4

I'm a math student taking philosophy classes. So I have some math background but am absolutely unfamiliar with metaphysics&c. Recently I've come around the "Nothing Is Impossible" paper as a part of that course. The author is only constrained by S5. Here's the part I'm having trouble with:

Assume that it is possible for there to be something. If this is so, there are possible worlds (accessible from the actual world) in which beings exist.

The question basically is: why need't be accessible from the actual world?

I looked it up and found this SE answer about accessibility relation. It says:

The axioms of S5, for instance, require the relation to be an equivalence relation and so have the result that every world is accessible from every other world.

I'd rather conclude: "every world is accessible from every other world within its equivalence class" and I don't see why there would be exactly one class. Of course I might be missing some point in the definitions.

Some possible explanations I see:

  • It might be meaningful to treat "$P$ is possible" as "$P$ is true is some accessible world". Doesn't explain the proposition from the SE answer I mentioned.
  • Perhaps in S5 there can only be a single equivalence class and I missed it? Then why is it so? What's the purpose of such a trivial accessibility relation?
4
  • 1
    Every world is accessible from every other world indeed implies that there is a single equivalence class. Why is it trivial? This semantics only aims at capturing talk of possibility and necessity, and talk of things being possible in worlds that are impossible (=inaccessible from the actual world) seems irrelevant. Jan 3, 2018 at 20:56
  • @QuentinRuyant, I meant "trivial" as a relation (i.e. the relation is simply the Cartesian product). Jan 3, 2018 at 22:01
  • @QuentinRuyant Which axiom of S5 then implies that every world is accessible from every other? Referring to en.wikipedia.org/wiki/S5_(modal_logic)#The_axioms_of_S5 Jan 3, 2018 at 22:02
  • none, but I guess that as in the case of arithmetic one can assume that there are non standard models of the axioms. Jan 6, 2018 at 12:55

2 Answers 2

2

Your two 'possible explanations' are correct. In Kripke semantics we may say informally that a proposition is possibly true in world w iff it is true in some world accessible to w, and a proposition is necessarily true in world w iff it is true in all worlds accessible to w. The axioms of S5 are sufficiently strong to ensure that the accessibility relation is serial, reflexive, symmetric and transitive. As a result, all possible worlds in S5 collapse to a single equivalence class.

This does not make S5 trivial. Indeed, many logicians have argued that this is a valuable property. For one thing, it follows that S5 correctly captures the relation of validity in classical propositional logic. In other words, if we wished to express "A entails B" as a modal statement of the form ◻(A → B) then S5 would be the correct logic for the purpose.

The Wikipedia article on Kripke semantics contains a good description of how the axioms of the various modal logics correspond with their frame conditions.

This paper by Burgess has a useful account of the relation between validity and modal logic. John Burgess Which Modal Logic is the Right One? Notre Dame Journal of Formal Logic 40, 1 (1999).

1

Question:

Assume that it is possible for there to be something. If this is so, there are possible worlds (accessible from the actual world) in which beings exist.

The question basically is: why need't be accessible from the actual world?

Answer:

Because when they said "possible for there to be something" they meant "possible, relative to the actual world". If something occurs in a world that's in a different S5 equivalence class from the actual world, it is not really "possible".

Different equivalence classes are basically different models of S5, so one might as well assume there is only one equivalence class in each model.

1
  • 1
    Would you have a reference the reader can go to for more information? +1 Dec 11, 2018 at 11:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .