Download Reflexive families of closed sets

Reflexive families of closed sets
Zhongqiang Yang and Dongsheng Zhao
Abstract. Let S(X) denote the set of all closed subsets of a topological
space X, and C(X) denote the set of all continuous mappings f : X →
X. A family A ⊆ S(X) is called reflexive if there exists C ⊆ C(X) such
that A = {A ∈ S(X) : f (A) ⊆ A for every f ∈ C}. In this paper, we
investigate the conditions for a family of closed sets to be reflexive.
Recall [3] that a collection A of closed subspaces of a Hilbert space H is
called reflexive if there exists a collection F of continuous operators on H
such that
A = Lat(F) = {A| A is a closed subspace of H with T (A) ⊆ A, ∀T ∈ F}.
Reflexive families of continuous operators is defined in a dual way. See [2][3]
[4] [5] [6][7] for more about the characterizations of such families. In [9],
the second author considered the reflexive families in concrete categories.
For the category SET of sets, a complete characterization for both reflexive
1991 Mathematics Subject Classification. 47A46,54C08,54D05,54E35.
Key words and phrases. reflexive families of closed sets, s-reflexive topological space,
strongly zero-dimensional metric space, countable metric space, hereditarily disconnected
space, reflexive closed set.
The first author was supported for this work by National Natural Science Foundation
of China (No. 10471084) and by Guangdong Provincial Natural Science Foundation (No.
04010985).
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Z. YANG AND D. ZHAO
families of sets and reflexive families of mappings were obtained in [9]. In
the present paper we investigate the reflexive families in the context of
topological spaces. Given a topological space X, let S(X) be the set of all
closed sets in a topological space X, and C(X) the set of all continuous
mappings f : X → X. For any A ⊆ S(X) and F ⊆ C(X) define Alg(A) =
{f ∈ C(X) : f (A) ⊆ A for every A ∈ A} and Lat(F) = {A ∈ S(X) :
f (A) ⊆ A for every f ∈ F}. A family A ⊆ S(X) is reflexive if there exists
F ⊆ C(X) such that A = Lat(F), or equivalently if Lat(Alg(A)) = A. A
topological space X is s-reflexive whenever every family A of closed subsets
of X satisfying conditions
(a) X, ∅ ∈ A,
(b) B ⊆ A implies
T
B ∈ A, and
S
(c) B ⊆ A implies cl( B) ∈ A
is reflexive.
The main results in this paper are: every strongly zero-dimensional complete metric space is s-reflexive; every countable metric space is s-reflexive;
every Hausdorff s-reflexive space is hereditarily disconnected. From these it
is deduced that a locally compact metric space is s-reflexive if and only if it
is zero-dimensional.
We end the paper by listing some problems for possible further work.
Let S(X) be the set of all closed subsets in a topological space X, and
C(X) be the set of all continuous mappings f : X → X. For any A ⊆ S(X)
and F ⊆ C(X), define
Alg(A) = {f ∈ C(X) : f (A) ⊆ A, ∀A ∈ A},
Lat(F) = {A ∈ S(X) : f (A) ⊆ A, ∀f ∈ F}.
REFLEXIVE FAMILIES OF CLOSED SETS
3
The two mappings Alg : P(S(X)) → P(C(X)) and Lat : P(C(X)) →
P(S(X)) form a Galois connection between P(S(X)) and P(C(X)). Thus
as in the general case, for any A ⊆ S(X) and F ⊆ C(X) we have
(i) Lat(Alg(A)) ⊇ A, Alg(Lat(F)) ⊇ F;
(ii) Alg(Lat(Alg(A))) = Alg(A), Lat(Alg(Lat(F))) = Lat(F).
A family A ⊆ S(X) is called reflexive if A = Lat(Alg(A)). Similarly,
F ⊆ C(X) is reflexive if F = Alg(Lat(F)).
As in the general case [9], A ⊆ S(X) is reflexive if and only if there
exists F ⊆ C(X) such that A = Lat(F). Also by the above condition (i),
A is reflexive if and only if
Lat(Alg(A)) ⊆ A.
Lemma 1. If A ⊆ S(X) is reflexive, then the following conditions are
satisfied:
(a) X, ∅ ∈ A.
(b) B ⊆ A implies
T
B ∈ A.
S
(c) B ⊆ A implies cl( B) ∈ A.
(d) If D is a connect component of A ∈ A and B ⊆ D for some nonempty B in A, then D ∈ A.
