## Wednesday, September 2, 2020

### An argument that there are more positive odd numbers than positive even ones

Suppose that we embrace these intuitive theses about sets of natural numbers:

1. If A is a proper subset of B, then B has more members than A.

2. If A has at least as many members as B, then 1 + A has at least as many members as 1 + B, where 1 + C = {1 + c : c ∈ C} is C shifted over by one to the right.

3. Either there are more odd positive numbers than even positive numbers or there are at least as many even positive numbers as odd positive numbers.

Let O be the positive odd numbers and E be the positive even numbers. Write A ≲ B to mean that B has at least as many members as A, and write A ∼ B to mean that they have the same number of members. For a reductio, suppose that O ≲ E. Then 1 + O ≲ 1 + E by (2). But 1 + O = E. Thus, O ≲ E ≲ 1 + E. But 1 + E is all the odd numbers starting with three, which is a proper subset of O, which contradicts (1). So we do not have O ≲ E, and hence by (3):

1. There are more odd positive numbers than even positive numbers!

Of course, the usual Cantorian way of comparing sizes of sets rejects (1).

I think the non-Cantorian’s best bet is either to embrace the conclusion (4) or to deny the special case of totality of comparison in (3). In either case, the non-Cantorian needs to deny the intuitive claim that:

1. There are equally many odd positive numbers and even positive numbers.

Note that (2) does not say or imply that 1 + A has the same number of members as A. Since that would imply that {1, 2, 3, ...} and {2, 3, 4, ...} have the same number of members, that would beg the question against the typical non-Cantorian for whom (1) is a central intuition. One might also wonder whether there is a way of comparing sets of natural numbers that satisfies (1)–(3): the answer is yes (even with (3) generalized to all pairs of sets of naturals).