MATEMATIQKI VESNIK 49 (1997), 81–84

UDK 515.122.23 originalni nauqni rad

research paper

CORRECTIONS TO “SOME GENERALIZATIONS OF T_{D}-SPACES” AND “A GENERALIZATION OF NORMAL SPACES”

Bhamini M. P. Nayar

Abstract. Some corrections to the papers “Some Generalizations of T_{D}-Spaces (Mat. Vesnik 34 (1982), 221–230)” and “A Generalization of Normal Spaces (ibid. 35 (1983), 1–10)” are given.

1. Semi-T_{D} spaces [1]

The first part of the proofs of Theorems 1.6 and 2.4 of [1] are incorrect. The following two Theorems and their proofs provide the statements and proofs of the first parts of the Theorems 1.6 and 2.4 of [1]. The proofs of Theorems 1.6 and 2.4 [1] given in the paper prove the second parts of the statements of the Theorems.

Theorem 1.1. A semi-T_{1} space is semi-T_{D}.

Proof. Let X be a semi-T_{1} space and x ∈ X. Then either {x} is nowhere dense or {x} ⊆ int(cl{x}) [3] where int A and cl A respectively denote the interior and closure of a set A. If {x} is nowhere dense, cl(X − cl{x}) = X and therefore, X − cl{x} ⊆ X − d{x} ⊆ cl(X − cl{x}). So, X − d{x} is semi-open, d{x} being the derived set of {x}. If {x} is not nowhere dense, {x} ⊆ int(cl{x}) = scl{x}, the semiclosure of {x} [3]. Since X is semi-T_{1}, {x} = int(cl{x}). Hence d{x} = cl{x}−{x}, the boundary of an open set and hence a nowhere dense set. Therfore X − d{x} is semi-open. Thus d{x} is semi-closed.

Theorem 1.2. A pairwise semi-T_{1} space is pairwise semi-T_{D}.

Proof. Let (X, T_{1}, T_{2}) be pairwise semi-T_{1}. Then (X, T_{1}) and (X, T_{2}) are semi-T_{1} and hence both are semi-T_{D}, in view of the above Theorem. For x ∈ X, T_{1}- d{x} is T_{1}-semi-closed and T_{2}-d{x} is T_{2}-semi-closed. Suppose y ∉ (T_{1}-d{x}∩T_{2}- d{x}). If y ∉ (T_{1}-d{x} ∪ T_{2}-d{x}), then there is a T_{1}-semi-open set U and a T_{2}-semi-open set V, each containing y, and each having empty intersection with (T_{1}-d{x}∩T_{2}-d{x}). Now suppose y ∈ (T_{1}-d{x}∪T_{2}-d{x}) and y ∉ (T_{1}-d{x}∩T_{2}- d{x}). Suppose y ∈ T_{1}-d{x} and y ∉ T_{2}-d{x}. Then there is a T_{2}-semi-open set V_{2} containing y such that V_{2} ∩ (T_{1}-d{x} ∩ T_{2}-d{x}) = ∅. Since y ∈ T_{1}-d{x}, (X − T_{1}-d{x}) ∪ {y} is a T_{1}-semi-open set containing y. For, X − T_{1}-d{x} is T_{1}-semi-open and every T_{1}-open set containing y contains x and x ∈ (X − T_{1}- d{x}). Hence y ∈ T_{1}-cl(X − T_{1}-d{x}). Thus (X − T_{1}-d{x}) ∪ {y} is T_{1}-semi-open and ((X − T_{1}-d{x}) ∪ {y}) ∩ (T_{1}-d{x} ∩ T_{2}-d{x}) = ∅, since y ∉ T_{2}-d{x} and (X − T_{1}-d{x}) ∩ T_{1}-d{x} = ∅. The case when y ∈ T_{2}-d{x} and y ∉ T_{1}-d{x} can be similarly handled. Therefore T_{1}-d{x} ∩ T_{2}-d{x} is both T_{1}-semi-closed as well as T_{2}-semi-closed. Thus (X, T_{1}, T_{2}) is semi-T_{D}.

AMS Subject Classification: 54 D 15

The following example of Prof. Hindman of the Depatrment of Mathematics, Howard University, Washington, D. C. shows that arbitrary product of semi-T_{D} spaces need not be semi-T_{D} and hence Theorem 1.15 and its analogue in bitopological spaces, Theorem 2.8, are incorrect.

Let X = {0, 1}, and T = {∅, {0}, X}. For each n ∈ N, let X_{n} = X and let Y = ∏ X_{n}. X is T_{D} and hence is a semi-T_{D} space. The derived set of {(0)} is Y − {(0)} which is not semi-closed in Y.

However, as we shall prove in the following theorem, the product of finitely many semi-T_{D} spaces is semi-T_{D}. We shall state the following Lemma without proof.

Lemma 1.3. If A, B, C and D are any four sets, then A × B − C × D = ((A − C) × B) ∪ (A × (B − D)).

Theorem 1.4. If X and Y are two semi-T_{D} spaces, then X × Y is semi-T_{D}.

