of X ×ad 2 and it is an open refinement of U ∩ B0. Let
V1 = {{(x, 1)} : (x, 1) 6∈ ∪V0}. Then V1 is a discrete family of open sets, since
X × {0} ⊆ ∪V0. Thus V = V0 ∪ V1 is an open refinement of U which is locally
finite. �


The Alexandroff Duplicate and its subspaces 195

Proposition 2.8. If X is an hereditarily paracompact space, then also X ×ad 2
is an hereditarily paracompact space.

Proof. It is sufficient to show that any open subspace of X×ad 2 is paracompact
(see [4]). Let A be an open subspace of X ×ad 2 and U an open cover of A,
without loss of generality, assume that U ⊂ B. Then U = U0 ∪ U1 where
U0 = {Û \ (FU × {1}) : U ∈ U

′

∧ FU ∈ [X]
<ω} and U

′

is an open covering
of π0(A ∩ X × {0}), and U1 ⊆ {{(x, 1)} : (x, 1) ∈ A}. Since X is hereditarily

paracompact, there exists V
′

open refinement of U
′

which is locally finite at
the points of π0(A ∩ X × {0}) . Now, for each V ∈ V

′

, choose U ∈ U
′

such

that V ⊆ U and define FV = FU . Let V0 = {V̂ \ (FV × {1}) : V ∈ V
′

} and
V1 = {{(x, 1)} : (x, 1) 6∈ ∪V0} ∩ U1. Then V1 is a discrete family of open sets
and V = V0 ∪ V1 is an open refinement of U which is locally finite. Then A is
paracompact in X ×ad 2. �

Corollary 2.9 (Alexandroff, Urysohn [1]). [0, 1]×ad 2 is a first countable com-
pact Hausdorff, hereditarily paracompact space which has an uncountable dis-
joint family of open sets.

From the definition of Menger-Urysohn dimension, it follows that indX ≤ n
with n ≥ 0 if and only if there exists a base B of X such that ind∂B ≤ n− 1 for
each B ∈ B. In particular indX = 0 if and only if the space is 0-dimensional
and for every subspace Y ⊂ X, and we have that indY ≤ indX.
In order to calculate the small inductive dimension of the Alexandroff duplicate
and its subspaces, we need the following lemmas.

Lemma 2.10. Let (X,τ) be a topological space, and U,V ∈ τ. Then

(i) V̂ ⊆ V̂ and V̂ \ V̂ = ∂V × {1}

(ii) If V ⊆ U, then V̂ ⊆ Û

Proof. Observe that V × {0} = V × {0} and V × {1} = V × {1} ∪ V
′

× {0}.

Then V̂ = V × {0} ∪ V × {1} = V × {0} ∪ V × {1} ⊆ V × {0} ∪ V × {1} = V̂ .

Moreover V̂ \V̂ = (V ×{0}∪V ×{1})\(V ×{0}∪V × {1}) = V ×{1}\V ×{1} =
∂V × {1}.

If V ⊆ U, by (i), V̂ ⊆ V̂ ⊆ Û. �

Lemma 2.11. Let Z ⊆ X ×ad 2. If W ⊆ Z, then

clZ (W) ∩ (A × {1}) = W ∩ (A × {1}) = intZ (W) ∩ (A × {1})

Proof. For each (x, 1) ∈ A×{1}, (x, 1) ∈ clZ (W) if and only if (x, 1) ∈ W , and
(x, 1) ∈ intZ (W) if and only if (x, 1) ∈ W . �

Lemma 2.12. Let Z ⊆ X ×ad 2 and D ⊆ B. Then

(i) clZ (D̂) ∩ (B × {0}) = clB(D) × {0};

(ii) intZ (D̂) ∩ (B × {0})= intB(D) × {0};

(iii) ∂Z (D̂ ∩ (A × {1})) = ((D ∩ A) × {0})
′

;

(iv) clZ (D̂ ∩ Z) = (clB(D) × {0}) ∪ (D ∩ (A × {1}));


196 A. Caserta and S. Watson

(v) intZ (D̂ ∩ Z) = (intB(D) × {0}) ∪ (D ∩ (A × {1}));

(vi) ∂B×{0}(D × {0}) = ∂Z (D̂ ∩ Z).

Proof. (i). The point (x, 0) belongs to clZ (D̂)∩(B×{0}) if and only if for each

U∗ ∈ τZ (x, 0), U
∗ = (Û \{(x, 1)})∩Z , U∗∩(D̂) 6= ∅ hence U∗∩(D×{0}) 6= ∅.

