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

Steve Awodey awodey at cmu.edu
Mon Nov 11 12:18:24 EST 2002

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.

interior plainly preserves subset inclusions (let's write them  p |-  q ),
and intersections (p & q), so:

p => q  |-  p => q
(p => q) & p  |-  q
int((p => q) & p)  |- int(q)
int(p => q) & int(p)  |- int(q)
int(p => q)  |- int(p) => int(q)
 |- int(p => q) => (int(p) => int(q))

Steve Awodey

More information about the FOM mailing list