[FOM] Is axiom R sound on a topological space?

[Todd Wilson]

On Fri, 8 Nov 2002, abuchan at mail.unomaha.edu wrote:
> It is now popular to interpret the box operator in modal logic S4 as the
> interior operator on a topological space and the diamond as the closure
> following Tarski and McKinsey's paper 'The Algebra of Topology'.  If we
> pick the right axioms for S4 that correspond nicely with Kuratowski's
> axioms for closure and interior then it is a relatively exercise to show
> that S4 is sound.  However, it is not at all clear that the
> axiom (R):= BOX(p->q)->BOX(p)->BOX(q) is sound.  Does anyone see the proof
> for this?

Write subset as <=, intersection as /\, union as \/, define

   A -> B = int((X - A) \/ B),

and note that int (is monotone and) preserves finite intersections: 

   int(A /\ B) = int(A) /\ int(B).

Since all intuitionistic identities hold under this interpretation,

   int(A->B) -> (int(A) -> int(B)) = (int(A->B) /\ int(A)) -> int(B)
                                   = int((A->B) /\ A) -> int(B)
                                   = int(A /\ B) -> int(B)
                                   = X.

-- 
Todd Wilson                               A smile is not an individual
Computer Science Department               product; it is a co-product.
California State University, Fresno                 -- Thich Nhat Hanh