DruzhininaAGT.dvi


@
Applied General Topology

c© Universidad Politécnica de Valencia

Volume 7, No. 2, 2006

pp. 139-150

Connected metrizable subtopologies and
partitions into copies of the Cantor set

Irina Druzhinina

Abstract. We prove under Martin’s Axiom that every separa-

ble metrizable space represented as the union of less than 2
ω

zero-

dimensional compact subsets is zero-dimensional. On the other hand,

we show in ZF C that every separable completely metrizable space with-

out isolated points is the union of 2
ω

pairwise disjoint copies of the

Cantor set.

2000 AMS Classification: 54B15, 54C05, 54C10, 54D05, 54D30, 54E35,
54E50

Keywords: Metrizable space, Completely metrizable space, Condensation,
Connected metrizable subtopology, Cantor set, Zero-dimensional space, Mar-
tin’s Axiom

1. Introduction

We say that a space X condenses onto a space Y if there exists a continuous
bijection f : X → Y . A space X has a weaker connected topology (also called
connected subtopology) if X condenses onto a connected space. In the article [6]
on connected subtopologies, the following two results were proved:

(a) Let Cα be the Cantor set, for each α ∈ A, and X =
⊕

{Cα : α ∈ A} the
disjoint topological union of the spaces Cα. The space X has a weaker
T3 connected topology iff it has a weaker Tychonoff connected topology iff
|A| ≥ ω1.

(b) If Iα is the unit segment [0, 1], for each α ∈ A, and X =
⊕

{Iα : α ∈ A} is
the disjoint topological union of the spaces Iα, then space X has a weaker
T3 connected topology iff |A| ≥ ω.

In Section 3 of this article, we study the problem of when the spaces of
the form

⊕
{Cα : α ∈ A} or

⊕
{Iα : α ∈ A} admit connected metrizable

subtopologies. Here we mention three more results on connected subtopologies
presented in [4] that will be used in the sequel.



140 I. Druzhinina

Theorem 1.1. Let X be a metrizable space of weight κ ≤ 2ω. If there exists a
closed set P ⊆ X which admits a condensation onto a connected non-compact
metrizable space, then X condenses onto a connected separable metrizable space.

Theorem 1.2. Every metrizable space of weight 2ω admits a condensation
onto a connected separable metrizable space.

Theorem 1.3. Every metrizable space of weight κ ≥ 2ω with achievable extent
admits a weaker connected metrizable topology.

Recall that a metrizable space X has an achievable extent if X contains a
closed discrete subset of the size equal to the weight of X.

The authors of [6] rose the following problem:

Problem 1.4. Is it true in ZF C that the topological sum of ω1 copies of the
Cantor set condenses onto a connected compact space?

Delay and Just in [3] gave the negative answer to the above problem under
the assumption that the real line cannot be covered by ω1 nowhere dense sets.

In Section 2 of the article we show under M A that if separable metrizable
space X is a union of less than 2ω zero-dimensional compact subspaces, then
X is also zero-dimensional. In particular, under M A + ¬CH, all metrizable
subtopologies on the space

⊕
{Cα : α ∈ ω1} are zero-dimensional. This result

and Theorem 1.2 together imply that the existence of a connected metrizable
subtopology on the topological sum of ω1 copies of the Cantor set does not
depend on ZF C.

We also prove in Section 2 that the topological sum
⊕

{Iα : α ∈ A}, where
each Iα is a copy of the unit segment [0, 1], admits a connected metrizable
subtopology iff |A| ≥ ω.

In connection with Problem 1.4 we show in Section 3 that the topological sum
of 2ω copies of the Cantor set condenses onto the closed unit interval and, even
more, it condenses onto every compact metrizable space without isolated points.
Finally, we generalize the latter fact and prove that every separable complete
metrizable space without isolated points can be represented as a disjoint union
of 2ω copies of the Cantor set.

The reader can consult Kunen’s book [5] for details about Martin’s Axiom
(for short, M A).

2. Connected metrizable subtopologies

Let Cα be the Cantor set for each α < ω1 and let X =
⊕

{Cα : α < ω1}.
We consider the question whether X admits a connected metrizable subtopol-
ogy. Notice that under CH, the answer is affirmative, since one can apply
Theorem 1.2.

The next theorem, combined with the well-known fact that every metrizable
space of weight ≤ 2ω admits a separable metrizable subtopology (see Lemma
2.5 in [4]), implies that the answer to the question is “no” under M A + ¬CH.

In what follows the family of all finite subsets of a set A is denoted by [A]<ω.



Connected metrizable subtopologies 141

Theorem 2.1. Suppose that M A holds. Let X be a separable metrizable
space represented as X =

⋃
α<κ

Kα, where κ < 2
ω and every Kα is a zero-

dimensional compact subspace of X. Then X is also zero-dimensional.