Thus (x, 0) ∈ clZ (D × {0}) ∩ B × {0} = clB(D) × {0}.
(ii). Clearly if (x, 0) ∈ intX (D ∩ B) × {0} = intB(D) × {0}, then (x, 0) ∈

intZ (D̂). Moreover, if (x, 0) ∈ intZ (D̂), there exists U
∗ ∈ τZ (x, 0), U

∗ =

(Û \ {(x, 1)}) ∩ Z, U∗ ⊆ D̂, so (U ∩ B) × {0} ⊆ D × {0}. Hence (x, 0) ∈
intX ((D ∩ B) × {0}) = intB (D) × {0}.

(iii). Since D̂ ∩ (A × {1}) is an open subset of Z, ∂Z (D̂ ∩ A × {1}) =

D̂ ∩ (A × {1}) \ (D̂ ∩ (A × {1})). Moreover, for each (x, 1) ∈ Z \ (D̂ ∩ A ×

{1}), (x, 1) /∈ D̂ ∩ (A × {1}), thus for each (x, 1) ∈ Z \ (D̂ ∩ (A × {1})),

(x, 1) /∈ ∂Z (D̂ ∩ (A × {1})). For each (x, 0) ∈ Z \ (D̂ ∩ (A × {1})), (x, 0) ∈

(D̂ ∩ (A × {1})) if and only if (x, 0) ∈ (D ∩ A) × {1} ∪ ((D ∩ A) × {0})
′

, since

(D̂ ∩ (A × {1})) = (D∩A)×{1}∪((D∩A)×{0})
′

. Then ∂Z (D̂ ∩ (A × {1})) =

((D ∩ A) × {0})
′

.

(iv). From (i), it follows that clZ (D̂ ∩ Z) ⊆ clB (D) × {0} ∪ (D ∩ (A × {1})).

On the other hand, clZ (D̂∩Z) ⊇ clZ (D×{0}) = clB(D)×{0} and clZ (D̂∩Z) ⊇

(D̂ ∩ Z) ⊇ (D ∩ (A × {1})).

(v). From (ii), it follows that intZ (D̂∩Z) ⊆ intB (D)×{0}∪(D∩(A×{1})).

On the other hand, we have that intZ (D̂ ∩ Z) ⊇ intZ (D̂ ∩ Z) ∩ (A × {1}) =

(D∩A)×{1} and intZ (D̂∩Z) ⊇ intZ (întB∩Z) = întB∩Z ⊆ (întB∩Z)∩(B×{0})
= intB(D) × {0}. �

Theorem 2.13. Let Z ⊆ X ×ad 2 with B 6= ∅, then indZ = indB.

Proof. Since B ≃ B ×{0} ⊂ Z, then indB ≤ indZ. It is sufficient to show that
indZ ≤ indB, using induction on indB.

For indB = 0, there exists a base G for B, such that for each G ∈ G, G is
clopen in B, then for each G ∈ G, also Ĝ ∩ Z is clopen in Z. Now, for each
G ∈ G and F ∈ [A]<ω, (Ĝ \ F × {1}) ∩ Z is a basic open sets in Z. They are
also a clopen sets, because the boundaries never contain isolated points. Then
{(a, 1)}a∈A ∪ {(Ĝ \ F × {1}) ∩ Z : F ∈ [A]

<ω,G ∈ G} is a base of clopen sets
for Z and indZ = 0.

Let n ∈ ω and indB = n, we want to prove that indZ ≤ n. Since indB = n
there exists a base G for B, such that for each G ∈ G, ind∂BG ≤ n − 1.
Thus ∂B×{0}G × {0} ≤ n − 1. Let (Ĝ \ F × {1}) ∩ Z the corresponding basic

open set in Z, then from Lemma 2.12 we have that ∂Z ((Ĝ \ F × {1}) ∩ Z) =

∂Z (Ĝ ∩ Z) = ∂B×{0}G × {0}, then it follows that ind∂Z ((Ĝ \ F × {1}) ∩ Z) =
ind∂B×{0}G × {0} ≤ n − 1. Thus indZ ≤ n − 1. �


The Alexandroff Duplicate and its subspaces 197

We conclude this section analyzing the Baire property of the Alexandroff
duplicate and its subspaces. We first characterize dense sets of subspaces of
the Alexandroff duplicate.

Lemma 2.14. A topological space (X,τ) has a dense set of isolated points,
if and only if any {Di}i∈I arbitrary family of dense subsets of X is such that⋂

i∈I Di is dense too.

Proof. Let D a dense set in X consisting of isolated points. Notice that if Di is
a dense set in X, then D ⊆ Di, otherwise there exist x ∈ D \ Di and {x} is an
open set which does not intersect Di, contradiction. Therefore D ⊆

⋂
i∈I Di.

