Journal Name:
- European Journal of Pure and Applied Mathematics
Author Name | University of Author | Faculty of Author |
---|---|---|
Abstract (2. Language):
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) n F)/(U n (X/F) c c1(A)/(U U (X/F))
is not true in general as shown by the following example.
Example 2. Let (X, T) and I as be as in [1, Example 1], where X = [a,b,c},r = {0, {a}, {a,c},X} and I = {0, {b}, {c}, {b, c}}. Then the set of all Ig-closed in X is {0, {a}, {b}, {c}, {a,b}, {a,c}, {b,c},X}. Let A = {c},U = {a, c} and F = {b, c}. Then (cl(A) n F)/(U n (X/F)) = {b,c} and c1(A)/(U 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, T), then A n F is an Ig-closed set in (X,T). Proof. Let A n F c U and U is
open. Then A c U U (X/F). Since A is Ig-closed, we have c1(A)/(U U (X/F)) e I. Now, c1(A n F) c cl(A) n F = (cl(A) n F)/(X/F). Therefore,
cl(A n F)/U c (cl(A) n F)/U = cl(A) n F n (X/U )
= cl(A) n (X/(U U (X/F))) = c1(A)/(U U (X/F)) e I.
Hence c1(A n F)/U e I and A n F is Ig-closed in (X, T).
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