Proof. Let B be a countable base for X. Pick a point x0 ∈ X and take any
element W ∈ B such that x0 ∈ W . We have to show that there exists a clopen
set U in X such that x0 ∈ U ⊆ W .

Notice that for every α < κ, the open subspace Kα ∩ W of Kα is σ-compact
since Kα is compact and metrizable. Hence, we can represent each Kα in the
form Kα =

⋃
n∈ω Kα,n ∪(Kα\W ), where the sets Kα,n are compact and satisfy

Kα,n ⊆ W . Therefore, we can assume from the very beginning that Kα ⊆ W
or Kα ∩ W = ∅ for each α < κ.

Let us consider the following family:

P =
{

(F, γ, λ) : F ∈ [κ]<ω, γ, λ ∈ [B]<ω, x0 ∈
⋃

γ,

⋃
γ ⊆ W,

⋃
γ ∩

⋃
λ = ∅,

⋃

α∈F

Kα ⊆ (
⋃

γ) ∪ (
⋃

λ)

}
.

It is easy to see that the family P is not empty. Now we introduce a partial
order ≤ in P by the following rule:

(F1, γ1, λ1) ≤ (F2, γ2, λ2) ⇐⇒ F2 ⊆ F1 & γ2 ⊆ γ1 & λ2 ⊆ λ1.

We claim that the poset (P, ≤) satisfies the countable chain condition. In-
deed, let Q ⊆ P be a subset of cardinality ℵ1. Since |B| = ℵ0 there exist
Q∗ ⊆ Q, γ∗ ∈ [B]<ω, and λ∗ ∈ [B]<ω such that |Q∗| = ℵ1 and all elements of Q

∗

have the form (F, γ∗, λ∗) with F ∈ [κ]<ω. Take two elements P1 = (F1, γ
∗, λ∗)

and P2 = (F2, γ
∗, λ∗) of Q∗. It is easy to see that P = (F1 ∪ F2, γ

∗, λ∗) is in P,
P ≤ P1 and P ≤ P2. This proves our claim. In fact, almost the same argument
shows that (P, ≤) is σ-centered.

For every α ∈ κ, put Dα = {(F, γ, λ) ∈ P : α ∈ F }. Let us verify that
Dα is dense in (P, ≤). Take an element P = (F, γ, λ) ∈ P. It suffices to
find an element P ∗ ∈ Dα such that P

∗ ≤ P . If Kα ⊆
⋃

γ ∪
⋃

λ, then P ∗ =

(F ∪{α}, γ, λ) ∈ Dα and, clearly, P
∗ ≤ P . Otherwise, we put T1 = Kα∩

⋃
γ and

T2 = Kα
⋂ ⋃

λ. Then T1 and T2 are disjoint, and we may assume that T1 6= ∅.
Then, clearly, Kα ⊆ W . Since Kα is zero-dimensional, there exist clopen sets
A and B in Kα such that A ∪ B = Kα, A ∩ B = ∅, and T1 ⊆ A, T2 ⊆ B.
Evidently, the sets A and B are closed in X. Therefore, the sets A ∪

⋃
γ and

B ∪
⋃

λ are closed in X and disjoint. We choose disjoint open sets O1 and O2
in X such that A ∪

⋃
γ ⊆ O1 and B ∪

⋃
λ ⊆ O2. In addition, we can take O1

with O1 ⊆ W , since A ∪
⋃

γ ⊆ W . Let γ′ be an open cover of A by elements

of the base B such that
⋃

γ′ ⊆ O1, and λ
′ an open cover of B by elements of

B such that
⋃

λ′ ⊆ O2. It is easy to see that P
∗ = (F ∪ {α}, γ ∪ γ′, λ ∪ λ′) is

an element of P. In addition, P ∗ ∈ Dα and P
∗ ≤ P . In the case when T1 = ∅

we take P ∗ = (F ∪ {α}, γ, λ ∪ λ′). Then again P ∗ ∈ Dα and P
∗ ≤ P .



142 I. Druzhinina

We have thus shown that Dα is dense in (P, ≤) for each α < κ. It is also
clear that the cardinality of the family D = {Dα : α ∈ κ} is not greater than
κ < 2ω. Hence, M A implies that there exists a D-generic filter G in (P, ≤),
that is, a filter G ⊆ P such that G ∩ Dα 6= ∅ for each α < ω1. Now we define
two open subsets U and V of X by

U =
⋃

{
⋃

γ : ∃ F ∈ [ κ]<ω ∃ λ ∈ [B]<ω such that (F, γ, λ) ∈ G },

V =
⋃

{
⋃

λ : ∃ F ∈ [ κ]<ω ∃ γ ∈ [B]<ω such that (F, γ, λ) ∈ G }.

