Metrizable DH-spaces with a dense complete subset

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by  Sergey Medvedev

METRIZABLE DH-SPACES WITH A DENSE COMPLETE SUBSET
SERGEY MEDVEDEV
Abstract. It is proved that for an h-homogeneous space X the following conditions are
equivalent: 1) X is a densely homogeneous space with a dense complete subspace; 2) X is
6 σ-discretely controlled.
1
0
2

n All spaces under discussion are metrizable and strongly zero-dimensional.
a In 2005, M. Hruˇs´ak and B. Z. Avil´es [1, Theorem 2.3] proved that every analytic CDH-
J
5 space is completely Baire. It is therefore of interest to construct a non-complete CDH-
1
space with a dense complete subset. It was realized by R. Hern´andez-Guti´errez, M. Hruˇs´ak,

]
N and J. van Mill [2, Corollary 4.6] and A. Medini [3, Theorem 7.3]. They shown that if a
G homogeneous space X with a dense complete subset can be embedded in the Cantor set
h. 2ω by a special way, then X is CDH. Moreover, Medini [3, Theorem 9.4] proved that Xω
at is CDH if X is a countably controlled space. We suggest a criterion for checking the CDH-
m
property for a space without its embedding into the Cantor set. A similar result is obtained
[
for non-separable spaces. Notice that the following characterization of non-Baire densely

1
v homogeneous h-homogeneous spaces was given by Medvedev [4, Theorem 6]:
8
Theorem 1. Let X be a non-σ-discrete h-homogeneous space. Then X is a densely
9
7 homogeneous space of the first category if and only if every σ-discrete subset of X is a G -set
δ
3
0 in X.
.
1 For all undefined terms and notations see [5].
0
6 We shall say that a space X is h-homogeneous if IndX = 0 and every non-empty clopen
1
subset of X is homeomorphic to X (this term was first used by Ostrovsky [6]). By a complete
:
v
space we mean a completely metrizable space, or an absolute G -space. A separable topo-
i δ
X
logical space X is countable dense homogeneous (briefly, CDH) if, given any two countable
r
a
dense subsets A and B of X, there is a homeomorphism h: X → X such that h(A) = B.
Similarly, a space X is densely homogeneous (briefly, DH) if, given any two σ-discrete dense
subsets A and B of X, there is a homeomorphism h: X → X such that h(A) = B.
Medini [3] says that a separable space X is countably controlled if for every countable
D ⊂ X there exists a Polish subspace G of X such that D ⊂ G. Similarly, we shall say that
a space X is σ-discretely controlled if for every σ-discrete D ⊂ X there exists a completely
metrizable subspace G of X such that D ⊂ G. Clearly, every σ-discretely controlled space
2010 Mathematics Subject Classification. 54H05, 54E52.
Key words and phrases. h-homogeneous space, densely homogeneous space, CDH-space, KR-cover.
1
2 SERGEY MEDVEDEV
has a dense complete subspace because a metrizable space always contains a dense σ-discrete
subspace.
If U is a family of sets, then U = {U: U ∈ U}.
We shall use technique from S[7] whicSh based on the following lemma.
Lemma 1. Let X be a metrizable space with IndX = 0 and F is a nowhere dense closed
i i i
subset of X , where i ∈ {1,2}. Let f: F → F be a homeomorphism.
i 1 2
Then there exist a σ-discrete (in X ) cover V of X \F by non-empty pairwise disjoint
i i i i
clopen subsets of X , i ∈ {1,2}, and a bijection ψ: V → V such that for any subsets
i 1 2
D ⊆ V and D ⊆ V and any bijection g: D → D satisfying g(D ∩V) = D ∩ψ(V)
1 1 2 2 1 2 1 2
for eveSry V ∈ V , theScombination f▽g: F ∪D → F ∪D is continuous at each point of
1 1 1 2 2
F and its inverse (f▽g)−1 is continuous at each point of F .
1 2
Under the conditions of Lemma , we say [8] that:
1) the cover V forms a residual family with respect to F , where i ∈ {1,2},
i i
2) the bijection ψ: V → V is agreed to f.
1 2
Remark 1. A construction similar to Lemma was used by B. Knaster and M. Reichbach
[9], R. Hern´andez-Guti´errez, M. Hruˇs´ak, and J. van Mill [2], and A. Medini [3] for separable
spaces and by A. V. Ostrovsky [6] and the author [10] for non-separable spaces. The last
two papers were written in Russian, therefore the proof of Lemma was also given in [8].
