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

abuchan@mail.unomaha.edu abuchan at mail.unomaha.edu
Fri Nov 8 11:31:26 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.

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