Viceversa, if X does not have a dense set of isolated points, then there exists
an open set U such that U has no isolated point. Now, for each x ∈ U, X \ {x}
is dense, so

⋂
x∈U X \ {x} is dense too. We have a contradiction, because

X \ U =
⋂

x∈U X \ {x} is not dense in X. �

Lemma 2.15. Let (X,τ) be a topological space. If {Di}i∈I is an arbitrary
family of dense subsets of X ×ad 2, then

⋂
i∈I Di is dense too.

Proof. The set X × {1} ⊆ X ×ad 2 is a dense set of isolated points. The result
follows from Lemma 2.14. �

Corollary 2.16. X ×ad 2 is Baire for each topological space X.

Lemma 2.17. Let Z ⊆ X ×ad 2. An open set U is a dense subset of Z if and
only if there exists D dense open in B \ A

′

and U = (A × {1}) ∪ (D × {0}).

Proof. Let U = (A × {1}) ∪ (D × {0}) be a subset of Z. Since D is dense

in B \ A
′

, we have D × {0} = D × {0} = (B \ A
′

) × {0}. Then U =

(A × {1}) ∪ (D × {0}) = (A×{1}) ∪D × {0} = A×{1}∪ (B ∩A
′

) ×{0}∪ (B \

A
′

) × {0} = Z. Since B \A
′

is an open set we have that U is an open set in Z.
Let U be a dense open set in Z, then A×{1} ⊆ U. Assume by contradiction

that for every dense open in B\A
′

we have that either U ( (A×{1})∪(D×{0})

or U ) (A × {1}) ∪ (D × {0}). Let D = U ∩ (B \ A
′

). Since (B \ A
′

) × {0} is

open in Z, i.e., D is dense open in B \ A
′

. Then V = (A × {1}) ∪ (D × {0})
is dense open set in Z such that V ⊆ U. If U ) (A × {1}) ∪ (D × {0}) then
V ) U which is a contradiction. If U ( (A × {1}) ∪ (D × {0}) we have that
U ( V ⊆ U, a contradiction. �

Lemma 2.18. Let {Dα}α∈κ be a family of dense open sets in B \ A
′

. Then⋂
α∈κ A × {1} ∪ Dα × {0} is dense in Z if and only if

⋂
α∈κ Dα is dense in

B \ A
′

.

Proof.
⋂

α∈κ(A × {1}) ∪ (Dα × {0}) = Z if and only if A × {1}∪
⋂

α∈κ Dα × {0} ⊇

Z if and only if A × {1} ∪ (B ∩ A
′

) × {0} ∪
⋂

α∈κ Dα × {0} ⊇ Z if and only if⋂
α∈κ Dα × {0} ⊇ (B \ A

′

) × {0}, i.e.,
⋂

α∈κ Dα is dense in B \ A
′

. �

Definition 2.19. A topological space X is κ-Baire if the intersection of less
then κ dense open sets is dense; in this case, we write Baire(X) = κ. Further,


198 A. Caserta and S. Watson

we set Baire(X) = ∞ if for all dense open sets their intersection is dense, and
assume that Baire(∅) = ∞.

Theorem 2.20. Let Z ⊆ X ×ad 2. Then Baire(Z)= Baire(B \ A
′

).

Proof. To prove the equality, we first observe that if (B \ A
′

) × {0} = ∅, then

A
′

⊇ B, i.e., A × {1} = Z. Thus Z has a dense set of isolated points and

so Baire(Z) = Baire((B \ A
′

) × {0}) = ∞. Assume that (B \ A
′

) × {0} 6=
∅ and Baire(Z) = κ. Since Z has κ many dense open sets with non-dense
intersection, by Lemma 2.17, without loss of generality, dense open sets have
form A×{1}∪Dα with α ∈ κ, and Dα‘s are dense open sets in (B\A

′

)×{0}. By

Lemma 2.17,
⋂

α∈k Dα × {0} is not dense in (B \ A
′

) × {0}. So (B \ A
′

) × {0}
has κ many dense open sets with non-dense intersection. Viceversa, assume
that Baire((B \A

′

) ×{0}) = κ, so B \A
′

has κ many dense open sets {Dα}α∈k
with non-dense intersection. By Lemma 2.17 and Lemma 2.18, {A×{1}∪Dα ×
{0}}α∈κ is a family of dense open sets with non-dense intersection. Thus Z has
κ many dense open sets with non-dense intersection, i.e., Baire(Z) ≤ κ. �

3. Metrizability and Lindelöf property

Before extracting the Michael line and its relatives from [0, 1]×ad 2, we prove
some useful characterizations of those subspaces of X ×ad 2 that are metrizable
or Lindelöf.

