DISCONNECTED SPACES
ZBIGNIEW DUSZY ´NSKI
We give several characterizations of extremally disconnected spaces by utilizingb- open and regular open sets. The paper is closely related to [D. Sivaraj,A note on extremally disconnected spaces. Indian J. Pure Appl. Math. 17(12) (1986), 1373–
1375] and [T. Noiri,Characterization of extremally disconnected spaces. Indian J.
Pure Appl. Math. 19(4) (1988), 325–329].
AMS 2010 Subject Classification: 54C08.
Key words: α-open, preopen,b-open, semi-preopen, regular open sets, extremally disconnected space.
1. INTRODUCTION
Exremally disconnected spaces, studied in this paper, are topological spaces in which the closure of every open subset is still open [10] (no separa- tion axiom is assumed). For a brief discussion on the role of such spaces in various branches of mathematics, the reader is referred (for instance) to [5, Introduction]. Over the last 25 years several characterizations of extremally disconnected spaces were given, see [11, Theorem 2.9], [12, Proposition 4.1], [23], [21], [2, Section 5], [8]. Characterizations for extremal disconnectedness are known also for the case of bitopological spaces [5, Section 4].
Our purpose is to give further such characterizations.
2. PRELIMINARIES
Throughout this paper, (X, τ) will denote a topological space and int (·) (resp. cl (·)) will stand for the interior (resp. the closure) operator with respect toτ [for a space (X, τα) (see below) these operators will be denoted by intτα(·) and clτα(·) respectively]. Let S be a subset of a space (X, τ). The S is said to be regular open (resp. regular closed) if S = int (cl (S)) (resp. S = cl (int (S))). TheS is said to beα-open[18] [resp. semi-open [13];preopen [17]; b-open[4] (equiv. γ-open [9] orsp-open [7]); semi-preopen [3] (equiv.
β-open [1])] in (X, τ), if S ⊂ int (cl (int (S))) [resp. S ⊂ cl (int (S)); S ⊂
REV. ROUMAINE MATH. PURES APPL.,56(2011),4, 295–302
int (cl (S)); S ⊂cl (int (S))∪int (cl (S)); S ⊂cl (int (cl (S)))]. The collection of all regular open (resp. regular closed, closed,α-open, semi-open, preopen,b- open, semi-preopen) subsets of (X, τ) is denoted by RO (X, τ) (resp. RC (X, τ), C (X, τ), τα, SO (X, τ), PO (X, τ), BO (X, τ), SPO (X, τ)). It is well-known that for any space (X, τ) the family τα forms a topology on X [18]. Also, the following inclusions hold for each (X, τ): τ ⊂τα= SO (X, τ)∩PO (X, τ) [20], SO (X, τ)∪PO (X, τ) ⊂ BO (X, τ) ⊂ SPO (X, τ) [4]. A set S is said to be semi-closed [6] (resp. preclosed, b-closed [4], semi-preclosed [3]) in (X, τ), ifX\S∈SO (X, τ) (resp. X\S ∈PO (X, τ),X\S∈BO (X, τ),X\S∈ SPO (X, τ)). The family of all semi-closed (resp. semi-preclosed) subsets of (X, τ) is denoted by SC (X, τ) (resp. SPC (X, τ)). The operators of semi- closure and semi-interior [6] (denoted as scl (·) and sint (·), respectively), preclosure (pcl (·)), b-closureand b-interior [4] (bcl (·) andbint (·), resp.), semi-preclosure and semi-preinterior[3] (spcl (·) and spint (·), resp.) are defined in a manner similar to the definitions of ordinary closure and interior – compare [14] and [15]. The set of all points x of (X, τ) such that S ∩ int (cl (G))6=∅ for any neigbourhoodG∈τ of x, is called the δ-closure ofS [24] and is denoted as clδ(S). We recall below some relations from [3] that are to be used in the sequel: intτα(S) =S∩int (cl (int (S))) [3, Theorem 1.5(d)], scl (S) = S ∪int (cl (S)) [3, Theorem 1.5(a)], pcl (S) = S ∪cl (int (S)) [3, Theorem 1.5(e)], spcl (S) =S∪int (cl (int (S))) [3, Theorem 2.15].
LetXbe a nonempty set. A subfamilyM ⊂2X is said to be aminimal structure [22] (see [14] and [15]) if ∅, X ∈ M. A space (X, τ) is said to be extremally disconnected [10] if cl (S)∈τ for each S∈τ.
3. RESULTS
In [4, Proposition 2.5(1)] it has been shown that for any subset S of an (X, τ),
bcl (S) = scl (S)∩pcl (S).