Independently Lemma was obtained by F. van Engelen [11, Theorem 2.3] for weight-
homogeneous spaces. In his notation, the triple hV ,V ,ψi is called a Knaster-Reichbach
1 2
cover, or a KR-cover, for hX \F ,X \F ,fi.
1 1 2 2
The main theorem
Theorem 2. Let X be an h-homogeneous space. Then X is a densely homogeneous space
with a dense complete subspace if and only if X is σ-discretely controlled.
Proof. We first verify that if X is a densely homogeneous space, then X is σ-discretely
∗
controlled. Take a σ-discrete set D ⊂ X and expand it to a dense σ-discrete set D ⊂ X.
Under the conditions of the theorem X has a dense complete subspace G; hence there is
a dense σ-discrete set A ⊂ G. Let us choose a homeomorphism f: X → X such that
∗
f(A) = D . Then D lies entirely within the completely metrizable space f(G).
Now we shall prove that a σ-discretely controlled space X is densely homogeneous.
To begin, assume additionally that X is a completely metrizable space.
Clearly, every completely metrizable space is σ-discretely controlled. The statement is
trivial if X is a point. By virtue of [5, Exercise 4.3.H], the theorem holds if X is the Cantor
set or the space of irrationals. According to [10, Theorem 5.4] (see also [7, Corollary 2]) the
Baire space B(k) of weight k is densely homogeneous. There is no another h-homogeneous
complete spaces.
METRIZABLE DH-SPACES WITH A DENSE COMPLETE SUBSET 3
Next, consider the main case when X is not completely metrizable.
Fix a metric ̺ on X which is bounded by 1.
Take two σ-discrete dense subsets A and B of X.
According to the Dowker-Hurewicz theorem (see [5, Theorem 7.3.1]) there exists a se-
quence U ,U ,... of discrete clopen covers of X such that U is a refinement of U for
0 1 n+1 n
each n and the family U = {U ∈ U : n ∈ ω} forms a base for X. For each U ∈ U fix a
n
∗
homeomorphism ϕ : X → U. One can verify that the sets A = A ∪ {ϕ (A): U ∈ U}
U U
and B∗ = B ∪ {ϕ (B): U ∈ U} are σ-discrete. Then the set A∗ ∪B∗ Sis σ-discrete. Since
U
X is σ-discreteSly controlled, there exists a complete subspace G ⊂ X containing A∗ ∪ B∗.
Clearly, G is dense in the h-homogeneous space X. The Stone theorem [12, Theorem 1]
implies that there is a homeomorphism χ: B(τ) → G, where τ = w(X). Fix the standard
metric on the Baire space B(τ). By construction, ϕ (F)∩(A∪B) = ∅ for any U ∈ U and
U
∗ ∗
F ⊂ X \(A ∪B ).
Let A = {A : n ∈ ω} and B = {B : n ∈ ω}, where each A , B is a closed discrete
n n n n
subset of X,Sand X \G = {X : n ∈Sω}, where each X is a closed nowhere dense subset
n n
of X. S
We will construct a homeomorphism f: X → X satisfying f(A) = B by induction on n.
For each n ∈ ω we will define a homeomorphism f : X → X, closed nowhere dense sets F
n n
and E , a KR-cover hV ,W ,ψ i for hX \F ,X \E ,f | i such that:
n n n n n n n Fn
(a) f (F ) = E ,
n n n
(b) F ∪X ∪A ⊂ F ,
n n n n+1
(c) E ∪X ∪B ⊂ E ,
n n n n+1
(d) f (A∩F ) = B ∩E ,
n n n
(e)f (x) = f (x) for each x ∈ F whenever m ≥ n,
m n n
(f) V refines V and W refines W ,
n+1 n n+1 n
∗ ∗ ∗
(g) if V ∈ V and V ∈ V for n < m, then V ⊂ V if and only if ψ (V ) ⊂ ψ (V),
n m m n
(h) mesh(V ) ≤ 1/(n+1) and mesh(W ) ≤ 1/(n+1),
n n
(i) diam(χ−1(U ∩G)) ≤ 1/(n+1) for each U ∈ V ∪W .
n n
For n = 0, let F = E = ∅, V = W = {X}, ψ (X) = X, and f (x) = x.
0 0 0 0 0 0
Now let us consider the induction step n+1 for some n ∈ ω.
