3. Complete Domains
A poset is a complete domain if X is complete and is a basis. For instance, is a complete domain (for any nonempty set Y). From now on, we consider that is a complete domain.
Proposition 1. .
Proof. Let us consider . Since is a basis, . Since , there exists such that . Thus, . Now let . Since , it follows that , and there exists . Since , and , it follows that , and so . In conclusion, . □
Corollary 1. Any complete domain is an algebraic domain.
Corollary 2. Let and . Then if and only if .
Proof. Let . Since , we have , and so there is such that . Since , then . □
Let be the function defined by for all . Just to simplify the notation, we use instead of .
Proposition 2. for all .
Proof. Let and such that . If , since and , we have , which is a contradiction! Therefore, . Then , and .
On the other hand, = = . However, = , and so . Thus, . □
If , then = , and .
Proposition 3. If then .
Proof. If then , meaning that . □
Proposition 4. for all .
Proof. If we assume , there is such that . Since , we have , and because , there exists such that and . Then , which is a contradiction. □
Proposition 5. for all .
Proof. Let . Then . □
Proposition 6. If and , then either or .
Proof. . Since , we have if and only if if and only if or if and only if or . □
Corollary 3. If , then .
Proof. Let such that . Since , it follows that or . Since , then , and so . Thus, . □
Proposition 7. for all .
Proof. Let . We have and . Let such that and , and let such that . Since and , then either or . Since , we have or , and so . Thus, , and . □
Proposition 8. for all .
Proof. Let . Then for all for all . Let such that for all , and let such that . Then, for all , there exists such that and . Since and , it follows that . Then ; hence, for all , and so . Thus, . It follows that . Thus, . □
Proposition 9. If then .
Proof. We have and . Let with . Since , there exists such that and . Then , and . It follows that , which is equivalent to . Then . □
Proposition 10. .
Proof. Let . Since , we have . Let with , and let with . Since , . Then (because ). It follows that . Thus, for each with we have , and so . Thus, for all .
On the other hand, if then . Then . Since , we obtain . □
According to these results, the mapping c is a bijection.
Proposition 11. For all we have .
Proof. Let . Then . Thus, . □
4. Interior Operators over Complete Domains
Definition 1. Let be a complete domain.
A function is called an interior operator (over X) if:
- 1.
;
- 2.
;
- 3.
for all finite subsets A of X.
Let i be an interior operator. From the third condition applied for , we have . In fact, the third condition is equivalent to and for all . In addition, if such that , then . Therefore, i is isotone. Since , it follows that .
Let
I be a nonempty subset of
X. For each
, we define the set
Then, the function
given by
is well-defined.
Proposition 12. If I is a lower subset, then is antitone.
Proof. Given such that , if then . We suppose that , and . Then there exists such that and . If then , and so . Thus, and . Hence, . □
In what follows, we assume that I is an ideal of , i.e., .
Proposition 13. for all .
Proof. Let . Since , we have and so . If , then , and so .
We suppose that ; let such that . Since and , from Corollary 2 we have that , and so there exists such that and . Since , there is such that and . Let ; then , and so . In addition, . Since and , we have . Consequently, and . Thus, . □
Proposition 14. .
Proof. Let and . Then there is such that and . If , there exists such that and . However, if and only if , hence for all such that , we have . Since such that , then , which is a contradiction. Hence , and then . Therefore, . Since , we have , i.e., . □
Proposition 15. .
Proof. Let and we suppose that there exists such that and . Since if and only if , there is such that and . Since if and only if , it follows that for all with we have . Then , and so there exists such that . Since , we have and . Because and , it follows that , which is a contradiction. To conclude, for all such that , we obtain . Thus, . □
Proposition 16. .
Proof. Let , and assume that there is such that and . Since if and only if , we have . Thus, for all with , we have . Since and , we obtain . On the other hand, if then , which is a contradiction. Therefore, for all such that we have . Thus, . □
Corollary 4. We have and .
Proof. Since , it follows that . Since i is isotone, it follows that . □
From the previous result, it follows that , and so . More generally, we have the following result.
Proposition 17. .
Proof. Let . For all there is such that and . Then , and so . It follows that . □
Proposition 18. for all and .
