(2) IfX is a space withe(X)≥ω1, thenA(X) is not discretely absolutely star-Lindel¨of

MATEMATIQKI VESNIK

63, 3 (2011), 219–222 September 2011

originalni nauqni rad research paper

A NOTE ON DISCRETELY ABSOLUTELY STAR-LINDEL ¨OF SPACES

Yan-Kui Song

Abstract. In this paper, we prove the following statements:

(1) If X is a normal discretely absolutely star-Lindel¨of space withe(X) < ω1, then the Alexandroff duplicateA(X) ofX is discretely absolutely star-Lindel¨of.

(2) IfX is a space withe(X)≥ω1, thenA(X) is not discretely absolutely star-Lindel¨of.

The two statements answer a question raised by Song.

1. Introduction

By a space, we mean aT1 topological space. Recall that a spaceX is star- Lindel¨of (see [2, 4] under different names) (discretely star-Lindel¨of) (see [7, 9]) if for every open coverU ofX, there exists a countable subset (a countable discrete closed subset, respectively) F of X such that St(F,U) = X, where St(F,U) = S{U ∈ U :U∩F 6=∅}. It is clear that every separable space and every discretely star-Lindel¨of space is star-Lindel¨of as well as every space of countable extent, in particular, every countably compact space or every Lindel¨of space.

A space X is absolutely star-Lindel¨of (see [1, 5]) (discretely absolutely star- Lindel¨of) (see [6, 8]) if for every open coverU ofX and every dense subspaceDof X, there exists a countable subsetF of D such that St(F,U) =X (F is discrete and closed inX such thatSt(F,U) =X, respectively).

From the above definitions, it is not difficult to see that every absolutely star- Lindel¨of space is star-Lindel¨of, every discretely absolutely star-Lindel¨of space is absolutely star-Lindel¨of and every discretely absolutely star-Lindel¨of space is dis- cretely star-Lindel¨of.

Song [8] constructed an example showing that there exists a Tychonoff dis- cretely absolutely star-Lindel¨of spaceX withe(X) = csuch that the Alexandroff

2010 AMS Subject Classification: 54A25, 54D20.

Keywords and phrases: Star-Lindel¨of; absolutely star-Lindel¨of; discretely absolutely star- Lindel¨of.

The author acknowledges support from the National Science Foundation of Jiangsu Higher Education Institutions of China (Grant No 07KJB110055)

219

220 Yan-Kui Song

duplicateA(X) of X is not discretely absolutely star-Lindel¨of, and asked the fol- lowing question:

Question. Does there exist a normal discretely absolutely star-Lindel¨of space X such thatA(X) is not discretely absolutely star-Lindel¨of?

The purpose of this paper is to answer the question by showing two statements stated in the abstract.

Moreover, the cardinality of a set Ais denoted by |A|. The extent e(X) of a spaceX is the smallest infinite cardinalκsuch that every discrete closed subset of a space X has cardinality at most κ. Letω denote the first infinite cardinal and ω1 the first uncountable cardinal. Other terms and symbols that we do not define will be used as in [3].

2. Some results on discretely absolutely star-Lindel¨of spaces For a space X, recall that the Alexandroff duplicate A(X) of a space X, de- noted byA(X), is constructed in the following way: The underlying set ofA(X) is X× {0,1}and each point ofX× {1}is isolated; a basic neighbourhood of a point hx,0i ∈ X× {0} is a set of the form (U × {0})∪((U × {1})\ {hx,1i}), whereU is a neighborhood of xof X. It is well-known that A(X) is compact (countably compact, Lindel¨of) if and only if so isX andA(X) is Hausdorff (regular, Tychonoff, normal) if and only if so isX.

Theorem 2.1. If X is a normal discretely absolutely star-Lindel¨of space X withe(X)< ω1, thenA(X)is discretely absolutely star-Lindel¨of.