Let us check that the sets U and V satisfy the following conditions:

(1) x0 ∈ U ⊆ W ;
(2) U ∩ V = ∅;
(3) U ∪ V = X.

Condition (1) follows from the definition of P. To show that (2) holds we
note that if P1 = (F1, γ1, λ1) ∈ G and P2 = (F2, γ2, λ2) ∈ G, then there exists
P = (F, γ, λ) ∈ G such that P ≤ P1 and P ≤ P2. Therefore, γ1 ∪ γ2 ⊆ γ

and λ1 ∪ λ2 ⊆ λ. Since
⋃

γ ∩
⋃

λ = ∅, we have that
⋃

γ1 ∩
⋃

λ2 = ∅ and⋃
γ2 ∩

⋃
λ1 = ∅. Hence,

⋃
γ ∩

⋃
λ = ∅ for arbitrary γ and λ that are used

to form U and V . To check (3) we note that for each α < κ, there exists
P = (F ∗, γ∗, λ∗) ∈ G ∩ Dα. Hence, α ∈ F

∗ and Kα ⊆ (
⋃

γ∗) ∪ (
⋃

λ∗). Since⋃
γ∗ ⊆ U and

⋃
λ∗ ⊆ V , we have Kα ⊆ U ∪ V . It follows from (1) and (3)

that V 6= ∅ and, hence, U and V are clopen sets in X. Thus, the clopen set
U ⊆ X satisfies x0 ∈ U ⊆ W . �

Corollary 2.2. The existence of a connected metrizable subtopology on the

space X =
⊕

{Cα : α ∈ ω1}, where each Cα is a copy of the Cantor set, does
not depend on ZF C.

In the second part of this section we study the problem whether there exists
a connected metrizable subtopology on the space X =

⊕
{Iα : α ∈ A}, where

each Iα is the unit segment.

Example 2.3. Let In be the unit segment [0, 1] for each n ∈ ω and X =⊕
{In : n ∈ ω}. Then the space X condenses onto a connected non-compact

subspace of the plane R2.

Proof. Let {rn : n ∈ N} be the set of rational numbers in [0, 1]. For each n ∈ N,
put

Jn = {(x, rn) ∈ R
2 : 1/n ≤ x ≤ 1}

and J0 = {0} × [0, 1]. Let us consider Y =
⋃
{Jn : n ∈ ω}. We claim that Y

is a connected subspace of R2. Indeed, let U be a clopen set of Y such that

J0 ⊆ U . Since J0 is compact, there exists ǫ > 0 such that if Ũ = {(x, y) ∈

R
2 : 0 ≤ x < ǫ, 0 ≤ y ≤ 1}, then J0 ⊆ Ũ ∩ Y ⊆ U . Take n0 ∈ N such that

1/n0 < ǫ. Then for every n ≥ n0, cn = (1/n, rn) ∈ Ũ ∩ Y ⊆ U . Since cn ∈ Jn,
Jn is connected, and U is a clopen set in Y , we conclude that Jn ⊆ U for every
n ≥ n0.



Connected metrizable subtopologies 143

Take m < n0 and let V be an open neighborhood of the point cm =
(1/m, rm) in Y . Evidently,

|V ∩ {(1/m, rn) : n ∈ N}| = ℵ0.

Therefore, there exists k > n0 such that (1/m, rk) ∈ Jk ∩ V ⊆ U ∩ V 6= ∅.
It follows that cm = (1/m, rm) ∈ clY (U ). Since cm ∈ Jm, we conclude that
Jm ⊆ U for every m < n0. Thus, U = Y and this proves that Y is connected.

For each n ∈ ω, take a continuous bijection fn : In → Jn. Then the sum of
the functions fn, say,

f = ∇n∈ωfn : X → Y

is a continuous bijection (see Proposition 2.1.11 in [2]). Hence, f is a conden-
sation of X onto the connected separable metrizable space Y ⊆ R2. Finally, Y
is not closed in R2 and is not compact. �

Corollary 2.4. Let X =
⊕

{Iα : α ∈ A}, where each Iα is the unit segment
[0, 1]. Then the space X admits a connected metrizable subtopology if and only
if |A | ≤ 1 or |A | ≥ ω.

Proof. If 1 < |A | < ω, then X is a disconnected compact space and, hence, it
does not admit a connected Hausdorff subtopology.

If |A | = ω, it follows from Example 2.3 that X has a connected separable
metrizable subtopology.

If ω < |A | ≤ 2ω, then X ⊇ P =
⊕

{In : n ∈ ω}. The set P is closed in X and
condenses onto a connected non-compact metrizable space. Apply Theorem 1.1
to conclude that X admits a connected separable metrizable subtopology.

