I find the idea that an actual infinity is impossible very counterintuitive, but sometimes arguments can establish something very counterintuitive. Actually, even if the following argument does not show that actual infinities are impossible, it will, I think, show that one cannot make any probabilistic inferences in an infinite multiverse. And that’s an interesting conclusion in its own right. (That said, I have some technical qualms about the argument that I can’t articulate.)
Begin with the Parity Principle: If it is almost certain (i.e., if it has probability one) that (a) the basic properties Q and R have exactly the same distribution at t1, and that (b) x is a substance existing at t1 and a member of basic kind K, and no other information is available about x, then the probability that x has Q is equal to the probability that x has R. (Here, I allow such probability values as “undefined”, “inscrutable” as well as intervals, vague values.)
For my argument I will need the assumption that it is possible to have indiscernibles–objects that have all the same basic properties. I actually think this assumption is false, but I am hoping that this assumption can be relaxed.
The possibility of indiscernibles and the Parity Principle are going to be the only potentially controversial assumptions in the argument. If you trust me on this point, you can stop reading the argument, and just argue against these two assumptions. Though you might want to read on to see how exactly I understand the “same distribution” condition in the Parity Principle.
Now, if actual infinities are possible, and indiscernibles are possible, the following is possible: At t0, there are countably infinitely many indiscernible Ks. Then, two independent random processes independently happen to each K. The first process has a probability 1/3 of making the object have Q. The second process has probability 1/2 of making the object have R. Nothing else in the world has Q or R. Between t0 and t1, nothing else changes. It is certain that x0 is a K, and that the Ks are substances.
The following claim is obviously true:
Thesis 1: Given the above information, the probability that x0 has Q at t1 is 1/2 and the probability that x0 has R at t1 is 1/3.
However, the following claim is a consequence of the Parity Principle (argument will be given shortly):
Thesis 2: Given the above information, the probability that x0 has Q at t1 is equal o the probability that x0 has R at t1.
It follows that 1/2 = 1/3, which is absurd. Hence, we have to reject the assumption that actual infinities are possible.
Now it’s time for the argument for Thesis 2. By the Parity Principle, all we need to show is that the distribution of R and Q at t1 is the same. I haven’t yet defined sameness of distribution. The following seems to be the right definition: there is an isomorphism f on the set of substances existing at t1 such that if P is any basic property other than Q and R, x has P iff f(x) has P, x has Q iff f(x) has R, x has R iff f(x) has Q, and if S is any basic relation (with two or more relata), then (x1,…,xn) stand in S iff (f(x1),…,f(xn)) stand in S. But now let’s think about what we know about the distribution of R and Q at t1. Almost certainly (i.e., with probability one), it is true that each of the following complex properties is had by infinitely many Ks: P&Q, P&~Q, ~P&Q and ~P&~Q. (It is an easy theorem that the probability that, say, only finitely many Ks have P&Q is zero.) Moreover, the Ks are otherwise indiscernible, and nothing but a K has P or Q. Let f be any one-to-one function defined as follows. If x is not a K, then f(x)=x. If x has both P and Q, then f(x)=x. If x has both ~P and ~Q, then f(x)=x. Moreover, f maps the set of x’s that have P&~Q one-to-one onto the set of x’s that have ~P&Q. This last condition can be satisfied as both sets are infinite, and at most countably so, and hence have the same cardinality. It is easy to see, using the fact that the Ks are indiscernible except with respect to P and Q, that f is the requisite isomorphism, and hence the sameness-of-distribution point holds.