Proof. We prove thatA(X) is discretely absolutely star-Lindel¨of. To this end, letU be an open cover ofA(X). Clearly every point ofX× {1}is isolated. LetE be the set of all isolated points ofX, and let

D= (X× {1})∪(E× {0}).

ThenD is a dense subspace of A(X) and every dense subset ofA(X) containsD.

Thus it is sufficient to show that there exists a countable subsetF ⊆Dsuch thatF is discrete closed inA(X) andSt(F,U) =A(X). For everyx∈X, we pick an open neighborhood (Ux× {0,1})\ {hx,1i}ofhx,0isuch that (Ux× {0,1})\ {hx,1i} ⊆U for someU ∈ U, where Uxis an open subset ofX containingx. If we put

V={U_x:x∈X}.

ThenV is an open cover ofX. Hence there exists a countable subsetF0⊆X such thatF0is discrete closed inX andSt(F0,V) =X, sinceX is discretely absolutely star-Lindel¨of and every discretely absolutely star-Lindel¨of space is discretely star- Lindel¨of. For the collection{Ux:x∈F0} ofX, sinceF0is a discrete closed subset of a normal spaceX, then there exists a pairwise disjoint open family{Vx:x∈F0} in X such that x∈ Vx ⊆Ux for each x∈F0. By normality of X there exists an open subsetV ofX such that

F0⊆V ⊆V ⊆ [

x∈F0

Vx.

Obviously,{V ∩Vx:x∈F0} is a discrete family of nonempty open subsets ofX.

A note on discretely absolutely star-Lindel¨of spaces 221 Let

F₁⁰ ={x∈F0:xis not isolated inX}.

For eachx∈F₁⁰, we pickyx∈V ∩Vxsuch thatx6=yx. Then {x:x∈F0} ∪ {yx:x∈F₁⁰} is discrete closed inX and

hyx,1i,hx,0i ∈(Ux× {0,1})\ {hx,1i}for eachx∈F₁⁰.

LetF1=F0× {1}. For eachx∈X\(F0∪ {Ux:x∈F₁⁰}), there existsx⁰∈X such thatx∈Ux⁰ andUx⁰∩F06=∅, hence (Ux⁰× {0,1} \ {hx⁰,1i} ∩F16=∅. Let

F2=F1∪ {hyx,1i:x∈F₁⁰} ∪ {(F0\F₁⁰)× {0}).

Then F2 is a countable discrete closed (in A(X)) subset of D and X × {0} ⊆ St(F2,U). Let F3 =A(X)\St(F1,U). ThenF3 is a discrete and closed subset of A(X) andF3⊆D. Sincee(X)< ω1, then e(A(X))< ω1. ThusF3is countable.

If we putF =F2∪F3, thenF is a countable discrete closed (inA(X)) subset ofD such thatA(X) =St(F,U), which completes the proof.

Theorem 2.2. If X is a space withe(X)≥ω1, then A(X) is not discretely absolutely star-Lindel¨of.

Proof. Since e(X) ≥ ω₁, then there exists a discrete closed subset B of X such that|B| ≥ω1, henceB× {1} is a closed and open subset ofA(X) and every point ofB× {1}is an isolated point ofA(X). To show thatA(X) is not discretely absolutely star-Lindel¨of. LetC be the set all isolated points ofX. Let us consider the open cover

U ={A(X)\(B× {1})} ∪ {hx,1i:x∈B}

and the dense subset

D= (C× {0})∪(X× {1})

of A(X). Then, for any countable subset F ofD there exists a point x∈B such that hx,1i∈/ F, since |B| ≥ω1. Hence hx,1i∈/ St(F,U), since{hx,1i} is the only element ofU containinghx,1ifor eachx∈B, which completes the proof.

Acknowledgements. The author would like to thank Prof. R. Li for his kind help and valuable suggestions.

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222 Yan-Kui Song

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(received 05.06.2010; in revised form 24.09.2010)

School of Mathematical Science, Nanjing Normal University, Nanjing 210046, P.R. China E-mail:[email protected]