First, fix V ∈ V and W = ψ (V) ∈ W . Let Y = V ∩X , where m is the least i with
n n n m
V ∩X 6= ∅, and Z = W ∩X , where l is the least i with W ∩X 6= ∅.
i l i
The open set V \Y contains the set V ∩G. Then for each point x ∈ V ∩G we can find its
neighborhood U ∈ U such that U ⊂ V \Y, diam(U ) ≤ 1/(n+2), diam(χ−1(U ∩G)) ≤
x ℓ x x x
∗
1/(n + 2), and ℓ is the least suitable index. The family V = {U : x ∈ V ∩ G} consists
V x
∗ ∗
of pairwise disjoint clopen subsets of X, V ⊂ U, and V ∩ G ⊂ V ⊂ V \ Y. Clearly,
V V
S
4 SERGEY MEDVEDEV
∗ ∗
Y ⊂ V \ V . Note that V \ V is a closed nowhere dense subset of V because the set
V V
V ∩G is dSense in V. S
∗ ∗
Likewise, choose a family W = {U : x ∈ W ∩ G} such that W consists of pairwise
V x V
disjoint clopen subsets of X, W∗ ⊂ U, diam(U) ≤ 1/(n+2), diam(χ−1(U ∩G)) ≤ 1/(n+2)
V
∗ ∗
for each U ∈ W , and W ∩G ⊂ W ⊂ W \Z.
V V
Since X is locally of weight τ,Sthere exists a closed discrete set A′ ⊂ A∩V of cardinality τ
′ ′
such that A ∩V ⊂ A, where m is the least i with V ∩A 6= ∅. The union Y ∪A is a closed
m i
∗ ′
nowhere dense subset of V. Hence V contains a set U ∈ V such that U ∩ (Y ∪ A) = ∅.
V
′ ′
Similarly, choose a closed discrete set B ⊂ B ∩W of cardinality τ such that B ∩W ⊂ B ,
l
∗ ′
where l is the least i with W ∩B 6= ∅. Then W contains a set O ∈ W which misses Z∪B .
i V
′ ′
Fix a bijection g: A → B .
′ ∗ ∗
By construction, A, V \ V , and ϕ (W \ W ) are disjoint closed nowhere dense
V U V
subsets of V. Similarly, B′, WS\ W∗, and ϕ (VS\ V∗) are disjoint closed nowhere dense
V O V
subsets of W. Put S S
′ ∗ ∗
F = A ∪(V \ V )∪ϕ (W \ W ),
V V U V
′ ∗ ∗
EV = B ∪(W \S WV)∪ϕO(V \S VV).
′ S ′ S
Notice that A∩F = A and B ∩E = B .
V V
Define the homeomorphism f : F → E by setting
V V V
′
g(x) if x ∈ A,
∗
f (x) =  ϕ (x) if x ∈ V \ V ,
V O V
 (ϕU)−1(x) if x ∈ ϕU(WS \ WV∗).
By Lemma , there exists a KR-cover hV ,W ,ψ i for hV \SF ,W \E ,f i. By construc-
V V V V V V
∗ ′ ∗ ′
tion, V = V \A and W = W \B . By reducing V and W to fragments if
V V V V V V
it is nSecessaryS, we can obtainSthat diaSm(χ−1(U ∩G)) ≤ 1/(n+ 2) for each U ∈ V ∪W ,
V V
V ⊂ U, and W ⊂ U.
V V
Repeat this construction for each V ∈ V . It is not hard to check that F = F ∪
n n+1 n
{F : V ∈ V } and E = E ∪ {E : V ∈ V } are closed nowhere dense subsets of X.
V n n+1 n V n
SFrom Lemma it follows that the mSapping
f (x) if x ∈ F for some V ∈ V ,
f (x) = V V n
n+1 (cid:26) fn(x) if x ∈ Fn
is a homeomorphism between F and E .
n+1 n+1
Define V = {U ∈ V : V ∈ V } and W = {W : V ∈ V }. Then ψ =
n+1 V n n+1 V n n+1
{ψ : V ∈ V } is a bijection between V and W which is agreed to f . Thus,
V n n+1 n+1 n+1
ShV ,W ,ψ i is a KR-cover for hX \F ,X \E ,f i.
n+1 n+1 n+1 n+1 n+1 n+1
It remains to extend the mapping f from F to the space X. Both sets U ∈ V
n+1 n+1 n+1
and ψ (U) ∈ W are clopen in the h-homogeneous space X. Choose a homeomorphism
n+1 n+1
METRIZABLE DH-SPACES WITH A DENSE COMPLETE SUBSET 5
g : U → ψ (U) and put f (x) = g (x) if x ∈ U ∈ V . From Lemma it follows that
U n+1 n+1 U n+1
f : X → X is a homeomorphism.
n+1
One can check that all conditions (a)–(i) are satisfied. This completes the induction step.
