B entails that a conjunction of determinate truths is determinateOctober 26, 2010
I know it’s been quiet for a while around here. I have finally finished a paper on higher order vagueness which has been a long time coming, and since I expect it to be in review for quite a while longer I decided to put it online. (Note: I’ll come back to the title of this post in a bit, after I’ve filled in some of the background.)
The paper is concerned with a number of arguments that purport to show that it is always a precise matter whether something is determinate at every finite order. This would entail, for example, that it was always a precise matter whether someone was determinately a child at every order, and thus, presumably, that this is also a knowable matter. But it seems just as bad to be able to know things like “I stopped being a determinate child at every order after 123098309851248 nanoseconds from my birth” as to know the corresponding kinds of things about being a child.
What could the premisses be that give such a paradoxical conclusion? One of the principles, distributivity, says that a (possibly infinite) conjunction of determinate truths is determinate, the other, B, says . If is the conjunction of and so on, distributivity easily gives us (1) . Given a logic of K for determinacy we quickly get , which combined with (an instance of B) gives (2) . Excluded middle and (1) and (2) gives us , which is the bad conclusion.
In the paper I argue that B is the culprit.* The main moving part in Field’s solution to this problem, by contrast, is the rejection of distributivity. I think I finally have a conclusive argument that it is B that is responsible, and that is that B actually *entails* distributivity! In other words, no matter how you block the paradox you’ve got to deny B.
I think this is quite surprising and the argument is quite cute, so I’ve written it up in a note. I’ve put it in a pdf rather than post it up here, but it’s only two pages and the argument is actually only a few lines. Comments would be very welcome.
* Actually a whole chain of principles weaker than B can cause problems, the weakest which I consider being , which corresponds to the frame condition: if x can see y, there is a finite chain of steps from y back to x each step of which x can see.