Proof. Only (d) needs verification. For any f ∈ Alg(A), f (B) ⊆ B
because B ∈ A. Also f (B) ⊆ f (D) ⊆ f (A) ⊆ A, and f (D) ∩ D ⊇
f (B) ∩ B = f (B) 6= ∅. Thus f (D) is a connected set contained in A and
has non-empty intersection with the connected component D of A, hence
f (D) ⊆ D. Therefore D ∈ Lat(Alg(A)) = A.
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Z. YANG AND D. ZHAO
If X is a space with the discrete topology, then S(X) = P(X) and
C(X) is the set of all mappings from X to X. By Theorem 1 of [9], a family
A ⊆ P(X) is reflexive if and only if A is closed under arbitrary union
and intersection, so every family A of closed subsets of a discrete space
satisfying conditions (a),(b) and (c) in Lemma 1 is reflexive. A natural
question is: beside the discrete topological spaces, which else spaces also
have this property?
Definition 1. A topological space X is called s-reflexive if every family
A of closed subsets of X satisfying the conditions (a),(b) and (c) in Lemma
1 is reflexive.
Lemma 2. For every topological space X, the following conditions are
equivalent:
(1) For any two points a and b in X, the following mapping fa,b : X → X
is continuous:
fa,b (x) =


x,
if x 6= a,

b,
if x = a.
(2) For any finite set D ⊆ X and any element b ∈ X the following
mapping
fD,b : X → X is continuous:
fD,b (x) =


x,
if x 6∈ D,

b,
if x ∈ D.
(3) For any non-empty proper closed set B and finite set D, both B \ D
and B ∪ D are closed.
REFLEXIVE FAMILIES OF CLOSED SETS
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Proof. (1) implies (2). This follows from that fD,b is actually the composition of mappings in {fa,b : a ∈ D}.
(2) implies (3). Let B be a non-empty proper closed set and D be a
−1
finite set. Choose any b ∈ B, then fD,b
(B) = B ∪ D, so B ∪ D is closed.
−1
(B) = B \ D is also closed.
Choose any c ∈ X \ B, then fD,c
(3) implies (1). If a = b, then fa,b = idX , the identity mapping, which is
continuous. If a 6= b, for any non-empty proper closed set B,


B ∪ {a}, if b ∈ B,
−1
fa,b (B) =

B \ {a}, if b 6∈ B.
−1
(B) is always closed and hence fa,b is continuous.
So fa,b
Corollary 1. Let X satisfy the equivalent conditions in Lemma 2 with
|X| =
6 2 . If there exists one closed singleton {a}, then X is T1 .
In fact, assume that X is not a singleton. For any point x ∈ X, {x, a} =
{a} ∪ {x} is a proper closed set. Thus {x, a} \ {a} = {x} is also closed.
The set {a, b} with the non-T1 topology satisfies the conditions in Lemma
2. Thus the requirement |X| =
6 2 is needed.
Let A ⊆ S(X) be a collection of closed subsets of X satisfying conditions
T
(a),(b) and (c)in Lemma 1. For each Y ∈ S(X), let φA (Y ) = {A ∈ A :
Y ⊆ A}. If no confusion occurs, we simply write φ(Y ) for φA (Y ). On this
mapping, we have the following simple lemma.
Lemma 3. Let A ⊆ S(X) satisfy conditions (a),(b) and (c).
(1) For each Y ⊆ X, φ(Y ) ∈ A. And Y ∈ A if and only if Y = φ(Y ).
(2) For any B ∈ S(X), B ∈ A if and only if φ({x}) ⊆ B for all x ∈ B.
S
(3) For any B ∈ S(X), B ∈ A if and only if B = {φ({x}) : x ∈ B}.
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Z. YANG AND D. ZHAO
Lemma 4. Let A ⊆ S(X) satisfy conditions (a),(b) and (c). For each
U ⊆ X define κ(U ) = {x ∈ X : φ({x}) ∩ U 6= ∅}. Then
(1) κ(V ) ⊇ V for all V ⊆ X;
S
S
(2) κ( i∈I Ui ) = i∈I κ(Ui );
(3) if U is an open set of X, then κ(U ) is also open.
Proof. (1) and (2) are trivial. We only prove (3). Let U be open and
(xλ )λ∈D be a net in X \ κ(U ) which converges to the point a ∈ X. For each
S
λ ∈ D, φ({xλ }) ∩ U = ∅, Thus cl λ∈D φ({xλ }) ∩ U = ∅. By condition (c),
S
the set B = cl λ∈D φ({xλ }) is in A. Now a ∈ B, so φ({a}) ⊆ B ⊆ X \ U .
Hence φ({a}) ∩ U = ∅, which implies a ∈ X \ κ(U ). Therefore, κ(U ) is
open.
