Arpad Szaz Hello, dear Arpad!

Using another notation, replacing from your notation A with M, C with A and denoting /\ =intersected with; C(P) notation for the complement of the set P), we have

M=A\int(cl(A))=A/\C(int(cl(A)))=A/\cl(C(cl(A)))

A < cl(A) => C(cl(A)) cl(C(cl(A)))

Dear Dinu,

I am very greateful for your intention to help me.

Your computation is correct. However, it can be shortened.

Note that

( A setminus int(cl(A))) subseteq ( A setminus int(A)) = bor(A)

since (A subseteq cl(A)), and thus (int(A) subseteq int(cl(A)) .

(This is true in every reflexive relator space.)

However, I know only of the theorem that

res(A)=(cl(A) setminus A)

is rare (nowhere dense) whenever A is open.

(This true in every topological, topologically filtered relator space.)

Note also that

bnd(A)=(cl(A) setminus int(A)) = res(A) union bor(A).

(This true in every reflexive relator space.)

There is also another important fact that

cl(A) intersection B =cl(A intersection B) intersection B

whenever B is open.

(This is true in every topologically filtered relator space.)

Otherwise, the book of Kuratowski should be used to find a proof.

(Moreover, couterexamples in relator spaces should constructed.)

Relators (generalized uniformities) are more powerful and flexible tools,

than topologies, and their obvious generalizations. Therefore, topologies,

and their immediate generalizations, should not be studied sepatately.

Namely, relators are arbitrary families of relations on a set. While, generalized topologies are certain families of subsets of a set.

And, for a subset A of a set X, one can at once consider the

Pervin preorder relation

R=(A times A) union (A^{c} times X).

This makes superfluous not only the separate study of generalized topologies, but also that of stacks (ascending families) such as filters

and grills for instance.

With best regards, and wishes for the New Year,

Arpad

Dear Arpad, best wishes for the New Year 2024!

Many, many years ago, when I was student, I've studied intensively topological spaces, traditionally, but also using filters and ultrafilters. My opinion is that the approaches are completely equivalent, only that there are cases in which one approach is more suggestive and easier to apply. It depends on case by case.

It seems that the situation is the same when relator spaces are used, things are equivalent, but some results can be obtained more easily and suggestively.

Anyway, the knowledge in the field of topology was extremely useful to me!

With all my best regards,

Dinu

Dear Dinu,

Thank for your letter. However, the various approaches to

"Topology" are not at all equivalent.

"Generalized open sets" are weaker tools than " generalized closures" , "generalized closures" are weaker tools than "generalized proximities",

"generalized proximities" are weaker tools than "generalized convergences of nets to nets", and "generalized convergences of nets to nets" are weaker tools than " relators (generalized uniformities)" .

Moreover, "generalized neighbourhoods of point or sets " and

" collections of covers" are also very strong tools. There are also some kinds of "nearness" for families of sets.

Furthermore, "super relators" and" bilrelators" are much stronger tools, than "ordinary relators".

These make "syntopogeous structures", "bitopologies" and "ideal

topologies" superfluous.

Of course, nobody dealing with deeply with one of the classical

structures wants to acknowledge the truth of the above statements.

With many thanks for your interest,

Arpad

Dear Arpad,

Many thanks for your interesting description regarding all these notions!

All the best,

Dinu

The requested proof: Let E = int(cl(A)). Since A is a subset of C, E is an open subset of int(cl(C)) as well. By the definition of A, for any element y of E, y is not an element of A. So, if x is an element of E, then E is a neighborhood of x that does not intersect A. Hence x does not belong to cl(A), thus it cannot belong to E = int(cl(A)), a contradiction. Therefore, E must be empty.

Zoltán Boros Arpad Szaz Dear Zoltan, you affirm "for any element y of E, y is not an element of A". But that's not true! Because intA is included in E = int(cl(A)), so, if we take an arbitrary y in E, then y can be very well in intA(therefore y being in A).

Dinu Teodorescu This is a proof by contradiction. By definition, A contains those elements of C that do not belong to int(cl(C)). This means, on one hand, that A is a subset of C and thus E = int(cl(A)) is a subset of int(cl(C)).

On the other hand, elements of int(cl(C)) do not belong to A due to the definition of A. Hence E and A are disjoint. So, if x is an element of E, then E is a neighborhood of x that does not intersect A. Hence x does not belong to cl(A), thus it cannot belong to E = int(cl(A)), a contradiction. Therefore, E must be empty.

It is natural that the elements of the empty set can fulfill a statetement and its opposite at the same time. A proof by contradiction works like this.

Zoltán Boros Sorry, but I hadn't read your first comment carefully! Very clear now! Congrats for your nice and simple proof! All the best!

Dear Dinu,

Meantime, we have obtained some better results.

However, counterexamples and proximal versions are still

needed.

For the details, see the attached Extract-2.pdf.

With best wishes,

Arpad

Dear Arpad,

Many thanks! I will read carefully!

All the best,

Dinu