Let |A | > 2ω. In each Iα, take a point xα. Then P = {xα : α ∈ A} is a
closed discrete set in X whose size equal to the weight of X, that is, X has
achievable extent. It follows from Theorem 1.3 that X admits a connected
metrizable subtopology. �

3. Condensations of the disjoint topological union of 2ω copies
of the Cantor set

In Theorem 3.3 we prove that the topological sum of 2ω copies of the Cantor
set condenses onto the unit interval I = [0, 1]. Then we extend this fact to
compact metrizable spaces without isolated points (Theorem 3.8). We finish
with Theorem 3.10 that shows that the conclusion is valid for every separable
complete metrizable space without isolated points.

We will use the following notation. The Cantor set is C = {0, 1}ω. If A ⊆ ω,
then πA : C → {0, 1}

A is the projection and, for n ∈ ω, πn is the projection of
C onto the nth factor. If A ⊆ B ⊆ ω, then πBA is the projection of {0, 1}

B onto
{0, 1}A. For an open canonical set U ⊆ {0, 1}A, we put

coord(U ) = {i ∈ A : |πi(U )| = 1}.

If A ⊆ B ⊆ ω and f ∈ {0, 1}A, then the set O(f ) = (πBA )
−1(f ) is called the

cylinder over f in {0, 1}B.



144 I. Druzhinina

Let X be a space. We say that S ⊆ X is a first category set in X if
S =

⋃∞
n=1

Sn, where each Sn is a nowhere dense subset of X.

We start with a lemma.

Lemma 3.1. Let S be a subset of the Cantor set C. If there exists a family
{An : n ∈ ω} of pairwise disjoint infinite subsets of ω such that ω =

⋃
{An :

n ∈ ω} and |{0, 1} An \ πAn (S)| = c for each n ∈ ω, then C is the union of
c = 2ω pairwise disjoint copies of the Cantor set such that S intersects each of
these copies in at most one point.

Proof. Let Cn = {0, 1}
An be a copy of the Cantor set for each n ∈ ω. Clearly,

C ∼=
∏∞

n=0
Cn. For every n ∈ ω, put Sn = πAn (S). Since |Cn \ Sn| = c, it is

possible to represent each Cn as the union Cn =
⋃
{Pn,α : α ∈ Γn}, where each

Pn,α is a doubleton, Pn,α ∩ Pn,β = ∅ if α 6= β, |Sn ∩ Pn,α| ≤ 1, and Γn is an
index set of cardinality c.

For an element γ = (γn)n∈ω of Γ =
∏∞

n=0
Γn, put Cγ =

∏∞
n=0

Pn,γn . It is
routine to verify that:

(i) Cγ ⊆ C is a copy of the Cantor set for every γ ∈ Γ;
(ii) C =

⋃
{Cγ : γ ∈ Γ} and |Γ| = 2

ω;
(iii) Cγ ∩ Cλ = ∅ if γ 6= λ;
(iv) |S ∩ Cγ| ≤ 1 for every γ ∈ Γ.

Our lemma is proved. �

Corollary 3.2. Let S be a countable subset of the Cantor set C. Then C is
the union of 2 ω pairwise disjoint copies of the Cantor set such that S intersects
each of these copies in at most one point.

Proof. Clearly, there exists a family {An : n ∈ ω} of disjoint infinite sets
such that ω =

⋃∞
n=0

An. We have that |πAn (S)| = ℵ0 and, therefore, the
complement {0, 1}An \ πAn (S) has cardinality c, for each n ∈ ω. Now apply
Lemma 3.1. �

Theorem 3.3. The unit segment I = [0, 1] is the union of 2 ω pairwise disjoint
copies of the Cantor set.

Proof. We take the Cantor set C = {0, 1} ω and consider the mapping f : C → I
defined by the formula

f (x) =

∞∑

i=0

xi
2i+1

,

where xi is ith coordinate of the point x ∈ C.
It is known that f is a continuous mapping onto I (see 3.2.b of [2]). Although

f is not one-to-one, it is easy to see that there is a countable infinite subset
S ⊂ C such that |f −1(f (x))| = 2 for every x ∈ S and |f −1(f (x))| = 1 for
x ∈ C \ S. It follows from Corollary 3.2 that there exists a representation
C =

⋃
{Cγ : γ ∈ Γ} such that the family {Cγ : γ ∈ Γ} and the set S satisfy

conditions (i)–(iv) in the proof of Lemma 3.1. Put Kγ = f (Cγ ). Since |Cγ ∩
S| ≤ 1, it is evident that fγ = f ↾Cγ : Cγ → Kγ is a homeomorphism; therefore,



Connected metrizable subtopologies 145

Kγ is a copy of the Cantor set for every γ ∈ Γ. Notice that I =
⋃
{Kγ : γ ∈ Γ}.