Definition 3.1. Let (X,τ) be a topological space and A and B subsets of X.
A is κ-discrete in A∪B, if A is the union of κ many sets with no accumulation
points in B.

Lemma 3.2. Let Z ⊆ X ×ad 2. If every closed set of Z is Gκ, then A is
κ-discrete in A ∪ B.

Proof. Since A × {1} is open set in Z, then it is an Fκ set. Let A = ∪{Kα :
α ∈ κ} where each Kα × {1} is closed in Z. By contradiction, if Kα had an
accumulation point b ∈ B, then Kα × {1} would have an accumulation point
at (b, 0) ∈ Z. �

Lemma 3.3. Let Z ⊆ X ×ad 2 and K ⊆ A such that K has no accumulation
points in B. Then K × {1} is closed in Z.

Proof. Since K has no accumulation points in B, for each b ∈ B there exists
U ∈ τX (b) such that U ∩ K ⊆ {b}. For each (b, 0) ∈ B × {0} there exists

UZ = (Û \ {(b, 1)}) ∩ Z such that UZ ∩ (K × {1}) = ∅. Then K × {1} has no
accumulation points in B×{0}. On the other hand, since every point in A×{1}
is an isolated point, K ×{1} cannot have accumulation points in A×{1}. Thus
K × {1} has no accumulation points, so K × {1} is closed in Z. �

Definition 3.4. Let (X,τ) be a topological space and A ⊆ X. We say that B
is a base for the points of A in X, if each B ∈ B is an open set in X and for
each x ∈ A and U ∈ τX (x) there exist B ∈ B such that x ∈ B ⊆ U.


The Alexandroff Duplicate and its subspaces 199

Theorem 3.5. Let Z ⊆ X ×ad 2 and κ be an infinite cardinal. Then Z has a
κ-discrete base if and only if B has a κ-discrete base in X and A is κ- discrete
in A ∪ B.
Furthermore we have:

(i) If B is a κ-discrete base for B in X and A =
⋃

β<κ
Kβ with each Kβ

having no accumulation points in B, then B∗ = {V̂ \(
⋃

β∈F Kβ ×{1}) :

F ∈ [κ]<ω,V ∈ B} is a κ-discrete base for B × {0} in Z.
(ii) If B∗ =

⋃
γ∈κ Bγ

∗ is a κ discrete base for Z, then BB×{0} =
⋃

γ∈κ{U ∩

(B × {0}) : U ∈ B∗γ} is a κ-discrete base for B × {0} in Z and A =⋃
γ∈κ{a : {(a, 1)} ∈ B

∗
γ}.

Proof. Assume that B is a κ-discrete base for B in X and A =
⋃

α∈κ Kα where
each Kα has no accumulation points in B. Then Kα = {{(k, 1)} : k ∈ Kα} is
a discrete family of open sets in Z and K = ∪{Kα : α ∈ κ} is a κ-discrete open
family which is a base for A × {1}.

Next we want to find a κ-discrete base for B×{0} in Z. Let (b, 0) ∈ B×{0},

and (Û \ {(b, 1)}) ∩Z a neighborhood of (b, 0) in Z. If b ∈ A, there exists γ ∈ κ
such that (b, 1) ∈ Kγ ×{1} which is closed in Z by Lemma 3.3. Moreover there

exists V ∈ B such that b ∈ V ⊆ U. Let F = {γ} hence (b, 0) ∈ V̂ \ (
⋃

β∈F Kβ ×

{1}) ∩ Z ⊆ V̂ \ {(b, 1)} ∩ Z ⊆ Û \ {(b, 1)} ∩ Z. Then B∗ is a base for B × {0}
in Z. Since B is a κ-discrete base for B, we have that B =

⋃
γ∈κ Bγ . Then

B∗ =
⋃

γ∈κ Bγ
∗ where Bγ

∗ = {V̂ \ (
⋃

β∈F Kβ × {1}) : F ∈ [κ]
<ω,V ∈ Bγ}. It

remains to prove that for each γ ∈ κ, Bγ
∗ is a κ-discrete family in Z. Since for

every γ ∈ κ, Bγ is discrete family, then {V̂ : V ∈ Bγ} is discrete too. Let γ ∈ κ

and F ∈ [κ]<ω be fixed, then{V̂ \ (
⋃

ξ∈F Kξ × {1}) : V ∈ Bγ} ⊆ {V̂ : V ∈ Bγ}

is a discrete family. Hence B∗ =
⋃

γ∈κ(
⋃
{V̂ \(

⋃
ξ∈F Kξ ×{1}) : F ∈ [κ]

<ω,V ∈

Bγ}) is κ-discrete too. Further, B
∗ ∪ K is a κ-discrete base for Z.