Proof. If and , then . Since , . Let such that . Then , and so there is such that and . Since and , we obtain . Thus, and so . □
Proposition 19. Let , , , and .
For every function we have:
- 1.
;
- 2.
For all with , we have if and only if for all with , there exist such that .
Proof. 1. Since , we have and . Therefore, . Let and such that . If , then for all there exist and such that . Let . Then and . Since , there is such that . Thus, , and so . Therefore, .
2. “⇒”: Let and such that . Let . Then and B is directed. Hence, . Since , there is such that . Let us suppose that , where . Then , and so .
“⇐”: Let and such that . From the hypothesis, it follows that there exist such that . Since A is directed, there is such that . Then . Therefore, . □
Since , from it follows that . If the converse is true, we say that I has the property . Therefore, implies ; also, .
Proposition 20. The following statements are equivalent:
- 1.
,
- 2.
for all .
Proof. We suppose that , and consider . Since , then .
Conversely, suppose that for all . If , then . Therefore, . □
Remark 1. If , then .
Since is antitone, for every . A special case is when . We say that I has the property if . Thus, we have .
Proposition 21. The following statements are equivalent:
- 1.
;
- 2.
.
Proof. if and only if if and only if if and only if for all and such that we have if and only if for all such that we have . □
Proposition 22. If , then over .
Proof. Let and such that . Then , and so there is such that and . Since and , we have . Then, since , it follows that and . Thus, and so . Since , then . According to Proposition 16, it follows that . Therefore, . □
Theorem 1. is an interior operator on X such that .
Proof. By definition, . Let . , since (according to Proposition 15). Hence, .
Let . Then . In addition, , and so is an interior operator. Since and , it follows that . □
To simplify the notation, the operator is denoted also by .
Proposition 23. .
Proof. Since , for all . Thus, . □
Proposition 24. .
Proof. From Corollary 4, we have and . Thus, . In addition, . □
Proposition 25. - 1.
;
- 2.
if and only if for all .
Proof. 1. Let . Then if and only if if and only if . Therefore, , and so , where and .
2. Let and . Then if and only if if and only if if and only if if and only if if and only if . Thus, if and only if
“⇒”: Let . Since , it follows from the hypothesis that . Then, according to , we have . However, that implies that implies . Then we obtain . Thus, .
“⇐”: Let and . From the hypothesis we have , and then . From it follows that . □
Corollary 5. If , then .
From Corollary 5, Propositions 19 and 22, the next result follows.
Proposition 26. Let , and be a mapping.
- 1.
If , then .
- 2.
If , then .
Let be two interior operators over X, and .
Remark 2. If and , then for all .
Moreover, and .
Proposition 27. If and , then .
Proof. Let and . Then and there is such that and . From , it follows that . Therefore, there is and such that , and and . Since , and , then , and so . Hence, . It follows that , and so . □
Corollary 6. If , and , then.
Proof. From Theorem 1, the inequality is generally true (without the conditions of the hypothesis). Since , it follows also that . □
Corollary 7. If , then .
Corollary 8. If and , then .
Corollary 9. If , then .
Corollary 10. If , , and , then .
Corollary 11. If , then .
We present some examples of -ideals having the property.
Let us consider a complete domain, a cardinal number, and . We define by . If satisfies and is a complete ideal, then . Consequently, if with , then . In addition, if is a cardinal number with , is a complete ideal such that and , then . Thus, if is a cardinal number with and , then .
We also present some examples of -ideals with the property.
Let be a complete domain, and be a topology. For , we denote by the interior of A relative to . The function defined by is an interior operator on X. Let be a cardinal number, be an -ideal relative to inclusion and . Then and if and only if .
Let us consider ideals
with the property
. From Baire’s theorem, each complete metric space is a Baire space [
4]. We get the following result.
Theorem 2. Let d be a metric on such that is a complete metric space, let τ be the metric topology and an -ideal of meagre sets. Then .
An interesting situation is when
can be organized as an algebraic domain. In such a case, based on the results presented in [
5], we obtain the following result.
Theorem 3. Let ⊑ be a partial order on such that is an algebraic domain, τ be the density topology (according to [5]), be a cardinal number and be the ideal of all α-unions of τ-nowhere dense subsets of . Then .