Theorem 1. If X satisfies the equivalent conditions in Lemma 2, then
X is s-reflexive.
Proof. Suppose A ⊆ S(X) satisfies conditions (a),(b) and (c). Let
B be a closed set not in A. By Lemma 3 (2), there is a ∈ B such that
φ({x}) 6⊆ B. Choose a point b ∈ φ({a}) \ B and define the mapping
f : X → X by
f (x) =


x,
if x 6= a,

b,
if x = a.
By the assumption on X, f is continuous. For every A ∈ A, if a ∈ A
then φ({a}) ⊆ A, so f (a) = b ∈ φ({a}) ⊆ A, from this it follows that
f (A) ⊆ A. If a 6∈ A, then f (A) = A. Thus f ∈ Alg(A). But f (B) 6⊆ B, so
A ⊇ Lat(Alg(A)) and hence A = Lat(Alg(A)). So A is reflexive.
Example 1. (1) Every discrete and indiscrete space satisfies the conditions in Lemma 2, so they are s-reflexive.
REFLEXIVE FAMILIES OF CLOSED SETS
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(2) Let λ be an infinite cardinal and X be any set. Define τ to be
the topology consisting all subsets whose complements are either X or has
cardinal less than λ. Then (X, τ ) satisfies the conditions in Lemma 2, it is
therefore s-reflexive.
(3) Suppose E is a subset of X. Define the topology τE such that a subset
A of X is closed if and only if either A = X, or A\E is finite. Then (X, τE )
obviously satisfies the conditions in Lemma 2. So (X, τE ) is s-reflexive.
(4) The Euclidean interval X = [−1, 1] is not s-reflexive. As a matter of
fact, let A = {A ∈ S(X) : A ⊆ [0, 1] or A 3 1}. Then it is easy to verify that
A satisfies conditions (a),(b) and (c) in Lemma 1. For every f ∈ Alg(A)
and every x ∈ X, since {x, 1} ∈ A, f ({x, 1}) ⊆ {x, 1}, so f (x) = x or
f (x) = 1. Let A = {x ∈ [−1, 1) : f (x) = x} and B = {x ∈ [−1, 1) : f (x) =
1}. Then A and B are closed in [−1, 1) and are disjoint, and A∪B = [−1, 1).
Hence A = ∅ or B = ∅ because [−1, 1) is connected. For every x ∈ [0, 1],
since {x} ∈ A we have f (x) = x. It follows that A ⊃ [0, 1) 6= ∅. This
implies A = [−1, 1). Notice that f (1) = 1 for every f ∈ Alg(A) because
{1} ∈ A. Thus Alg(A) = {idX }. But Lat({idX }) = S(X) 6= A. Hence A
is not reflexive.
Since the interval [−1, 1] equipped with the discrete topology is s-reflexive
and the Euclidean interval is a continuous image of it, so the continuous image of s-reflexive spaces need not be s-reflexive.
In the following we investigate the s-reflexivity of some special spaces .
Lemma 5. [1, Theorem 7.3.1(iii)] If (X, d) is a strongly zero-dimensional
metric space, then there is a sequence W1 , W2 , · · · , of open covers of the
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Z. YANG AND D. ZHAO
space X such that Wi consists of clopen sets of X, diam(W ) < 1/i for each
W ∈ Wi , and Wi+1 is a refinement of Wi .
Lemma 6. Let U be an open cover of a strongly zero-dimensional metric
space X. Then there is a refinement V of U consisting of pairwise disjoint
clopen sets in X.
This lemma is a corollary of [8, Lemma 3 ]. The following direct proof
was provided by the referee.
Proof. Since X is paracompact, we can assume that the family U =
{Uα : α < λ} is locally finite and cl(Uα ) ⊆ U for every α < λ. At first
we construct a family {Wα : α < λ} of clopen sets of the subspace U by
transfinite induction such that, for every α < λ,
(i) Wα ⊆ Uα ;
(ii) {Wξ : ξ ≤ α} ∪ {Uξ : α < ξ < λ} is a cover of X.
In fact, if Wβ is constructed for every β < α in such a way that the
conditions (i) and (ii) are fulfilled for {Wβ : β < α}, we consider the closed
set
[
[
Aα = X \ ( {Wξ : ξ < α} ∪ {Uξ : α < ξ < λ}).
It follows from the local finiteness of the family U that Aα ⊆ Uα . Since X
is strongly zero-dimensional, we can choose a clopen subset Wα such that
Aα ⊆ Wα ⊆ Uα . Then the family {Wβ : β ≤ α} satisfies the conditions
(i) and (ii). The transfinite induction is completed. Since the family {Wα :
α < λ} is locally finite, the family V = {Vα : α < λ}, where,
V0 = W0 and Vα = Wα \
is as required.