Let F = {F : n ∈ ω} and E = {E : n ∈ ω}. If x ∈ F for some n ∈ ω, set f(x) =
n n n
f (x). CondSition (e) implies that thisSrule define a bijection between F and E. Moreover,
n
from conditions (c) and (d) it follows that f(A) = B. For every point x ∈ X \ F ⊂ G
and every n ∈ ω there exists the unique clopen set Ux ∈ V that contains x. Clearly,
n n
x = {Ux: n ∈ ω}. By virtue of the Cantor theorem [5, Theorem 4.3.8] and condition (i),
n
theinTtersection {χ−1(G∩ψ (Ux)) : n ∈ ω}is non-empty and consists of a point z ∈ B(τ).
n n
Condition (h) imTplies that the intersection {ψ (Ux): n ∈ ω} coincides with the point χ(z).
n n
Clearly, χ(z) ∈ X\E. Setf(x) = χ(z). LikTewise, everypointy ∈ X\E canberepresented as
{Uy: n ∈ ω}, where each Uy ∈ W . Then y = f(x ) for x = {ψ−1(Uy): n ∈ ω} ∈ X\F.
n n n 0 0 n n
HTence, we obtain the mapping f: X → X such that f(X) = XT. One easily verifies that f
is a bijection.
FromthedefinitionofaKR-coveritfollowsthatf iscontinuousonF andf−1 iscontinuous
on E. Condition (h) implies that the sequence {Ux: n ∈ ω} is a base at the point x ∈ X\F
n
and the sequence {Uy: n ∈ ω} is a base at the point y ∈ X\E. By construction, if y = f(x)
n
then f(Ux) = ϕ (Ux) = Uy. From this it follows that f is continuous on X \F and f−1 is
n n n n
continuous on X \E. Hence, f is a homeomorphism. (cid:3)
Corollary 1. Every countably controlled h-homogeneous space is CDH.
Let us show that an h-homogeneous σ-discretely controlled space satisfies a strengthening
version of dense homogeneity.
Theorem 3. Let X be an h-homogeneous σ-discretely controlled space. Suppose that
{A : n ∈ ω} and {B : n ∈ ω} are two families of σ-discrete dense subsets of X such that
n n
A ∩A = ∅ and B ∩B = ∅ whenever i 6= j. Then there is a homeomorphism f: X → X
i j i j
such that f(A ) = B for each n ∈ ω.
n n
Proof. The proof is similar to the corresponding part of the proof of Theorem . We only
point out the changes and omit like details. Use the notation from the proof of Theorem .
For each n ∈ ω let A = {Ai : i ∈ ω} and B = {Bi : i ∈ ω}, where Ai and Bi are
n n n n n n
closed discrete subsets of X.SDefine A = {A : n ∈ ωS} and B = {B : n ∈ ω}. Put
n n
S S
∗ ∗
A ∪B = (A∪B)∪ {ϕ (A∪B): U ∈ U}.
U
[
∗ ∗
Fix a complete subspace G ⊂ X containing the σ-discrete set A ∪B .
For each n ∈ ω we will construct a homeomorphism f : X → X satisfying conditions
n
(a)–(i), but conditions (b)–(d) will be replaced by the following ones:
(b*) F ∪X ∪ {Ai : i ≤ n} ⊂ F ,
n n n n+1
(c*) E ∪X ∪S{Bi : i ≤ n} ⊂ E .
n n n n+1
S
6 SERGEY MEDVEDEV
(d*) f (A ∩F ) = B ∩E for each j ≤ n.
n j n j n
For this, given V ∈ V , W = ψ (V) ∈ W , and j ≤ n, choose closed discrete sets
n n n
A′(j) ⊂ A ∩V and B′(j) ⊂ B ∩W each of cardinality τ such that (Ai ∩V)\F ⊂ A′(j)
j j j n
and (Bi ∩W)\E ⊂ B′(j) for every i ≤ n. Fix a bijection g : A′(j) → B′(j). Under the
j n j
′ ′ ′ ′
conditions of the theorem, A(j )∩A(j ) = ∅ and B (j )∩B (j ) = ∅ whenever j 6= j . We
1 2 1 2 1 2
′ ′ ′ ′ ′ ′
thus get a bijection g: A → B , where A = {A(j): j ≤ n} and B = {B (j): j ≤ n}.
We omit further details of the proof. (cid:3) S S
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South Ural State University, Chelyabinsk, 454080 Russia
E-mail address: [email protected]