It is easy to see that if Cγ ∩ S = ∅, then Kγ does not meet any other Kλ, and
if Cγ ∩ S = {x}, then there is a unique λ ∈ Γ \ {γ} such that Kγ ∩ Kλ 6= ∅. In
this case, Kγ ∩ Kλ = {y}, where y = f (x).

Hence, for every y ∈ f (S) there exists a unique pair (αy , βy) ∈ [Γ]
2 such

that Kαy ∩ Kβy = {y}. Put Ly = Kαy ∪ Kβy . Clearly, each Ly is a copy of the
Cantor set and Ly ∩ Lz = ∅ if y 6= z. We obtain that

I =
⋃

{Kγ : γ ∈ Γ, Cγ ∩ S = ∅} ∪
⋃

{Ly : y ∈ f (S)},

that is, I is the union of 2ω disjoint copies of the Cantor set. �

In the sequel we will show that Theorem 3.3 remains valid if the unit interval
I is replaced by a compact metrizable space without isolated points. First we
need some lemmas. The following result is trivial.

Lemma 3.4. If F ⊆ {0, 1}ω is a nowhere dense set and A ⊆ ω is finite, then
πω\A(F ) is a nowhere dense subset of {0, 1}

ω\A, where πω\A is the projection

of {0, 1}ω onto {0, 1}ω\A.

Lemma 3.5. Let S ⊆ C = {0, 1}ω be a first category set. Then there are
infinite disjoint sets A and B such that ω = A ∪ B and the sets πA(S) and
πB(S) are of the first category in {0, 1}

A and in {0, 1}B, respectively.

Proof. Let S =
⋃∞

m=0
Fm, where every Fm is a nowhere dense subset of C. We

will construct by induction two families {An : n ∈ ω} and {Bn : n ∈ ω} which
satisfy following conditions for each n ∈ ω:

(i) An, Bn ∈ [ω]
<ω;

(ii) An ∩ Bn = ∅;
(iii) An ⊆ An+1, An+1 \ An 6= ∅;
(iv) Bn ⊆ Bn+1, Bn+1 \ Bn 6= ∅;
(v) For every f ∈ {0, 1} An , there exists a family {U0f , U1f , . . . , Unf } of open

canonical subsets of {0, 1} ω\Bn such that π
ω\Bn
An

(Uif ) = {f}, Uif ∩πω\Bn (Fi) =
∅ and An ⊆ coord(Uif ) ⊆ An+1, for each i ≤ n;

(vi) For every g ∈ {0, 1} Bn , there exists a family {U0g, U1g, . . . , Ung} of open

canonical subsets of {0, 1} ω\An+1 such that π
ω\An+1
Bn

(Uig) = {g}, Uig ∩

πω\An+1 (Fi) = ∅ and Bn ⊆ coord(Uig ) ⊆ Bn+1, for each i ≤ n;
(vii) n ∈ An ∪ Bn.

We start with A0 = {0} and B0 = {1}. Now suppose that for some n ∈ ω
we have constructed {Ak : k ≤ n} and {Bk : k ≤ n} which satisfy (i)–(vii).

Let us construct An+1. Take f ∈ {0, 1}
An and let O(f ) be the cylinder

over f in {0, 1} ω\Bn. For each i = 0, 1, . . . , n, the set πω\Bn (Fi) is nowhere

dense in {0, 1} ω\Bn (see Lemma 3.4). Hence, for each i ≤ n there exists an open
canonical set Uif ⊆ {0, 1}

ω\Bn such that Uif ⊆ O(f ) and Uif ∩πω\Bn (Fi) = ∅.



146 I. Druzhinina

Clearly, π
ω\Bn
An

(Uif ) = {f}, An ⊆ coord(Uif ) and coord(Uif ) ∩ Bn = ∅. Put

Ãn+1 =
⋃

{

n⋃

i=0

coord(Uif ) : f ∈ {0, 1}
An} and an+1 = min(ω \ (An ∪ Bn)).

We set An+1 = Ãn+1 ∪{an+1}. It is easy to see that An+1 is finite, An ⊆ An+1 ,
An+1 \ An 6= ∅, and An+1 ∩ Bn = ∅. The choice of an+1 and (vii) together
imply that n + 1 ∈ An+1 ∪ Bn. In addition, for every f ∈ {0, 1}

An and each
i ≤ n we have that coord(Uif ) ⊆ An+1.