Assume now that Z has a κ-discrete base B∗ =
⋃

γ∈κ B
∗
γ . Set BB×{0} =⋃

γ∈κ{U ∩ (B × {0}) : U ∈ B
∗
γ}. Then BB×{0} is a κ-discrete base for B × {0}

in Z. Since A × {1} is a set of isolated points in Z, then for each a ∈ A, the
set {(a, 1)} ∈ B∗. Hence Kγ = {a : {(a, 1)} ∈ B

∗
γ} is a discrete family of points

such that A =
⋃

γ<κ
Kγ . Clearly every Kγ has no accumulation points in B,

otherwise Kγ would not be discrete. �

Definition 3.6. Let (X,τ) be a regular topological space and A ⊆ X. We say
that A is metrizable in X if there exists a σ-discrete open family in X which
is a base for the points of A in X.

Corollary 3.7. Let X be a regular space. The subspace Z ⊆ X ×ad 2 is
metrizable if and only if B has a countable base in X, and A is the union of
countable many sets with no accumulation points in B.

Proof. Since X ×ad 2 is regular, it follows from Theorem 3.5. �

In the following we characterize the Lindelöf property of a subspace Z ⊆
X ×ad 2. First we need the following lemma.


200 A. Caserta and S. Watson

Lemma 3.8. If Z ⊆ X ×ad 2 is Lindelöf, then every uncountable subset K of
A, with cf|K| > ℵ0, has a complete accumulation point in B.

Proof. Assume that Z is Lindelöf . By contradiction, assume that there exists
K ⊂ A such that cf|K| > ℵ0 and for each b ∈ B there exists Ub ∈ τX containing

b such that |Ub ∩ K| < |K|. Let Ub
∗ = Ûb \ {(b, 1)} ∩Z the corresponding open

set in Z. Then {Ub
∗}b∈B is an open cover of B × {0}, which is a closed subset

of a Lindelöf space. By Lindelöfness of B × {0} we can find a countable open
subcover {Ubi

∗}i∈ω of B × {0}. Let U
∗ =

⋃
i∈ω Ubi

∗. Then B × {0} ⊂ U∗ and

|U∗ ∩ K × {1}| = |
⋃

i∈ω

(Ubi
∗ ∩ K × {1})| ≤ Σi∈ω|Ubi

∗ ∩ K × {1}|.

Since |Ubi
∗∩K×{1}| < |K| and cf|K| > ℵ0, it follows that |U

∗∩K×{1}| < |K|.

Now, denote Ã = {a ∈ K : (a, 1) /∈ U∗} and let U = {U∗} ∪ {(a, 1) : a ∈ Ã}

be a cover of Z. Since {(a, 1) : a ∈ Ã} contain an uncountable subset of
isolated points we cannot find a countable subcover of U in contradiction with
the Lindelöfness of Z . �

Theorem 3.9. Let Z ⊆ X ×ad 2. The following statements are equivalent:

(i) Z is Lindelöf;
(ii) B is Lindelöf and every uncountable subset K of A, with cf|K| > ℵ0,

has a complete accumulation point in B;
(iii) B is Lindelöf and every uncountable subset K of A, with |K| regular

cardinal has a complete accumulation point in B;
(iv) B is Lindelöf and every uncountable subset K of A, with |K| = ℵ1 has

a complete accumulation point in B;
(v) B is Lindelöf and every uncountable subset K of A has an accumulation

point in B;
(vi) B is Lindelöf and for every open subset U in X containing B, we have

|A \ U| ≤ ℵ0;
(vii) B is Lindelöf and for every closed subset F in X that misses B, we

have |A ∩ F | ≤ ℵ0.

Proof. Assume that (i) holds. Since B × {0} is closed subspace of Z, it follows
that B × {0} and thus B is Lindelöf. Hence (ii) follows from Lemma 3.8.

The implications (ii) ⇒ (iii) and (iii) ⇒ (iv) are immediate.
For (iv) ⇒ (v), let K be uncountable subset of A. For |K| = ℵ1 is immediate.

Assume that |K| > ℵ1. Assume that K has no accumulation points in B. Then

there exists K
′

⊂ K such that |K
′

| = ℵ1 with no complete accumulation points
in B, contradiction.

Now, assume that (v) holds, we want to prove (i). Let U be an open cover
of Z, and assume, without loss of generality, that it consists of basic open sets.
Since B × {0} is Lindelöf, we can cover it with countably many open sets from
the cover U. Denote such a cover UB×{0}. Let K ⊆ A such that K × {1} =
Z \

⋃
UB×{0} where

⋃
UB×{0} ⊇ B ×{0}. We claim that K ×{1}∪UB×{0} is a

countable subcover of U for Z. Indeed, K × {1} ∩ (
⋃
UB×{0}) = ∅, so K × {1}


The Alexandroff Duplicate and its subspaces 201

has no accumulation points in B × {0} and so K has no accumulation points
in B.

