Re: Cantor's circular "proof" that evens = integers

Okay, first I will hopefully correctly state Jesse's proof, and then show why it is actually just a circular argument.

Preliminary definitions:

Set N is an infinite set containing all the natural numbers (for these proofs, I still prefer the "old" naturals, which begin with 1).

Set E is an infinite set containing all the even natural numbers, meaning every element e in E = 2n for some n in N.

Set N x E is the Cartesian product of N and E, the set of all the ordered pairs that can be formed from sets N and E, where the first element in each pair is from N.

Set f is a subset of N x E containing every ordered pair from N x E with the form <x,y> where y = 2x.

Observations and deductions:

For any n in N, there exists an element 2n in N. Therefore, set N x E contains an ordered pair of the form <n,2n> for EVERY n in N.

Set f is:

1. A function, since for any two pairs <x,y> and <x,y'>, y = y'. In other words, for any x in f, there is one y.

2. One-to-one, since it is also true that for any two pairs <x,y> and <x',y>, x = x'.

3. Onto, since for any y, there exists one x, and (by 1. above) for any x, there exists one y.

Therefore, there is a bijection between the infinite sets N and E, making sets N and E equinumerous.

In order to understand why this "proof" is actually just a circular argument, i.e., an argument that merely ADDS the belief that N and E are equinumerous as an axiom to the already existing axioms of set theory (as opposed to proving that this conclusion NECESSARILY FOLLOWS from the existing axioms), we have to make use of some basic facts concerning proofs.

We will refer to the possibility that sets N and E are equinumerous as property "P." The possibility that they are not equinumerous is "not-P." Technically, not-P can have many values, but for convenience, we will assume that not-P means that set E has HALF as many elements as set N. Now suppose that property Q is the LOGICAL EQUIVALENT of P, meaning that Q is true if and only if P is true, and that once we either prove OR ASSUME either P or Q, we can prove the other. The important point is that we cannot ASSUME Q and use it to "prove" P.

The next step is to get a list of things that are true and false under both P and not-P. In general, not-P means that we can have infinite sets of the natural numbers of different sizes (I THINK this is compatible with non-standard analysis). We could, for example, have set N1 containing infinitely many naturals, and set N2 containing twice as many elements as N1. An INTUITIVE example would be (despite proofs to the contrary) line segments 1 and 2 units long; intuitively, a 1-unit segment contains twice as many points as a 2-unit segment, and if we equate the points in each segment with the numbers in N1 and N2 -- and again, I must stress that this is an INTUITIVE description, not a claim of fact -- then we can see how one set could have twice as many numbers as the other. Also, if we take the number x corresponding to position 3/4 on the 1-unit line segment, it becomes clear that 2x is in set N2, and corresponds to position 1-1/2, but that it is NOT in set N1. Again, these are not statements of fact, but rather statements that are consistent with not-P, and inconsistent with P.

The problem concerns that fact that under not-P, a number 2x does NOT always exist for any x in set N1. Therefore, IF set N1 = N2, i.e., if P is true, THEN we can safely ASSUME that for any x in N, 2x is also in N. However, IF N1 |= N2, then we CANNOT assume that for any x in N, 2x is also in N. Since the "proof" above claims that for any x in N, 2x is also in N, it is a circular argument, one which merely ADDS the belief that N and E are equinumerous to the existing axioms.

Now, if anyone can PROVE that whenever x is in N, then 2x is in N, then you have a true PROOF that N and E are equinumerous. However, simply claiming that "this is obvious" is not adequate, because we are working with inherently incomprehensible objects, namely INFINITE sets. And remember, IF the existence of two infinite sets of naturals N1 and N2 such that N2 = 2 * N1 is, in fact, compatible with the basic axioms of set theory, then AT BEST we have a situation similar to that found in geometry, where we can CHOOSE whether to add axiom P, or axiom not-P, to the other axioms of set theory. However, it MIGHT even turn out to be the case that not only is not-P compatible with the basic axioms, but that in addition, P is NOT compatible with the basic axioms.

Moe's proof (I don't actually know if that's his name but ...) is a bit more complicated, and at least can be viewed in a manner that makes it harder to deal with than Jesse's proof. I'll let anyone interested have a crack at it first, before I point out the circular argument.

Phil
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