3. On Neutrosophic -Closed Sets
Definition 8. A neutrosophic -closed (-closed) set is defined as if whenever and G is an -open set in . Its complement is called a neutrosophic -open (-open) set.
Definition 9. Let U be an NS in NTS X. Then
= is an OS in X and O is said to be a neutrosophic -interior of U;
= is an CS in X and O is said to be a neutrosophic -closure of U.
Theorem 1. All -closed sets and N-closed sets are -closed sets.
Proof. Let U be an -closed set, then . Let , where G is -open. Since U is -closed, . Thus, U is -closed. ☐
Theorem 2. Every Nsemi-closed set in a neutrosophic set is an -closed set.
Proof. Let U be an Nsemi-closed set in , then . Let , where G is -open in . Since U is Nsemi-closed, . This shows that U is -closed set.
The converses of the above theorems are not true, as can be seen by the following counter example. ☐
Example 1. Let and neutrosophic sets be defined by Let . Here, is an open set, and . Then is -closed in but is not -closed; thus, it is not N-closed and is -closed in , but not -closed.
Theorem 3. . an -, an -.
Proof. Let G be an -open set such that . Since , then . But U is -closed, so , since and and hence . Therefore V is an -closed set. ☐
Theorem 4. Let U be an -open set in X and , then V is -open.
Proof. Suppose U is -open in X and . Then is -closed and . Then is an -closed set by Theorem 3.5. Hence, V is an -open set in X. ☐
Theorem 5. - -.
Proof. Let U be an -open set and let V be an -closed set such that . Then and hence , since is -closed. But , so . Conversely, suppose that the condition is satisfied. Then whenever is an -open set and . This implies that , where G is -open and . Therefore, is -closed and hence U is -open. ☐
Theorem 6. Let U be an -closed subset of . Then does not contain any non-empty -closed set.
Proof. Assume that U is an -closed set. Let F be a non-empty -closed set, such that F ⊆ . i.e., and . Therefore, . Since is an -open set, . i.e., . Therefore, F is empty. ☐
Corollary 1. Let U be an -closed set of . Then -U does not contain anynon-empty N-closed set.
Proof. The proof follows from the Theorem 3.9. ☐
Theorem 7. If U is both -open and -closed, then U is -closed.
Proof. Since U is both an -open and -closed set in X, then . We also have . Thus, . Therefore, U is an -closed set in X. ☐
4. On Neutrosophic -Continuity and Neutrosophic Contra -Continuity
Definition 10. A function is said to be a neutrosophic -continuous (briefly, -continuous) function if the inverse image of every open set in Y is an -open set in X.
Theorem 8. Let be a function. Then the following conditions are equivalent.
- (i)
g is -continuous;
- (ii)
The inverse of each N-open set U in Y is -open set in X.
Proof. The proof is obvious, since for each N-open set U of Y. ☐
Theorem 9. If is an -continuous mapping, then the following statements hold:
- (i)
, for all neutrosophic sets U in X;
- (ii)
, for all neutrosophic sets V in Y.
Proof. - (i)
Since is a neutrosophic closed set in Y and g is -continuous, then is -closed in X. Now, since , . Therefore, .
- (ii)
By replacing U with V in (i), we obtain . Hence, .
☐
Theorem 10. Let g be a function from an NTS to an NTS . Then the following statements are equivalent.
- (i)
g is a neutrosophic -continuous function;
- (ii)
For every NP and each NN U of , there exists an -open set V such that .
- (iii)
For every NP and each NN U of , there exists an -open set V such that and .
Proof. . If
is an NP in
X and if
U is an NN of
, then there exists an NOS
W in
Y such that
. Thus,
g is neutrosophic
-continuous,
is an
, and
Thus, (ii) is a valid statement.
. Let be an NP in X and let U be an NN of . Then there exists an U such that by (ii). Thus, we have and . Hence, (iii) is valid.
. Let
V be an NO set in
Y and let
. Then
. Since
V is an NOS, it follows that
V is an NN of
. Therefore, from (iii), there exists an
U such that
and
. This implies that
Therefore, we know that is an in X. Thus, g is neutrosophic -continuous. ☐
Definition 11. A function is said to be a neutrosophic contra -continuous function if the inverse image of each NOS V in Y is an NC set in X.
Theorem 11. Let be a function. Then the following assertions are equivalent:
- (i)
g is a neutrosophic contra -continuous function;
- (ii)
is an N C set in X, for each NOS V in Y.
Proof. Let g be any neutrosophic contra -continuous function and let V be any NOS in Y. Then is an NCS in Y. Based on these assumptions, is an in X. Hence, is an in X.
The converse of the theorem can be proved in the same way. ☐
Theorem 12. . -.
Proof. Let V be any NCS in X. Then , and g is onto, by assumption, which shows that . Hence, . Since g is an into mapping, we have . Therefore, , so is an O set in X. Hence, g is a neutrosophic contra -continuous mapping. ☐
Theorem 13. . :
- (i)
g -;
- (ii)
;
- (iii)
.
Proof. Let g be a neutrosophic contra -continuous mapping, let V be any NCS in Y and let be an NP in X and such that . Then . Let . Then U is an and .
The results follow from evident relations .
Let V be any NCS in Y and let be an NP in X such that . Then . According to the assumption, there exists an in X such that and . Hence, . Therefore, . Since is an arbitrary NP and is the union of all NPs in , we obtain that . Thus, g is a neutrosophic contra -continuous mapping. ☐
Corollary 2. Let and be NTS sets, and are the projections of onto , . If is a neutrosophic contra -continuous, then -.
Proof. This proof follows from the fact that the projections are all neutrosophic continuous functions. ☐
Theorem 14. . : --
Proof. For every NOS, V in holds . Since h is a neutrosophic contra -continuous mapping and is an NOS in , is an in , so g is a neutrosophic contra -continuous mapping. ☐