For (i) ⇒ (vi), assume that Z is Lindelöf. Let U open set containing B. If

|A \ U| > ℵ0, let UB = (Û ∩ Z) ⊇ B × {0} be an open set in Z. Then UB with
the points of (A × {1}) \ UB has no countable subcover, a contradiction.

Assume that (vi) holds, we want to prove (v). Let K be uncountable subset
of A with no accumulation points in B, then there exists an open set U ⊃
B × {0} which misses K × {1}. Then |A \ U| > ℵ0.

The equivalence between (vi) and (vii) is immediate. �

Corollary 3.10. Let X be hereditarily Lindelöf and Z ⊆ X ×ad 2. The follow-
ing statements are equivalent:

(i) Z is Lindelöf;
(ii) every uncountable subset K of A, with cf|K| > ℵ0, has a complete

accumulation point in B;
(iii) every uncountable subset K of A, with |K| regular cardinal has a com-

plete accumulation point in B;
(iv) every uncountable subset K of A, with |K| = ℵ1 has a complete accu-

mulation point in B;
(v) every uncountable subset K of A has an accumulation point in B;
(vi) for every open subset U in X containing B, we have |A \ U| ≤ ℵ0
(vii) for every closed subset F in X that misses B, we have |A ∩ F | ≤ ℵ0.

Corollary 3.11. Let Z ⊆ X ×ad 2. If

(i) B is Lindelöf and
(ii) for every closed subset F of X that misses B, we have |A ∩ F | ≤ ℵ0,

then Z is Lindelöf.
Moreover, under the additional assumption that A is uncountable, Z is not
metrizable.

Proof. From Theorem 3.9 it follows that Z is Lindelöf, and since A is uncount-
able, any uncountable subset of A has accumulation points in B. Hence A is not
σ-discrete in A∪B. By Proposition 3.5, it follows that Z is not metrizable. �

The following example shows that the restriction on the cofinality is needed.

Example 3.12 (Watson [15]). Let X = [0, 1] and assume that 2ℵ0 > ℵω.
Consider subsets of [0, 1], An’s, such that |An| = ℵn and An ⊂ [

1
n+1

, 1
n
]. Take

A =
⋃

n∈ω An, B = (0, 1] and Z = (A × {1}) ∪ (B × {0}) ⊂ [0, 1] ×ad 2. Let

Zn = Z ∩ ([
1
n
, 1] ×{2}). Then for every n ∈ ω, Zn is compact, so Z =

⋃
n∈ω Zn

is σ-compact, thus Lindelöf. However A has no complete accumulation points
in B, but every uncountable subset of A has accumulation points in B.

Corollary 3.13 (Michael, Corson [3] (see olso [8])). Z ⊆ [0, 1]×ad2 is Lindelöf
if and only if every uncountable subset of A has an accumulation point in B.

The Michael line and its relatives are subspaces of [0, 1] ×ad 2 of a certain
kind:


202 A. Caserta and S. Watson

Definition 3.14. If Z ⊆ X ×ad 2 where A and B are disjoint, then we say
that Z is a Michael-type line. Assume X = [0, 1] unless stated otherwise.

Corollary 3.7 gives us a necessary and sufficient condition for the metriz-
ability of a subspace of Alexandroff Duplicate. In the next result we give a
different condition for metrizability, which takes into account only the size of
the subspace A.

Proposition 3.15. Under MAℵ1 , if A,B ⊂ [0, 1] such that |A| ≤ ℵ1, then the
Michael-type line is metrizable.

Proof. MAℵ1 implies that every subset of the reals of cardinality at most ℵ1 is
a Q-set. Thus A is the union of countably many subsets closed in A ∪ B, i.e.,
A =

⋃
n∈ω Cn. Since A ∩ B = ∅ and Cn’s are closed in A ∪ B, we have that

for each n ∈ ω, Cn has no accumulation point in B. By Corollary 3.7 it follows
that Z is metrizable. �

Example 3.16 (Michael, Corson [12]). Let A = P and B = Q. Call this
Michael-type line the Michael line LNmic.

Now we show some property of the Michael-type line LNmic.

Lemma 3.17. Let Z ⊆ X ×ad 2. If

(i) X hereditarily paracompact,
(ii) indB = 0,
(iii) A is not σ-discrete in A ∪ B,

then Z is zero-dimensional, hereditarily paracompact but not metrizable.

