[FOM] Is axiom R sound on a topological space?
Vladimir N.Krupski
kru at voll.math.msu.su
Fri Nov 8 19:34:29 EST 2002
> 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.
>
Sorry, how many logics do you call S4? If it is a single logic
which is represented by two formalizms, say T1 and T2, with the
same set of theorems and you find T1 to be sound with respect to
topological interpretations then T2 is sound too.
You should simply derive all the axioms of T2 in the T1 system
and apply the soundness result for T1. The soundness of T1
already means that all its theorems are valid under every
topological interpretation (of course you need to prove that all axioms
of T1 are valid and all the rules of T1 preserve the validity).
Suppose R is an axiom of T2. In order to obtain the special proof
of the soundness of R you have to derive R in T1 and then
to apply the soundness argument to each step of this derivation.
Use logic and not the topology.
It is designed for this purpose exactly!
V.Kru
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