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

[John Goodrick]

We just need to show that for any point x, if

1. x is in int(P->Q)

and

2. x is in int(P)

then

3. x is in int(Q).

So pick open neighborhoods U and V witnessing 1. and 2.; then U intersect
V will witness 3.

-John

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?  In a toplogical space  <X,C>, (R) says:
>
>   int(X \ P union Q) is a subset of  X\( int (P)) union int(Q)
> where \ is set complement and int is the interior operator.
>
>
> _______________________________________________
> FOM mailing list
> FOM at cs.nyu.edu
> http://www.cs.nyu.edu/mailman/listinfo/fom
>