Proof. Since indB = 0 applying Theorem 2.13, follow that Z is zero-dimensional.
Moreover A is not a countable union of sets with no accumulation points in B.
By Proposition 3.5, follow that Z is not metrizable. From Proposition 2.8 it
follows that Z is hereditarily paracompact. �

Corollary 3.18. LNmic is zero-dimensional, hereditarily paracompact but not
metrizable.

Note that Corollary 1.20 can be restate as follows:

Corollary 3.19 (Wille [16]). LNmic cannot be retracted onto Q × {0}.

Example 3.20 (Michael [12]). Choose a Bernstein partition [0, 1] = A ∪ B
(that is, neither A nor B contains an uncountable compact subset). Call this
Michael-type line LNmcb, the Michael-Bernstein line

Corollary 3.21 (see also Tanaka [13]). LNmcb is Lindelöf space but not metriz-
able.

Proof. Let F be a closed subset of X such that F ∩ B = ∅. Since F is a
compact contained in A, then |F | ≤ ℵ0. Lemma 3.11 ends the proof. �

Corollary 3.22 (Dow [5]). Under MAℵ1 , LN
mcb is a Lindelöf first countable

space which is not metrizable but all of whose subspaces of cardinality ℵ1 are
metrizable.


The Alexandroff Duplicate and its subspaces 203

Proof. Apply Proposition 2.4, Corollary 3.15 and Corollary 3.21. �

Next we provide a sufficient condition for Z ⊆ X ×ad 2 to have a product
with the irrationals that fails to be normal. In the following, unless stated
otherwise, we assume that the X is Tychonoff.

Proposition 3.23. Let A, B and C subsets of a topological space X. If

(i) C is separable,
(ii) A ⊆ C and B ∩ C = ∅,
(iii) A is not union of countably many subsets with no accumulation points

in B,

then Z not metrizable, and the product of the Michael-type line Z with C is
not normal.

Proof. Let K = {((a, 1),a) : a ∈ A} and L = (B×{0})×C be subsets in Z×C.
First of all we prove that Kc is an open set in Z ×C. Indeed, let b ∈ B, c ∈ C,
and ((b, 0),c) ∈ Kc. Since C ∩ B = ∅, then b 6= c, and so, there exists U and

V disjoint open sets in X such that b ∈ U and c ∈ V . Hence (Û ∩Z) × (V ∩C)
is an open set in Z × C contained in Kc. Now, let a ∈ A, c ∈ C \ {a}, and
((a, 1),c) ∈ Kc. Then there exists U ∈ τX such that c ∈ U and a 6∈ U. Then
{(a, 1)} × (U ∩ C) is an open set in Z × C containing ((a, 1),c) and contained
in Kc. Thus K is closed in Z × C, K and L are disjoint closed sets in Z × C.
Suppose these sets were separated by disjoint open sets U and V respectively.
Each element ((a, 1),a) has a neighborhood of the form {(a, 1)} × Ua ⊆ U
with Ua open set in C containing a. Let D be the countable dense set in
C. Take da ∈ Ua ∩ D. Now, A is not the union of countably many subsets
with no accumulation points in B, then we can find a net {aσ : σ ∈ Σ} ⊂ A
accumulating to b0 ∈ B and d ∈ D so that daσ = d for every σ ∈ Σ. Then
((aσ, 1),d) accumulates to ((b0, 0),d) with ((aσ, 1),d) ∈ U and ((b0, 0),d) ∈ V .
Thus we have found ((b0, 0),d) ∈ U ∩ V which is impossible. �

Corollary 3.24. Let A, B and C be subsets of a topological space X. If

(i) C is separable,
(ii) A ⊆ C and B ∩ C = ∅,
(iii) every uncountable subset K of A has an accumulation point in B,
(iv) A is uncountable
(v) B is Lindelöf,

then the Michael-type line Z is Lindelöf, not metrizable, and Z × C is not
normal.

Proof. It follows directly from Theorem 3.9 and Proposition 3.23. �

Corollary 3.25. Let A and B be subsets of a topological space X. If X \ B is
separable and every uncountable subset K of A has an accumulation point in
B, then Z × (X \ B) is not normal.

Moreover, if X is compact, A is uncountable and X \B Čech-complete, then
the Michael-type line Z is also Lindelöf and not metrizable.


204 A. Caserta and S. Watson

Proof. Since X\B is a Čech-complete in X, it is a Gδ in X. Thus B is an Fσ of a
compact space, hence Lindelöf. The statement follows from Corollary 3.24. �

Corollary 3.26 (based on Michael [11]). Let A and B be subsets of [0, 1]. If
A is contained in a copy of the irrationals disjoint from B, and A is not the
union of countably many subsets with no accumulation points in B, then the
product of the Michael-type line with the irrational is not normal.