[
{Vβ : β < α} for every 0 < α < λ,
REFLEXIVE FAMILIES OF CLOSED SETS
9
Lemma 7. Let U be an open cover of a strongly zero-dimensional metric
space X. Then for any > 0, there is a refinement V of U such that V
consists of pairwise disjoint clopen sets and diam(V ) < for all V ∈ V.
Proof. By Lemma 6, there exists a refinement W of U consisting of
pairwise disjoint clopen sets in X. It follows from Lemma 5 that, for every
W ∈ W, there exists an open cover U(W ) of W such that diam(A) < for
each A ∈ U(W ). Using again Lemma 6, there exists a refinement V(W ) of
U(W ) consisting of pairwise disjoint clopen sets in X. Then the family
V=
[
{V(W ) : W ∈ W}
is as required.
Theorem 2. Every strongly zero-dimensional complete metric space
(X, d) is s-reflexive.
Proof. Let A be a family of closed sets in X satisfying conditions
(a),(b) and (c) of Lemma 1. We show A = Lat(Alg(A)). To this end, it is
enough to show that if B 6∈ A then there is f ∈ AlgA so that f (B) 6⊆ B.
Suppose B ∈ S(X) and B 6∈ A. Then there is b ∈ B such that φ({b}) 6⊆
B. Choose an element c ∈ φ({b})\B and a clopen set U0 with diam(U0 ) < 1
and c ∈ U0 ⊆ X \ B. By Lemma 4(3), κ(U0 ) is open and b ∈ κ(U0 ), there is
a clopen set V0 such that diam(V0 ) < 1, b ∈ V0 ⊆ κ(U0 ) and V0 ∩ U0 = ∅.
∞
Now we construct two sequences {Ui }∞
i=1 and {Vi }i=1 of collections of
pairwise disjoint non-empty clopen sets, and a mapping αn : Vn → Un for
each n, such that the following conditions are satisfied:
S
S
(i) Un = U0 , Vn = V0 for every n;
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Z. YANG AND D. ZHAO
(ii) diam(U ) ≤
1
n
for each U ∈ Un ;
(iii) Un+1 is a refinement of Un , and Vn+1 is a refinement of Vn ;
(iv) V ⊆ κ(αn (V )) for every V ∈ Vn ;
(v) if m ≤ n and V ∈ Vm , W ∈ Vn then either W ⊆ V or V ∩ W = ∅;
(vi) if m ≤ n, W ∈ Vn , V ∈ Vm and W ⊆ V , then αn (W ) ⊆ αm (V ).
First, let U1 = {U0 }, V1 = {V0 } and α1 (V0 ) = U0 . Then the above six
conditions are satisfied.
Now suppose Un , Vn and αn : Vn → Un have been defined and satisfy
conditions (i)-(vi). For any U ∈ Un , by Lemma 7 there exists a family
S
Un+1 (U ) of pairwise disjoint clopen sets in X such that Un+1 (U ) = U ,
S
1
diam(W ) ≤ n+1
for each W ∈ Un+1 (U ). Let Un+1 = {Un+1 (U ) : U ∈ Un }.
Obviously Un+1 is a refinement of Un . Next, for each V ∈ Vn , by (iv)
we have V ⊆ κ(αn (V )). Note that κ preserves union by Lemma 4(2), it
follows that {κ(W ) ∩ V : W ∈ Un+1 (αn (V ))} is an open cover of V . Again,
by Lemma 7, there is a cover Vn+1 (V ) of V consisting of pairwise disjoint
clopen sets and Vn+1 (V ) is finer than {κ(W ) ∩ V : W ∈ Un+1 (αn (V ))}. Put
S
Vn+1 = {Vn+1 (V ) : V ∈ Vn }. Then Vn+1 is a refinement of Vn .
To define αn+1 , for each A ∈ Vn+1 , there is a unique V ∈ Vn such that
A ∈ Vn+1 (V ). Then A ⊆ κ(E) ∩ V for some E ∈ Un+1 (αn (V )) ⊆ Un+1 .
Such a set E need not be unique. Choose any of them and let αn+1 (A) = E.
Thus we defined a mapping αn+1 : Vn+1 → Un+1 .
By induction we defined the sequences {Ui }, {Vi } and mapping αn for
each n. The conditions (i),(ii),(iii) and (iv) follow immediately from the
construction of these objects.