Now we construct Bn+1. Take g ∈ {0, 1}
Bn and let O(g) be the cylinder over

g in {0, 1}ω\An+1. For each i = 0, 1, . . . , n, the set πω\An+1 (Fi) is nowhere dense

in {0, 1}ω\An+1 (see Lemma 3.4). Hence, for each i ≤ n there exists an open
canonical set Uig ⊆ {0, 1}

ω\An+1 such that Uig ⊆ O(g) and Uig ∩πω\An+1 (Fi) =

∅. Clearly, π
ω\An+1
Bn

(Uig) = {g}, Bn ⊆ coord(Uig ) and coord(Uig ) ∩ An+1 = ∅.
Let

B̃n+1 =
⋃

{
n⋃

i=1

coord(Uig ) : g ∈ {0, 1}
Bn} and bn+1 ∈ (ω \ (An+1 ∪ Bn)).

We set Bn+1 = B̃n+1 ∪ {bn+1}. It follows from the construction that Bn+1
is finite, An+1 ∩ Bn+1 = ∅, Bn ⊆ Bn+1, and Bn+1 \ Bn 6= ∅. In addition,
coord(Uig ) ⊆ Bn+1, for every g ∈ {0, 1}

Bn and each i ≤ n. Therefore, An+1
and Bn+1 satisfy conditions (i)–(vii).

Thus we have constructed two families {An : n ∈ ω} and {Bn : n ∈ ω}.
Now we define A =

⋃
{An : n ∈ ω} and B =

⋃
{Bn : n ∈ ω}. It follows from

(i)–(iv) that A and B are disjoint infinite subset of ω. Condition (vii) implies
that A ∪ B = ω. It remains to verify that πA(S) and πB (S) are of the first
category in {0, 1}A and in {0, 1}B, respectively.

To prove that πA(S) =
⋃∞

m=0
πA(Fm) is of the first category in {0, 1}

A, it
suffices to verify that πA(Fm) is nowhere dense in {0, 1}

A, for each m ∈ ω.
Indeed, fix an open canonical non-empty set U ⊆ {0, 1}A and m ∈ ω. As
coord(U ) ⊆ A is finite, (iii) implies that there is k ∈ ω such that coord(U ) ⊆ Ak.
Let n = max{k, m}, then coord(U ) ⊆ An and m ≤ n. Take an arbitrary
f ∈ πAAn (U ) ⊆ {0, 1}

An . By (v), there exists an open canonical set Umf in

{0, 1}ω\Bn such that π
ω\Bn
An

(Umf ) = {f}, Umf ∩ πω\Bn (Fm) = ∅ and An ⊆

coord(Umf ) ⊆ An+1 ⊆ A. Put V = π
ω\Bn
A (Umf ) and let K = coord(U ). We

have:

K ⊆ An ⊆ coord(Umf ) = coord(V ) (1)

and

πAK (U ) = {π
An
K (f )} = π

ω\Bn
K (Umf ) = π

A
K (V ). (2)

It follows from (1) and (2) that V ⊆ U . To verify that V ∩ πA(Fm) = ∅,
we assume the contrary and choose an element g ∈ V ∩ πA(Fm). Since
coord(V ) ⊆ An+1 ⊆ A, we have that π

A
An+1

(g) ∈ πAAn+1 (V ) ∩ πAn+1 (Fm).

It follows from the definition of V that πAAn+1 (V ) = π
ω\Bn
An+1

(Umf ). Hence,



Connected metrizable subtopologies 147

πAAn+1 (g) ∈ π
ω\Bn
An+1

(Umf ) ∩ πAn+1 (Fm). Take an element g̃ ∈ πω\Bn (Fm) such

that π
ω\Bn
An+1

(g̃) = πAAn+1 (g). Then g̃ ∈ Umf , since coord(Umf ) ⊆ An+1. We

obtain a contradiction, as Umf ∩ πω\Bn (Fm) = ∅. We have thus proved that

πA(Fm) is nowhere dense in {0, 1}
A for each m ∈ ω and, hence, πA(S) =⋃∞

m=1
πA(Fm) is of the first category in {0, 1}

A. The verification that πB(S) =⋃∞
m=1

πB(Fm) is a first category set in {0, 1}
B is similar. �

Lemma 3.6. Let S be a set of the first category in the Cantor set C. There
is a decomposition ω =

⋃∞
n=0

An such that each An is infinite, An ∩ Am = ∅
if n 6= m, and πAn (S) is a first category set in {0, 1}

An for every n ∈ ω.

Proof. The proof consists of applying ω times the previous lemma. �

If f : X → Y is a continuous mapping and U ⊆ X, then we put f #(U ) =
Y \ f (X \ U ). In the following lemma we collect some well-known facts that
will be frequently used in the sequel.

Lemma 3.7.

(a) Let f : X → Y be a continuous closed and irreducible mapping. If U ⊆
X is open, then V = f −1(f #(U )) is an open dense subset of U (see [1,
Problem 109, Chapter VI]).

(b) The inverse image of a closed nowhere dense set under a continuous closed
irreducible mapping is a closed nowhere dense set.

