Dear Irakli,

Concerning your question, I can only note that if \Cal R is a relator

(family of relations) on set X then \Cal R and \Cal R^{-1} induce two

generalized topologies on X. So X becomes a generalized

bitopological space.

Moreover, if \Cal R is a relator on X to Y, then the relator space

(X, Y,)(\Cal R) has to be clalled properly compact if for each

R \in \Cal R there exists a finite subset A of X such that R[A]=Y.

This definition is, is in my opininion, quite constructive.

By using appropriate closure operations on relators, you can get

several old an new compatness concepts. For some ideas in this

respect, see my earlier papers:

1. Uniformly, proximally, and topologically compact relators,

Math. Pannon. 8(1997), 103--116.

2. Lebesgue relators, Monatsh. Mat. 110( 1990), 315--319.

Sincerely yours,

Arpad

Dear Dr. Arpad Szaz,

Thank you for your answer and suggested reference. Unfortunately, till now I don't have any information about relators. I will explore yours mentioned articles.

With my best wishes,

Irakli

For a study of topological spaces and the problem of proving compactness constructively, see

C.M. Fox, Point-Set and Point-Free Topology in Constructive Set Theory, Ph.D. thesis, University of Manchester, 2005:

http://www.mims.manchester.ac.uk/research/logic/dissertations/chris-fox.pdf

Thank you prof. Peters, you always find so interesting answers

Consider the Bitopological space (R, l, v) where l is lower limit topology and u is upper limit topology. R is not (u,v)-clopen compact as {[n, n+1): n\in Z} is (u,v)-clopen cover but no finite subcover. But any bounded closed/clopen interval possibly (u,v)-clopen compact (Check this).