REFLEXIVE FAMILIES OF CLOSED SETS
11
To show that condition (v) is also satisfied, let m ≤ n, then Vn is a
refinement of Vm . If V ∈ Vm , W ∈ Vn and W 6⊆ V , then W ⊆ V 0 for some
V 0 ∈ Vm , where V 0 6= V , thus W ∩ V ⊆ V 0 ∩ V = ∅.
To prove (vi) is valid it is enough to check the case where n = m + 1.
Let V ∈ Vm with W ∈ Vm+1 and W ⊆ V . By the definition of αm+1 ,
αm+1 (W ) = E for some E ∈ Vm+1 (αm (V )), hence αm+1 (W ) = E ⊆ αm (V ).
Define the mapping f : X → X as follows:


 {x},
∞
{f (x)} =
T

{αn (Vn ) : x ∈ Vn ∈ Vn },

if x ∈ X\V0 ,
if x ∈ V0 .
n=1
First, f is well defined. As a matter of fact, if x ∈ V0 , then for each n,
there is a unique Vn ∈ Vn with x ∈ Vn . If x ∈ Vn ∈ Vn , x ∈ Vm ∈ Vm and
m ≤ n, then it follows from the condition (v) that ∅ 6= αn (Vn ) ⊆ αm (Vm ).
Thus {αn (Vn ) : n ∈ N} is a sequence of closed sets whose every finite
subfamily has a non-empty intersection. Furthermore, diam(αn (Vn )) ≤
T
for each n and X is complete, so the set ∞
n=1 αn (Vn ) is a singleton.
1
n
The mapping f is clearly continuous on X \ V0 . For any x ∈ V0 and
every > 0, there exist n and Vn such that x ∈ Vn ∈ Vn and αn (Vn ) ⊆
B(f (x), ). For each y ∈ Vn , by the definition of f , f (y) ∈ αn (Vn ), hence
f (y) ∈ B(f (x), ). This shows that f is also continuous on X \ V0 .
For any x ∈ X, we show f (x) ∈ φ({x}). If x 6∈ V0 , f (x) = x ∈ φ({x}).
If x ∈ V0 , then for each n, there is a unique Vn ∈ Vn such that x ∈ Vn . But
Vn ⊆ κ(αn (Vn )), so φ({x}) ∩ αn (Vn ) 6= ∅. Choose any xn ∈ φ({x}) ∩ αn (Vn ).
Then the sequence {xn } converges to f (x), so f (x) ∈ φ({x}) because φ({x})
is closed.
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Z. YANG AND D. ZHAO
Now for each A ∈ A, if x ∈ A and x ∈ X \ V0 , then f (x) = x ∈ A;
if x ∈ A ∩ V0 , then f (x) ∈ φ({x}) ⊆ A, again f (x) ∈ A. Therefore
f ∈ Alg(A). However, b ∈ V0 ∩ B, f (b) ∈ α1 (V0 ) = U0 , and U0 ∩ B = ∅, so
f (b) 6∈ B. Hence f (B) 6⊆ B, which implies B 6∈ Lat(Alg(A)). The proof is
complete.
∞
Note that in constructing the sequences {Ui }∞
i=1 , {Vi }i=1 and mapping
αn we did not make use of the completeness of X.
By [1, Corollary 6.2.8] every countable metric space is strongly zerodimensional but is not necessarily complete.
Theorem 3. Every countable metric space is s-reflexive.
Proof. Let X be a countable metric space. Then X is strongly zerodimensional. We show that X is s-reflexive. The proof is similar to that of
Theorem 2. Again let A be a family of closed sets in X satisfying conditions
(a), (b) and (c), and B 6∈ A. Let b ∈ B such that φ({b}) 6⊆ B. Choose
an element c ∈ φ({b}) \ B and a clopen set U0 with diam(U0 ) < 1 and
c ∈ U0 ⊆ X \ B. As κ(U0 ) is open and b ∈ κ(U0 ), there is a clopen set V0
such that diam(V0 ) < 1, b ∈ V0 ⊆ κ(U0 ) and V0 ∩ U0 = ∅.
Arrange V0 as {x1 , x2 , · · · }, where x1 = b.
∞
We now define by induction two sequences {Un }∞
n=1 and {Vn }n=1 of pair-
wise disjoint clopen sets of X, a map αn : Vn → Un for each n, and f (xn )
(n ∈ N), such that the conditions (i)-(vi) in the proof of Theorem 2 and the
following (vii)-(viii) hold:
(vii) for any 1 ≤ i < j ≤ n, if xi ∈ V ∈ Vn and xj ∈ V 0 ∈ Vn , then
V ∩ V 0 = ∅;
(viii) if i ≤ n and xi ∈ V ∈ Vn , then f (xi ) ∈ αn (V ) ∩ φ(xi ).