Corollary 3.27 (Michael [12]). LNmic × P is not normal.

Corollary 3.28 (Michael [12]). LNmcb × A is not normal.

In order to give another example of a Lindelöf space whose product with the
irrationals is not normal, we first recall another class of Michael-type lines.

Example 3.29. (Michael [12]; Burke, Davis [2]; van Douwen [14];
Lawrence [10]) Under b = ω1, let A ⊂

ωω be unbounded and well-ordered
in type ω1 by <

∗. Let B = Q. Call the Michael-type line LNmbd.

Lemma 3.30. Under b = ω1, let C =
ωω ⊆ X with X compact and A ⊆ ωω

be unbounded and well-ordered in type ω1 by <
∗, B = X \ C. Then every

uncountable subset of A has accumulation points in X \ C.

Proof. Let f be the standard homeomorphism between P and ωω. The set A is
a Fσ in A∪(X\C) ⊂ X if and only if A ⊂

⋃
n∈ω Kn where Kn’s are closed in X

and (
⋃

n∈ω Kn)∩(X \C) = ∅, i.e., Kn is a compact subset of P for each n ∈ ω.
Since every compact subset of ωω is bounded in ωω, then for each Kn ⊂ P,
there exists fn ∈ ω

ω such that f(Kn) ≤
∗ fn. Since b = ω1, take g ∈ ω

ω such
that fn ≤

∗ g for each n ∈ ω . Then g bounds
⋃

n∈ω f(Kn) = f(
⋃

n∈ω Kn)
which contains f(A). Then f(A) ∼= A ⊂ ωω is bounded. Observe that any
uncountable K ⊂ A is unbounded and then suck K cannot be compact, i.e., K
cannot be closed in (X\C)∪K. Thus any uncountable K ⊂ A has accumulation
points in X \ C. �

Corollary 3.31. (Michael [12]; Burke, Davis [2]; van Douwen [14];
Lawrence [10]) LNmbd is a Lindelöf space whose product with the irrationals
is not normal.

Proof. Follows from Lemma 3.30, Corollary 3.25. �

Problem 3.32 (implicit in Michael [12]). Is there a Lindelöf space whose
product with the irrationals is not normal?

References

[1] P. S. Alexandroff and P. S. Urysohn, Mémoire sur les espaces topologiques compacts,
Verh. Akad. Wetensch. Amsterdam, 14 (1929).

[2] D. K. Burke and S. W. Davis, Subsets of ωω and generalized metric spaces, Pacific J.
Math. 110(2) (1984), 273–281.

[3] H. H. Corson and E. Michael, Metrizability of certain countable unions, Illinois J. Math.
8 (1964), 351–360.


The Alexandroff Duplicate and its subspaces 205

[4] J. Dieudonné, Une généralisation des espace compact, J. de Math. Pures et Appl. 23
(1944), 65–76.

[5] A. Dow, An empty class of nonmetric spaces, Proc. Amer. Math. Soc. 104(3) (1988),
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[6] R. Engelking, General Topology (Heldermann Verlag, Berlin 1989).
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and the irrationals, Proc. Amer. Math. Soc. 110(2) (1990), 535–542.
[11] E. Michael, Paracompactness and the Lindelöf property in finite and countable Cartesian

products, Compositio Math. 23(2) (1971), 199–214.
[12] E. Michael, The product of a normal space and a metric space need not be normal, Bull.

Amer. Math. Soc. 69 (1963), 375–376.
[13] Y. Tanaka, Decompositions of spaces determined by compact subsets, Proc. Amer. Math.

Soc. 97(3) (1986), 549–555.

[14] E. K. van Douwen, The integers and topology, In K. Kunen and J. Vaughan, (eds.),
Handbook of Set-Theoretic Topology 111–169 (North-Holland, Amsterdam, 1984).

[15] S. Watson, The Construction of Topological Spaces: Planks and Resolutions, In M.
Hus̄ek and J. van Mill (eds.), Recent Progress in General Topology, 673–757 ( North-
Holland 1992).

[16] R. J. Wille, Sur les espaces faiblement rétractiles, Ned. Akad. Weten. Proc. 57 (1954),
527–532.

Received November 2005

Accepted June 2006

A. Caserta (agata.caserta@unina2.it)
Dipartimento di Matematica, Seconda Università degli Studi di Napoli, Caserta
81100, Italia

S. Watson (watson@hilbert.math.yorku.ca)
Department of Mathematics and Statistics, York University, Toronto M3J1P3,
Canada