[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
CMU
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