REFLEXIVE FAMILIES OF CLOSED SETS
13
For n = 1 we let U1 = {U0 }, V1 = {V0 }, α1 (V0 ) = U0 , and f (x1 ) = c.
Suppose for each i ≤ n, Ui , Vi , αi and f (i) have been defined and satisfy
the conditions. To define these objects for n + 1, for each U ∈ Un choose
Un+1 (U ) as a collection of pairwise disjoint clopen sets with diameter less
S
S
1
and Un+1 (U ) = U . Put Un+1 = {Un+1 (U ) : U ∈ Un }. Let
than n+1
V ∈ Vn+1 . We consider three cases:
Case A. V ∩ {x1 , x2 , · · · , xn , xn+1 } = ∅ or V ∩ {x1 , x2 , · · · , xn , xn+1 } =
{xn+1 }. Then Vn+1 (V ) and the restriction of αn+1 on Vn+1 (V ) are defined
in the same way as in the proof of Theorem 2. When the intersection is
{xn+1 }, there is W ∈ Vn+1 (V ) with xn+1 ∈ W . Then define f (xn+1 ) be any
point in αn+1 (W ) ∩ φ(xn+1 ). Note as xn+1 ∈ W ⊆ κ(αn+1 (W )), such a point
f (xn+1 ) exists.
Case B. V ∩ {x1 , x2 , · · · , xn , xn+1 } = {xi , xn+1 } for some i ≤ n. Choose
two disjoint clopen sets C and D such that xi ∈ C, xn+1 ∈ D, and C ⊆
κ(U 0 ) ∩ V for some U 0 ∈ Un+1 (αn (V )) with f (xi ) ∈ U 0 ( note: by induction assumption xi ∈ V ∈ Vn imply f (xi ) ∈ αn (V ) ∩ φ(xi ), so there is
U 0 ∈ Un+1 (αn (V )) with f (xi ) ∈ U 0 ∩ φ(xi ), hence xi ∈ κ(U 0 )). Also
D ⊆ κ(W 0 ) ∩ V for some W 0 ∈ Un+1 (αn (V )). Now choose a refinement
Vn+1 (V ) of {κ(W ) ∩ V : W ∈ Un+1 (αn (V ))} consisting of pairwise disjoint clopen sets and containing both C and D as members. Define αn+1 on
Vn+1 (V ) by letting αn+1 (C) = U 0 , and αn+1 (F ) as before if F 6= C. Also define f (xn+1 ) be any point in αn+1 (D)∩φ(xn+1 ) (note: αn+1 (D)∩φ(xn+1 ) 6= ∅
because xn+1 ∈ D ⊆ κ(αn+1 (D))). .
Case C. V ∩ {x1 , x2 , · · · , xn , xn+1 } = {xi } for some i ≤ n. Then the
construction of Vn+1 (V ) is defined in the same way as in the proof of theorem
2.
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Z. YANG AND D. ZHAO
At last let Vn+1 =
S
{Vn+1 (V ) : V ∈ Vn }. Since there is a unique V ∈ Vn
that contains xn+1 and satisfies the condition in either Case A or Case B,
so f (xn+1 ) is defined.
Let g : X → X be the mapping defined by g(x) = x for x 6∈ V0 and
g(xi ) = f (xi )(i = 1, 2, · · · ). From the construction it is clear that g(x) ∈
φ(x) for all x ∈ X, thus g(A) ⊆ A for all A ∈ A. In addition g(b) = f (b) =
c 6∈ B, g(B) 6⊆ B. To complete the proof we only need to verify that g is
continuous, and it is enough to show that f is continuous on V0 . For any
x = xn ∈ V0 and any > 0, choose m ≥ n, m1 < . Let xn ∈ V ∈ Vm . Then
f (xn ) ∈ αm (V ). If i ≥ m, xi ∈ V , then there is W ∈ Vi with xi ∈ W ⊆ V .
Then f (xi ) ∈ αi (W ) ⊆ αm (V ). Thus d(f (xn ), f (xi )) < diam(V ) <
Hence f is continuous at xn .
1
m
< .
A space X is called hereditarily disconnected if X does not contain any
connected subset of cardinality larger than one.
Lemma 8. For every T1 connected space X with more than two elements,
there is a connected subset B ⊆ X such that 1 < |B| and B 6= X.
Proof. Suppose the only connected subsets of X are X and singletons.