(c) If X is a Gδ-set in a compact space and X does not have isolated points,
then |X| ≥ c (see [2, 3.12.11 (b)]).

(d) A non-empty open subset of the Cantor set can be represented as the union
of countably many pairwise disjoint copies of the Cantor set.

Theorem 3.8. Every compact metrizable space without isolated points can be

represented as the union of 2 ω pairwise disjoint copies of the Cantor set.

Proof. We consider the Cantor set C and a compact metrizable space without
isolated points Y . We can assume that Y ⊆ Iω, where I is the closed unit
interval.

Let f : C → I be the continuous mapping onto I considered in the proof of
Theorem 3.3. Now define g : Cω → Iω by g =

∏
{gi : i ∈ ω} where gi = f for

each i ∈ ω. Since g is a perfect mapping, there exists a closed subset C′ ⊆ Cω

such that g(C′) = Y and g↾C′: C
′ → Y is irreducible (see [2, 3.1.C (a)]).

Clearly, C′ does not have isolated points and hence is homeomorphic to C.
Therefore, we can assume from the very beginning that there exists a continuous
irreducible mapping of C onto Y which is denoted by the same letter f . Let
S = {x ∈ C : |f −1(f (x))| > 1}.

Our first step is to verify that S is of the first category in C. For each
m ∈ N, we choose a finite covering {Um1, . . . , Umkm} of C by clopen sets of
diameter less than or equal to 1/m with respect to a given metric on C. For
each m ∈ N and each i = 1, 2, . . . , km, the set Vmi = f

−1(f #(Umi)) is open
and dense in Umi by (a) of Lemma 3.7 and, hence, Fmi = Umi \ Vmi is closed



148 I. Druzhinina

and nowhere dense in C. Therefore, for each m ∈ N the set Fm =
⋃km

i=1
Fmi

is closed and nowhere dense in C. Put F =
⋃∞

m=1
Fm. We claim that S = F .

Indeed, for x ∈ C \ S we have that |f −1(f (x))| = 1. From this, if x ∈ Umi
for some m ∈ N and i ∈ {1, 2, . . . , km}, then x ∈ f

−1(f #(Uni)) = Vmi, that
is, x /∈ Fmi = Umi \ Vmi. Now, it is evident that x /∈ F . Thus, F ⊆ S.
For x ∈ S, let z = f (x). There exists y ∈ C \ {x} such that z = f (y). We
choose m ∈ N and i ∈ {1, 2, . . . , km} such that x ∈ Umi and y /∈ Umi. Then
z = f (y) ∈ f (Y \Umi), that is, z /∈ f

#(Umi). We have that x ∈ f
−1(z)∩Umi ⊆

Umi \ f
−1(f #(Umi)) = Fmi ⊆ Fm ⊆ F . Therefore, S ⊆ F . This implies that

S = F =
⋃∞

m=1
Fm , where every Fm is a closed nowhere dense subset of C.

Apply Lemma 3.6 to find a family {An : n ∈ ω} of infinite 1 disjoint subsets
of ω whose union is equal to ω and such that πAn (S) is a first category set in
Cn = {0, 1}

An for each n ∈ ω. It follows that the complement Cn \ πAn (S) is
a dense subset of the Cantor set Cn. In addition, Cn \ πAn (S) is a Gδ-set in
Cn, since every set πAn (Fm) is closed in Cn. Then by Lemma 3.7(c), we obtain
that |Cn \ πAn (S)| = c for each n ∈ ω.

Thus the family {An : n ∈ ω} and the set S ⊆ C satisfy the conditions of
Lemma 3.1 and, hence, we can represent C in the form C =

⋃
{Cγ : γ ∈ Γ},

where |Γ| = 2 ω and every Cγ is a copy of C. In addition, Cγ ∩ Cλ = ∅ if γ 6= λ
and |S ∩ Cγ| ≤ 1 for each γ ∈ Γ.

Let Kγ = f (Cγ ) for each γ ∈ Γ. Since f ↾Cγ is a homeomorphism, Kγ is a
copy of the Cantor set for each γ ∈ Γ and we have that Y =

⋃
{Kγ : γ ∈ Γ}.

Note that the family {Kγ : γ ∈ Γ} has the following properties:

(i) if Cγ ∩ S = ∅, then Kγ ∩ Kλ = ∅ for each λ ∈ Γ \ {γ};
(ii) if Cγ ∩ S 6= ∅, then Cγ ∩ S = {s} for some s ∈ S; in this case for each

λ ∈ Γ \ {γ} we have:
(ii1) Kγ ∩ Kλ = ∅ if Cλ ∩ f

−1(f (s)) = ∅;
(ii2) Kγ ∩ Kλ = {f (s)} if Cλ ∩ f

−1(f (s)) 6= ∅.

