## Thursday, January 19, 2012

### Presentist counting

In a posthumous paper, David Lewis shows that one can find a presentist paraphrase of sentences like "There have ever been, are or ever will be n Fs" for any finite n. But his method doesn't work for infinite counting.

It turns out that there is a solution that works for finite and infinite counts, using a bit of set theory. For any set S of times, say that an object x exactly occupies S provided that at every time in S it was, is or will be the case that x exists and at no time outside of S it was, is or will be the case that x exists. For any non-empty set S of times, let nF(S) be a cardinality such that at every time t in S it was, is or will be the case that there are exactly nF(S) objects exactly occupying S. This is a presentist-friendly definition. Let N be any set of abstracta with cardinality nF(S) (e.g., if we have the Axiom of Choice, we should have an ordinal of that cardinality) and let eF(S) be the set of ordered pairs { <S,x> : xN }. We can think of the members of eF(S) as the ersatz Fs exactly occupying S. Let eF be the union of all the eF(S) as S ranges over all subsets of times. (It's quite possible that I'm using the Axiom of Choice in the above constructions.) Then "There have ever been, are or ever will be n Fs" can be given the truth condition |eF|=n.

This ersatzist construction suggests a general way in which presentists can talk of ersatz past, present or future objects. For instance, "There were, are or ever will be more Fs than Gs" gets the truth condition: |eG|≤|eF|. "Most Fs that have ever been, are or will be were, are or will be Gs" gets the truth condition |eFG|>(1/2)|eF|, where FG is the conjunction of F with G. I don't know just how much can be paraphrased in such ways, but I think quite a lot. Consequently, just as I think the B-theory can't be rejected on linguistic grounds, it's going to be hard to reject presentism on linguistic grounds.