EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 3, 2017, 561-562
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Comments on generalized closed sets with respect to an
ideal [S. Jafari and N. Rajesh, Eur. J. Pure Appl.

Math., Vol. 4(2)(2011), 147-151]

H. S. Al-Saadi

Mathematics Department, Faculty of Applied Sciences, Umm Al-Qura University
Saudi Arabia

There is an error in the proof of Theorem 5 of [1]. In fact:
Remark 1. In the proof of [1, Theorem 5] the inclusion

(cl(A) ∩ F )/(U ∩ (X/F ) ⊂ cl(A)/(U ∪ (X/F ))

is not true in general as shown by the following example.
Example 2. Let (X, τ) and I as be as in [1, Example 1], where X = {a, b, c}, τ =

{φ, {a}, {a, c}, X} and I = {φ, {b}, {c}, {b, c}}. Then the set of all Ig-closed in X is
{φ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, X}. Let A = {c}, U = {a, c} and F = {b, c}. Then
(cl(A) ∩ F )/(U ∩ (X/F )) = {b, c} and cl(A)/(U ∪ (X/F )) = {b}. Hence the inclusion in
Remark 1 is not true.

Remark 3. We provide here an alternative prove:
Theorem 4. [1, Theorem 5] Let A be an Ig-closed set and F be a closed set in

(X, τ), then A ∩ F is an Ig-closed set in (X, τ). Proof. Let A ∩ F ⊂ U and U is

open. Then A ⊂ U ∪ (X/F ). Since A is Ig-closed, we have cl(A)/(U ∪ (X/F )) ∈ I. Now,
cl(A ∩ F ) ⊂ cl(A) ∩ F = (cl(A) ∩ F )/(X/F ). Therefore,

cl(A ∩ F )/U ⊂ (cl(A) ∩ F )/U
= cl(A) ∩ F ∩ (X/U)

= cl(A) ∩ (X/(U ∪ (X/F )))

= cl(A)/(U ∪ (X/F )) ∈ I.

Hence cl(A ∩ F )/U ∈ I and A ∩ F is Ig-closed in (X, τ).

Email addresses: hasa112@hotmail.com (H. S. Al-Saadi)

http://www.ejpam.com 561 c© 2017 EJPAM All rights reserved.



H. S. Al-Saadi / Eur. J. Pure Appl. Math, 10 (3) (2017), 561-562 562

Acknowledgement

The useful discussion with Prof. T. Noiri is appreciated.
References

[1 ] S. Jafari and N. Rajesh, Generalized closed Sets with Respect to an Ideal, Eur. J.
Pure Appl. Math., Vol. 4(2)(2011), 147-151.