Now for every y ∈ f (S), we define Ky = {Kγ : γ ∈ Γ, Cγ ∩ f
−1(y) 6= ∅}. It

is easy to see that
⋃
Ky ∩

⋃
Kz = ∅ if y 6= z and

Y =
⋃

{Kγ : γ ∈ Γ, Cγ ∩ S = ∅} ∪
⋃

{
⋃

Ky : y ∈ f (S)}. (∗)

Consider the family Ky for some y ∈ f (S). Choose an element γy ∈ Γ such that
Kγy ∈ Ky. For every λ ∈ Γ\{γy} such that Kλ ∈ Ky, put Lλ = Kλ\{y}. Every
Lλ is an open subset of the Cantor set, so it can be represented as the union
of countably many disjoint copies of the Cantor set (item (d) of Lemma 3.7).
Since ⋃

Ky =
⋃

{Lλ : Kλ ∈ Ky, λ 6= γy} ∪ Kγy ,

for every y ∈ f (S), the set
⋃
Ky can be represented as the union of some number

κy of disjoint copies of the Cantor set, where |Ky| ≤ κy ≤ |Ky| · ω. Taking into
account the equality (∗), we obtain the conclusion of the theorem. �

Remark 3.9. Let Y be a compact metrizable space without isolated points,
C the Cantor set and f : C → Y a continuous irreducible mapping. If S is a set



Connected metrizable subtopologies 149

of the first category in C and {x ∈ C : |f −1(f (x))| > 1} ⊆ S, then repeating
the argument in the proof of Theorem 3.8 one can represent Y as the union of
2ω pairwise disjoint copies of the Cantor set such that every copy meets the set
f (S) in at most one point.

Due to the above observation it is possible to extend Theorem 3.8 as follows:

Theorem 3.10. Every separable completely metrizable space without isolated

points can be represented as the union of 2ω pairwise disjoint copies of the
Cantor set.

Proof. Let X be a separable completely metrizable space without isolated
points. We can assume that X ⊆ Iω. Then Y = clIω (X) is a metrizable
compactification of X. Since X has no isolated points and is dense in Y , it
follows that Y is a compact metrizable space without isolated points. The
space X is Čech-complete, therefore, the remainder R = Y \ X is a Fσ-set in
Y . Hence, R =

⋃∞
n=1

Fn, where every Fn is a closed nowhere dense subset of
Y .

Consider a continuous irreducible mapping f : C → Y , where C is the Cantor
set (see the proof of Theorem 3.8). Then each f −1(Fn) is a closed nowhere dense
subset of C, by (b) of Lemma 3.7, and S′ = f −1(R) =

⋃∞
n=1

f −1(Fn). In the
proof of Theorem 3.8 we have established that S′′ = {x ∈ C : |f −1(f (x))| > 1}
is the union of countably many closed nowhere dense subsets of C. Put S =
S′ ∪ S′′ and apply Remark 3.9 to conclude that Y =

⋃
{Kγ : γ ∈ Γ}, where

|Γ| = 2ω, every Kγ is a copy of the Cantor set, Kγ ∩ Kλ = ∅ if γ 6= λ and
|Kγ ∩ f (S)| ≤ 1 for each γ ∈ Γ.

Since R ⊆ f (S), the set R ∩ Kγ is empty or consists of one point. For each

γ ∈ Γ, we define K̃γ = Kγ \ (Kγ ∩ R). It is evident that X =
⋃
{K̃γ : γ ∈ Γ}.

Every K̃γ is a copy the Cantor set or an open subset of this copy. In the

last case K̃γ can be represented as a union of ω pairwise disjoint copies of the
Cantor set, by (d) of Lemma 3.7. This finishes the proof. �

References

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Reidel Publishing Co., Dordrecht, 1984. xvi+415 pp.

[2] R. Engelking, General Topology, Helderman Verlag, Berlin 1989.
[3] P. Delaney and W. Just, Two remarks on weaker connected topologies, Comment. Math.

Univ. Carolin. 40 no. 2 (1999), 327–329.
[4] I. Druzhinina, Condensations onto connected metrizable spaces, Houston J. Math. 30

no. 3, (2004), 751–766.
[5] K. Kunen, Set Theory, North Holland, 1980.
[6] M. G. Tkachenko, V. V. Tkachuk, V. Uspenskij, and R. G. Wilson, In quest of weaker

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150 I. Druzhinina

Received May 2004

Accepted April 2005

Irina Druzhinina (mich@xanum.uam.mx)
Departamento de Matemáticas, UAM, Iztapalapa, Av. San Rafael Atlixco 186,
Col. Vicentina, Del. Iztapalapa, C.P. 09340, México D.F.