With the use of [3, Theorem 1.5(a),(e)], one easily obtains the lemma below.
Lemma 1. Let S⊂X, (X, τ) is arbitrary. Then, (1)bcl (S) =S∪ int (cl (S))∩cl (int (S))
, (2)bint (S) =S∩ int (cl (S))∪cl (int (S)) .
Lemma 2. Let (X, τ) be arbitrary. We have
(1)bcl (S) = scl (S) = spcl (S) for each S∈SO (X, τ), (2)bint (S) = sint (S) = spint (S) for each S ∈SC (X, τ).
Proof. (1) Since S ∈ SO (X, τ) if and only if cl (S) = cl (int (S)) [19, Lemma 2], we have by (1) of Lemma 1 that
bcl (S) =S∪ int (cl (S))∩cl (S)
= scl (S), and on the other hand,
bcl (S) =S∪ int (cl (int (S)))∩cl (int (S))
= spcl (S).
(2) follows immediately by (1).
Theorem 1. The following statements are equivalent in any (X, τ):
(1) (X, τ) is e.d.
(2)For every S ∈SO (X, τ), bcl (S)∈τ. (3)For every S ∈SO (X, τ), bcl (S)∈C (X, τ).
(4)For every S ∈SO (X, τ), spcl (S)∈τ. (5)For every S ∈SO (X, τ), spcl (S)∈C (X, τ).
(6)For every S ∈SO (X, τ), bcl (S)∈τ ∩C (X, τ).
(7)For every S ∈SO (X, τ), spcl (S)∈τ ∩C (X, τ).
Proof. This follows directly from [23, Theorem 2.1(iii),(viii)] and Lem-
ma 2(1).
Remark1. It is well-known that in every space (X, τ), scl (S) = int (cl (S)) for each S ∈ PO (X, τ) [12, Proposition 2.7(a)]. Thus, scl (S) ∈ τ for each S ∈ τ, or S ∈ τα, or S ∈ PO (X, τ), in any e.d. space (X, τ). But, none of the three latter conditions suffices to characterize e.d. spaces. Indeed, it is enough to look at the real line equipped with the Euclidean topology.
Remark 2. Since S ∈ SO (X, τ) if and only if sint (S) = S, and S ∈ SC (X, τ) if and only if scl (S) =S [6, Theorem 1.4], Lemma 2 may be refor- mulated equivalently as follows:
(10) bcl (sint (S)) = scl (sint (S)) = spcl (sint (S)) for eachS⊂X, (20) bint (scl (S)) = sint (scl (S)) = spint (scl (S)) for eachS ⊂X.
(compare (10) and (20) with [4, Proposition 2.6(3)–(4)]). Thus, by [23, Theo- rem 2.1(iii)] (see also Theorem 1(4)) it follows all the conditions below are equivalent:
(1) (X, τ) is e.d.
(2) For everyS⊂X, scl (sint (S))∈τ. (3) For everyS⊂X, spcl (sint (S))∈τ. (4) For everyS⊂X,bcl (sint (S))∈τ. (5) For everyS⊂X, sint (scl (S))∈C (X, τ).
(6) For everyS⊂X, spint (scl (S))∈C (X, τ).
(7) For everyS⊂X,bint (scl (S))∈C (X, τ).
Theorem 2. The following are equivalent in any space (X, τ).
(1) (X, τ) is e.d.
(2)For every S ∈SO (X, τ), bcl (S) = cl (S).
(3)For every S ∈SC (X, τ), bint (S) = int (S).
(4)For every S ∈SO (X, τ), bcl (S) = clδ(S).
(5)For every S ∈SO (X, τ), spcl (S) = cl (S).
(6)For every S ∈SC (X, τ), spint (S) = int (S).
(7)For every S ∈SO (X, τ), spcl (S) = clδ(S).
Proof. (1) ⇔ (2). Follows by [23, Theorem 2.1(vii)] and Lemma 2(1).
(2) ⇔ (4). Use [23, Lemma 1.4]. The other equivalences require a respective use of Lemma 2.
Lemma 3. S∈SPO (X, τ) if and only if cl (S) = cl (int (cl (S))).
Proof. Clear – compare the proof of [21, Theorem 3.1].
Lemma 4. Let S be a subset of (X, τ). Then:
(1) [4, Proposition 2.6(1)]. bcl (int (S)) = int (bcl (S)) = int (cl (int (S))), (2)bcl intτα(S)
= intτα(bcl (S)) = int (cl (int (S))),
(3) [4, Proposition 2.6(2)]. bint (cl (S)) = cl (bint (S)) = cl (int (cl (S))), (4)bint clτα(S)
= clτα(bint (S)) = cl (int (cl (S))).