Choose any a ∈ X. Then X \ {a} is not connected and hence it is a union
of two disjoint non-empty open sets U and V in X. Then cl(U ) = U ∪ {a},
otherwise, cl(U ) = U , which implies U is clopen and so X is not connected.
Similarly, cl(V ) = V ∪ {a}. By the assumptions, cl(U ) is not connected, so
cl(U ) is a union of disjoint non-empty closed sets E and F in X. Assume a 6∈
E, then E is a non-empty proper clopen subset of X as E = X \ (F ∪ cl(V )).
But this contradicts the connectedness of X. The contradiction completes
the proof.
REFLEXIVE FAMILIES OF CLOSED SETS
15
Lemma 9. Every s-reflexive Hausdorff space is hereditarily disconnected.
Proof. If X has at most two elements, it is clearly hereditary disconnected. Now assume X has more than two elements. Suppose X is not
hereditarily disconnected. If X is not connected, then as it is not hereditarily disconnected, one of its connected components, say B, is a proper
non-singleton connected subset. If X is connected, by Lemma 8 there exists
also a proper non-singleton connected subset B. Choose b1 , b2 ∈ B with
b1 6= b2 and x0 ∈ X \ B. Let
F = {F ∈ S(X) : F = {b1 } or F 3 x0 } ∪ {∅}.
Then F satisfies conditions (a),(b) and (c)in Lemma 1, and {b2 } 6∈ F.
However, {b2 } ∈ Lat(Alg(F)) which implies that X is not s-reflexive. In
fact, if f ∈ Alg(F) and b ∈ B, from {b, x0 } ∈ F it follows that f (b) = b
or f (b) = x0 . Thus B is the union of the disjoint closed sets E = {b ∈ B :
f (b) = b} and K = {b ∈ B : f (b) = x0 }. Trivially, b1 ∈ E, thus E = B
since B is connected. In particular, f (b2 ) = b2 . We are done.
Theorem 4. Let X be a locally compact metric space. Then X is sreflexive if and only X is zero-dimensional.
Proof. Noting that every metrizable locally compact space is completely metrizable (see [1, p609]) and the property of s-reflexive is topological, the theorem follows from Theorem 2, Lemma 5 and the fact that hereditarily disconnectedness, zero-dimensional and strong zero-dimensional properties are equivalent for locally compact paracompact spaces (see [1,Theorem
6.2.9]).
16
Z. YANG AND D. ZHAO
In the following we construct a locally compact countable complete metric space which has a one point extension that is not s-reflexive. We shall
define a non-reflexive family of closed sets in the space that satisfies all
conditions (a),(b) and (c), as well as (d) in Lemma 1.
Example 2. Let
1
Y1 = {( , m) : n, m = 1, 2, · · · }
n
and
1
Y0 = {( , 0) : n = 1, 2, · · · } ∪ {(0, 0)}.
n
As a subspace of R2 , Y = Y0 ∪ Y1 is a locally compact countable complete
space. Let X = Y ∪ {p} where p is an element not in Y , and define a local
base at p as follows:
for every map g : N \ D → N ∪ {0}, where D is a finite subset of N, let
1
U (g) = {p} ∪ {( , m) : n ∈ N \ D, m > g(n)}.
n
Thus ( n1 , m) 6∈ U (g) for any n ∈ D and m ∈ N ∪ {0}. Note that U (g) ∩
U (h) = U (max{g, h}), where g : N \ D1 → N ∪ {0}, h : N \ D2 → N ∪ {0}
and max{g, h} : N \ (D1 ∪ D2 ) → N ∪ {0} defined by max{g, h}(n) =
max{f (n), g(n)} for every n ∈ N \ (D1 ∪ D2 ). Thus all U (g)’s form a local
base at p. Assume Y is an open subspace of X, we thus define a topology
on X.
Now we show that X is not s-reflexive. Define a map q : X → Y0 by
q(( n1 , m)) = ( n1 , 0), q(p) = q(0, 0) = (0, 0) for any n ∈ N and m ∈ {0} ∪ N.
Clearly, q is continuous.
The family
A = {q −1 (A) : A is closed in Y0 }
REFLEXIVE FAMILIES OF CLOSED SETS
17
satisfies conditions (a),(b) and (c). In fact, (a) and (b) are clearly valid.
To see that (c) is also satisfied, consider any family {Ai : i ∈ I} of closed
S
S
S
subsets of Y0 . Then cl( i∈I q −1 (Ai )) = cl(q −1 ( i∈I Ai )) = q −1 (cl( i∈I Ai )).
But A is not reflexive. In fact, {(0, 0)} 6∈ A. For any f ∈ Alg(A), as
{(0, 0), p} = q −1 ({(0, 0)}) is in A, we have f ((0, 0)) = (0, 0) or f ((0, 0)) = p.