Proof. Let (x, y) ∈ X × Y. Then d{(x, y)} = cl{(x, y)} − {(x, y)} = cl({x} × {y}) − ({x} × {y}) = ((cl{x} − {x}) × cl{y}) ∪ cl{x} × (cl{y} − {y}), in view of the above Lemma. Therefore, d{(x, y)} = (d{x} × cl{y}) ∪ (cl{x} × d{y}). Now X × Y − d{(x, y)} = X × Y − ((d{x} × cl{y}) ∪ (cl{x} × d{y})) = [((X − d{x}) × Y) ∪ (X × (Y − cl{y}))] ∩ [((X − cl{x}) × Y) ∪ (X × (Y − d{y}))] = ((X − cl{x}) × Y) ∪ ((X − d{x}) × (Y − d{y})) ∪ ((X − cl{x}) × (Y − cl{y})) ∪ (X × (Y − cl{y})). This is a union of three open sets and one semi-open set and hence is semi-open. Thus d{(x, y)} is semi-closed.

Corollary 1.5. Product of a finite collection of semi-T_{D} spaces is semi-T_{D}.

2. s-Normal spaces

There are counterexamples to show that Lemma 7 in [2] is incorrect. To state the correct proposition we need the following definition.

Definition 2.1 [2] A real valued function f on X is said to be quasi-lower semi-continuous, denoted as q-l.s.c., (respectively, quasi-upper semi-continuous, denoted q-u.s.c.) if the set {x : f(x) > b} (respectively, {x : f(x) < b}) is a semi-open subset of X where b is a real number.

Lemma 2.2 Let D be any dense subset of the space of positive real numbers with relative usual topology. If to each t ∈ D there corresponds a semi-open subset U_{t} of a space X such that t < s in D implies that scl U_{t} ⊆ U_{s} and ⋃_{t∈D} U_{t} = X, then the function f defined as f(x) = inf{t : x ∈ U_{t}} is quasi-upper semmi-continuous and quasi-lower semi-continuous.

The proof of Lemma 7 in [2] proves that the function satisfying the hypothesis of the Lemma is in fact q-l.s.c. and q-u.s.c.

The purpose of Lemma 7 in [2] is to prove an analogue of Urysohn’s Lemma for s-normal spaces. However, here we shall prove analogues of Urysohn’s Lemma for s-normal spaces and for semi-normal spaces without using the above Lemma. Thus the change in Lemma 7 in [2] does not affect Theorem 8 or any result consequential to Theorem 8 in [2].

Theorem 2.3. A space X is s-normal if and only if for any two disjoint semiclosed subsets A and B of X, there exists a semi-continuous function f: X → [0, 1] such that f(x) = 0 for every x ∈ A and f(x) = 1 for every x ∈ B.

Proof. Let A and B be two disjoint semi-closed subsets of X and let X be s-normal. Then there are disjoint semi-open sets U and V such that A ⊆ U and B ⊆ V. Consider the characteristic function χ_{scl V}. χ_{scl V}^{−1} (a, 1] = scl V, a semi-open

set, being the semi-closure of a semi-open set. χ_{scl V}^{−1} [0, a) = X − scl V, a semi-open

set and χ_{scl V}^{−1} (a, b) = ∅ where 0 < a < b < 1. Also χ_{scl V} (A) = 0 and χ_{scl V} (B) = 1 and χ_{scl V} is semi-continuous.

The converse is easy to prove.

Theorem 2.4. A space X is semi-normal if and only if for any two disjoint closed subsets A and B of X, there is a semi-continuous function f: X → [0, 1] such that f(A) = 0 and f(B) = 1.

Proof. Let X be semi-normal and let A and B be two disjoint closed subsets of X. Then there are disjoint semi-open sets U and V such that A ⊆ U and B ⊆ V. Now, scl V and scl U are disjoint semi-open sets and if h is the characteristic function of scl V, then h is semi-continuous, h(A) = 0 and h(B) = 1.

Converse is easy to prove.

The proof of Theorem 19 in [2] proves only the following and hence the Theorem 19 should be stated as follows:

Theorem 2.5. If X is an s-normal space and f and g are functions on X such that f is q-l.s.c. and g is q-u.s.c. and g(x) ≤ f(x) for every x ∈ X, then there is a q-u.s.c. and q-l.s.c. function h on X such that g(x) ⩽ h(x) ⩽ f(x).

REFERENCES

[1] S. P. Arya and M. P. Bhamini, Some generalizations of T_{D}-spaces, Mat. Vesnik 6(19)(34) (1982), 221–230.

[2] S. P. Arya and M. P. Bhamini, A generalization of normal spaces, Mat. Vesnik 35 (1983), 1–10.

[3] D. S. Janković and I. L. Reilly, On semi-separation properties, Indian J. Pure and Appl. Math., 16 (9) (1985), 957–964.

(received 10.09.1994, in revised form 20.01.1997.)

Department of Mathematics, University of the District of Columbia, Washington, D. C. 20008, U.S.A.