Proof. (2). We shall show thatbcl intτα(S)
= int (cl (int (S))). By Lem- ma 1(1) we have
bcl intτα(S)
= intτα(S)∪
int cl intτα(S)
∩cl int intτα(S) . Using [3, Theorem 1.5(d)] and [20, Lemma 3.5] we obtain
int cl intτα(S)
= int cl S∩int (cl (int (S)))
= int (cl (S))∩int (cl (int (S))) = int (cl (int (S))).
Furthermore, we have cl int intτα(S)
= cl int (S)∩int (cl (int (S)))
= cl (int (S)).
Finally, one gets bcl intτα(S)
= intτα(S)∪ int (cl (int (S)))∩cl (int (S))
= intτα(S)∪int (cl (int (S))) = int (cl (int (S))).
Now, it suffices to show intτα(bcl (S)) = int (cl (S)). Notice that the family BO (X, τ) is a minimal structure, so by [14, Lemma 2.3(iii)] (or [16, Lemma 3.3];
see also [15, Lemma 2.3]) we havebcl (bcl (S)) =bcl (S) for any subsetS ⊂X.
Since int (bcl (S)) = int (cl (int (S))) (see (1)), from the latter observation we get
int (bcl (bcl (S))) = int (cl (int (bcl (S)))).
Hence bcl (S) ∩ int (bcl (S)) = bcl (S) ∩ int (cl (int (bcl (S)))) and therefore int (bcl (S)) = intτα(bcl (S)).
Theorem 3. The following are equivalent for any (X, τ):
(1) (X, τ) is e.d.
(2)For every S ∈SPO (X, τ), bint (cl (S))∈τ. (3)For every S ∈SPO (X, τ), cl (bint (S))∈τ. (4)For every S ∈SPO (X, τ), bint (clτα(S))∈τ. (5)For every S ∈SPO (X, τ), clτα(bint (S))∈τ. (6)For every S ∈SPC (X, τ), bcl (int (S))∈C (X, τ).
(7)For every S ∈SPC (X, τ), int (bcl (S))∈C (X, τ).
(8)For every S ∈SPC (X, τ), bcl (intτα(S))∈C (X, τ).
(9)For every S ∈SPC (X, τ), intτα(bcl (S))∈C (X, τ).
Proof. Due to Lemma 4 and with taking the complement X\S, it is enough to show (1)⇔(2). But this equivalence holds by Lemma 3, the identity bint (cl (S)) = cl (int (cl (S))) (Lemma 4(3)), and by [21, Theorem 3.1(b)].
Theorem 4. Let (X, τ) be an e.d. space. If S1, S2 ∈SO (X, τ) then (1) scl (S1∩S2) = scl (S1)∩scl (S2),
(2)bcl (S1∩S2) =bcl (S1)∩bcl (S2), (3) spcl (S1∩S2) = spcl (S1)∩spcl (S2), (4)S1∩S2=∅ implies scl (S1)∩scl (S2) =∅, (5)S1∩S2=∅ implies bcl (S1)∩bcl (S2) =∅, (6)S1∩S2=∅ implies spcl (S1)∩spcl (S2) =∅,
Proof. As in each e.d. space (X, τ),τα = SO (X, τ) [11, Theorem 2.9], by Lemma 2(1) only the case (1) requires our attention. By [11, Theorem 2.9(c)]
we have int (cl (S1∩S2)) = int cl (S1)∩cl (S2)
for any S1, S2 ∈ SO (X, τ).
So, utilizing [12, Proposition 2.7(a)] the identity follows.
Remark 3. It should be noticed that even if each of the properties (1) thru (6) of Theorem 4 holds for every couple of α-open sets S1 and S2, the space need not be e.d. Indeed, this is so for (X, τ) where X = {a, b, c} and τ =
∅, X,{a},{b},{a, b} (details left to the reader).
Theorem 5. For any space (X, τ) the following are equivalent:
(1) (X, τ) is e.d.
(2) cl (S)∈RO (X, τ) for each S∈SPO (X, τ).
(3) cl (S)∈RO (X, τ) for each S∈BO (X, τ).
(4) cl (S)∈RO (X, τ) for each S∈SO (X, τ).
(5) cl (S)∈RO (X, τ) for each S∈PO (X, τ).
(6) cl (S)∈RO (X, τ) for each S∈τα. (7) cl (S)∈RO (X, τ) for each S∈τ.
(8) cl (S)∈RO (X, τ) for each S∈RO (X, τ).