If the later holds, then p = lim f (( n1 , 0)). By the definition of A, f (( n1 , 0)) ∈
n→∞
q −1 ({( n1 , 0)}), suppose f (( n1 , 0)) = ( n1 , g(n)), n ∈ N. Then we have a mapping g : N → N ∪ {0}. But then f (( n1 , 0)) 6∈ U (g) for all n ∈ N, that contradicts p = lim f (( n1 , 0)). This contradiction indicates f ((0, 0)) = (0, 0),
n→∞
hence {(0, 0)} ∈ Lat(Alg(A)). Thus A is not reflexive.
Note that the only non-empty connected subsets of X are singletons, so
the family A constructed above also satisfies condition (d).
A closed set A of X is called reflexive if {∅, A, X} is reflexive. Since
{∅, X} is reflexive for every X, so both ∅ and X are reflexive.
Lemma 10. A closed set A of X is reflexive if and only if for each
B ∈ S(X) with A ⊂ B 6= X, there exists f ∈ C(X) such that f (A) ⊆ A but
f (B) 6⊆ B.
Proof. Assume A 6= ∅ and A 6= X. The necessity is trivial. To show
the sufficiency, suppose B ∈ S(X) and B 6∈ {∅, A, X}. We verify that there
exists f ∈ C(X) such that f (A) ⊆ A but f (B) 6⊆ B. If A 6⊂ B, choose
a ∈ A \ B and define f ∈ C(X) by f (x) = a for all x ∈ X. Then f
satisfies the requirement. If A ⊂ B, it follows from the assumption that
there exists f ∈ C(X) such that f (A) ⊆ A but f (B) 6⊆ B. Thus {∅, A, X}
is reflexive.
18
Z. YANG AND D. ZHAO
Proposition 1. If every path-connected component is dense in a Tychonoff space X, then every closed set of X is reflexive.
Proof. We only consider non-empty proper closed sets. Let A, B ∈
S(X) with ∅ 6= A ⊂ B 6= X. Choose a ∈ A and b ∈ B \ A. By the
assumption there exists x0 ∈ X \ B such that a and x0 are in the same
path-connected component. Define a continuous mapping g : X → [0, 1]
such that g(x) = 0 for x ∈ A and g(b) = 1. Moreover, there exists a
continuous mapping h : [0, 1] → X such that h(0) = a and h(1) = x0 since a
and x0 are in the same path-connected component. Now f = h ◦ g ∈ C(X)
satisfying f (A) ⊆ A. However, f (b) = x0 6∈ B, so f (B) 6⊆ B. Thus A is
reflexive.
Let X = [0, 1] ∪ [2, 3] be the subspace of R consisting of two disjoint
closed intervals. Then [0, 1] is a path-connect component which is not dense
in X. But one still can prove, using Proposition 1, that every closed set of
X is reflexive.
Proposition 2. Every closed set of a zero-dimensional space is reflexive.
Proof. Let X be a zero-dimensional space, and A, B ∈ S(X) with
∅ 6= A ⊂ B 6= X. Choose a ∈ A and x0 ∈ X \ A. There exists a clopen set
U 3 x0 such that U ∩ A = ∅. Consider f : X → X defined by f (x) = a
if x 6∈ U and f (x) = x0 if x ∈ U . Then f ∈ C(X) and f (A) ⊆ A but
f (B) 6⊆ B.
Remark 1. We end the paper by listing some problems.
1). If X and Y are both s-reflexive, is the cartesian product X × Y still
s-reflexive?
REFLEXIVE FAMILIES OF CLOSED SETS
19
2). In what spaces, every (finite) chain of closed sets is reflexive? We
still do not know whether every chain in R is reflexive.
3). In Example 2, we constructed a family that satisfies all conditions
(a)-(d) and is not reflexive. It would be more interesting and rich full to
investigate spaces in which every family satisfying conditions (a)-(d) is reflexive.
4). Another possible future work is to find conditions characterizing
reflexive families of continuous endomappings. Such work would be much
more complicated than that on reflexive families of closed sets.
Acknowledgement 1. The authors would like to express their sincere
thanks to the referees for their valuable comments and suggestions.
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20
Z. YANG AND D. ZHAO
(Z. Yang) Department of Mathematics, Shantou University, Shantou,
Guangdong, 515063, China P.R.
E-mail address: [email protected]
(D. Zhao) Mathematics and Mathematics Education, National Institute
of Education, Nanyang Technological University, 1 Nanyang Walk,Singapore
637616
E-mail address: [email protected]