[FOM] Is axiom R sound on a topological space?
John Goodrick
goodrick at math.berkeley.edu
Fri Nov 8 14:04:00 EST 2002
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.
>
>
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