(9) cl (S1∩S2) = cl (S1)∩cl (S2) for each S1, S2 ∈RO (X, τ).
(10) cl (S1∩S2) = cl (S1)∩cl (S2) for each S1, S2 ∈τ.
Proof. (1)⇒(2). LetS∈SPO (X, τ). By [21, Theorem 3.1(2)], cl (S) = int (cl (S)) = int (cl (cl (S))).
(2)⇒ (3)⇒ (4)⇒ (5) ⇒(6) ⇒ (7)⇒ (8). Obvious.
(8)⇒(9). LetS1, S2 ∈RO (X, τ). By hypothesis we have
cl (S1)∩cl (S2) = cl (cl (S1)∩cl (S2)) = cl (int (cl (S1))∩int (cl (S2))).
So, using for instance [20, Lemma 3.5] we get
cl (S1)∩cl (S2) = cl (int (cl (S1∩S2))) = cl (S1∩S2).
(9) ⇒ (10). Let S1, S2 ∈ τ. Then int (cl (S1)),int (cl (S2))∈ RO (X, τ) and by (9),
cl (int (cl (S1))∩int (cl (S2))) = cl (int (cl (S1)))∩cl (int (cl (S2))).
Hence, making use of Lemma 3 and again [20, Lemma 3.5] we obtain cl (S1∩S2)
= cl (int (cl (S1∩S2))) = cl (S1)∩cl (S2).
(10)⇒ (1). See [11, Theorem 2.9].
Remark 4. It is worth being remarked that recently Zorlutuna [25, p. 943]
redefined e.d. spaces as follows: (X, τ) is e.d. ifS ∈τ for everyS∈RC (X, τ).
We advise the reader to reframe conditions (2)–(9) of Theorem 5 for the class RC (X, τ) instead of RO (X, τ) (by takings complements ofS, like in Remark 2 for instance).
The last two results improve Noiri’s [21, Theorems 3.3 and 3.6].
Theorem 6. The following are equivalent for any (X, τ):
(a) (X, τ) is e.d.
(b)If S1∈SPO (X, τ), S2∈SO (X, τ), thenS1∩S2 =∅impliescl (S1)∩ cl (S2) =∅.
(c) If S1∈ F1, S2∈ F2, then S1∩S2 =∅ implies cl (S1)∩cl (S2) =∅.
(d)If S1, S2 ∈RO (X, τ), then S1∩S2=∅ implies cl (S1)∩cl (S2) =∅.
where for F1 and F2 the following couples can be put in: (1) BO (X, τ), SO (X, τ); (2) PO (X, τ),SO (X, τ);(3) SO (X, τ),SO (X, τ); (4)τα,SO (X, τ);
(5)τ,SO (X, τ); (6) RO (X, τ),SO (X, τ); (7) SPO (X, τ),τα; (8) BO (X, τ), τα; (9) PO (X, τ), τα; (10) τα, τα; (11) τ, τα; (12) RO (X, τ), τα; (13) SPO (X, τ),τ;(14) BO (X, τ),τ;(15) PO (X, τ),τ; (16)τ,τ; (17) RO (X, τ), τ; (18) SPO (X, τ), RO (X, τ); (19) BO (X, τ), RO (X, τ); (20) PO (X, τ), RO (X, τ).
Proof. (a) ⇒ (b). See the proof of [21, Theorem 3.3]. (b) ⇒ (c) ⇒ (d). These are obvious for any case (1)–(20). (d) ⇒ (a). Suppose (d) holds
and (X, τ) is not e.d. By Theorem 5 there exist sets A, B ∈ RO (X, τ) with cl (A∩B) cl (A)∩cl (B). But ifA∩B =∅, then by (d) cl (A)∩cl (B) =∅.
A contradiction.
Theorem 7. The following are equivalent for any (X, τ):
(a) (X, τ) is e.d.
(b) IfS1∈SPO (X, τ),S2 ∈SO (X, τ), thencl (S1∩S2) = cl (S1)∩cl (S2).
(c) If S1 ∈ F1, S2 ∈ F2, then cl (S1∩S2) = cl (S1)∩cl (S2).
(d) If S1, S2∈RO (X, τ), then cl (S1∩S2) = cl (S1)∩cl (S2).
where F1 and F2 can be any of the couples (1)–(20) of Theorem 6.
Proof. (a)⇒ (b). See the proof of [21, Theorem 3.6]. (b) ⇒ (c) ⇒(d).
Obvious for any case (1)–(20). (d) ⇔ (a). Theorem 5.
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Received 18 January 2012 “Casimirus the Great” University Institute of Mathematics
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