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

Jesse F. Hughes wrote:
Phil <toob-headman@xxxxxxxxxxxxx> writes:


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.


And that 2n is in E for each 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.


And *only* containing such ordered pairs.


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.


Quite right, and you're done now.

[...]


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.


You're right that we skipped this step. Before you saw the proof that
|N| = |E|, you did not doubt that if n is in N, then so is 2n.
Indeed, you doubt it now only because you are clinging to your
original claim that this theorem is circular.

That is completely and utterly unfair of you. You are saying that I would deliberately lie about what I originally believed, AND that I am too much of a *** to admit when I overlooked something. Neither of those is true. I have admitted to being wrong several times, and I will do so again. And not only did I NOT believe that it is an INHERENT property of ANY infinite set of natural numbers that begins with 0 or 1 and skips no numbers, that if n is in N, then so is 2n, I STILL don't believe it! EVERY "proof" that I have seen that claims to prove that this is an inherent property of infinite sets simply ADDS this belief as a new axiom, period. EVERY ONE! Do you think that if n is in N, then so is 2n? That's an axiom; prove otherwise Jesse! PROVE, from the simple fact that you have a set N1 beginning with 0 or 1, that skips no numbers, with INFINITELY many elements, that there is no such thing as a set N2 that (1) has twice as many elements as N1, (2) has every element in N1, and (3) has infinitely many elements that are NOT in N1, and do so WITHOUT simply adding that belief, or a LOGICAL EQUIVALENT of that belief, as an axiom. Go ahead. Because you will, I assure you, be the FIRST person in history to do so. Nobody else has EVER done that!

Let me see if I can help you to prove what you are claiming. Let's look at the line segment [0,1], and the point 0.75 on that segment. As you know, that number is both preceded and followed by infinitely many numbers, corresponding to all the other positions on [0,.75), and (.75,1]. However, 2 * 0.75 is NOT a member of the infinitely many numbers on the segment [0,1]! It IS a member of the segment [0,2]. I am NOT CLAIMING that infinite sets of the naturals have a similar structure, but I AM claiming that no one has PROVEN that there is some INHERENT feature about the infinite set of naturals that prevents the existence of sets N1 and N2 with properties similar to what I have just described for the line segments [0,1] and [0,2]. Of course, we could simply add, as an axiom, the "fact" that N is NOT like this. We could, for example, state that N and E are equinumerous. Or, we could state that whenever n (similar to 0.75 from [0,1]) is in set N, that 2n (similar to 1.5) is in N (similar to the segment [0,1]). Both of these AXIOMS would do the trick. But the mere fact that N has infinitely many elements does NOT, by any "proof" I have ever seen, decide the issue. In EVERY proof, some axiom that is LOGICALLY EQUIVALENT to N and E being equinumerous has been used IN ADDITION TO the fact that N has infinitely many elements to "prove" that N and E are equinumerous.

Think about it for a minute. People don't just say that we have added an AXIOM that states that N and a proper subset of N (like E) are equinumerous; they claim that this is an INHERENT FEATURE of infinite sets. Indeed, an infinite set is usually DEFINED as being a set which is equinumerous with a proper subset of itself. But in order for that to have any meaning, it MUST be the case that this is not true because we have merely added an axiom that makes infinite sets equinumerous with a proper subset! Prior to any PROOFS that N and E are equinumerous, we have to accept that there are two basic options, either they are equinumerous, or E has half as many elements as N. In the former case, for any n in N, 2n is in N. In the latter case, for any n in N, 2n may or may NOT be in N! Remember, if we map the naturals to the points on the segment [0,1], the naturals will "run out" an infinitesimal distance from zero, because there are infinitely many more points on [0,1] than there are naturals. If we insert infinitely many points between each natural -- a quantity of points that is DEFINED as being sufficient to distribute the naturals evenly on the segment [0,1] -- then taking the natural which corresponds to the point 0.75, and doubling it at least APPEARS to result in a number that is NOT in the segment [0,1], and therefore NOT in N.

Again, I am NOT claiming here that N and E are NOT equinumerous, but I AM claiming that so far, we have no proofs that do not simply use some logical equivalent of N and E being equinumerous as an axiom. But by all means, do what no one else has done, and prove, WITHOUT using some logical equivalent of N and E being equinumerous, that they are. And yes, I understand that if a basic property P of sets IS logically equivalent to this, and it is impossible to assume not-P, then you have your proof, but as an example of what I would find logically unacceptable, although we can ASSUME that for any n in N, 2n is also in N, that is not enough, because we MAY be able to ASSUME that for any n in N, 2n NOT be in N, just as 2 * 0.75 is NOT in the segment [0,1]. In other words, given property P which is logically equivalent to "N and E are equinumerous," we can indeed use P to "prove" that N and E are equinumerous, but by itself, this merely gives us a BRANCH of mathematics in which this is true. Unless not-P is false, and we CANNOT have a another branch of mathematics in which N and E are NOT equinumerous, any axiom that is logically equivalent to P merely SELECTS one of two valid branches of mathematics, rather like selecting Euclidean versus non-Euclidean geometry. And if you are limited to CHOOSING which axiom you wish to ADD to the properties that exist simply because a set is infinite, i.e., to the properties that are INHERENT to ANY infinite set, namely P or not-P (equivalent to Euclidean or non-Euclidean geometry), then you do NOT have a PROOF that infinite sets are inherently equinumerous with a proper subset; you merely have made a DECISION to use one of TWO viable branches of mathematics. And if that is the case, then we CANNOT DEFINE an "infinite set" as being a set that is equinumerous with a proper subset of itself, because that is in fact an attribute of a separate AND OPTIONAL axiom.


Nonetheless, if we merely claimed without proof that this fact holds,
then it would not be a proof. But there is a proof that each element
of N can be doubled. It is a bit technical and to be honest, I don't
want to waste time giving it. It is not all that illuminating.
Anyway, someone with more patience will surely offer it.

Instead, let's talk about some simpler questions.

(1) Do you agree that whenever n is in N, then so is n+1?

(2) And hence, whenever n and m are in N, then so is n + m?

(3) And do you agree that 2n = n + n?

Which of these are not obvious?

Jesse, "obvious" doesn't cut it when dealing with infinite sets. Which is more obvious, that eliminating half a set's elements reduces its size, or that it leaves its size untouched? Yes, regardless of which option we choose, when dealing with infinite sets, we are going to have SOME results that appear to be contradictory. However, you CANNOT, as a true mathematician, arbitrarily say that "Well, THIS set of apparently contradictory features is just fine, but THAT set is not." If axiom P leads to a self-consistent -- but not OBVIOUSLY so -- branch of mathematics, and so does not-P, then they are BOTH viable options, true?

Has anyone PROVED that not-P, meaning N and E are NOT equinumerous, leads to a branch of mathematics that is inherently more contradictory than P? Hey, here's your chance! Prove that we have a situation similar to geometry, with two viable options P and not-P, and you'll go down in history! Or, be the first to truly PROVE, rather than ASSUME, that not-P does not, in fact, lead to a viable branch of mathematics, and you will still go down in history as the first man to PROVE, rather then ASSUME in a circular proof, that axioms logically equivalent to not-P are NOT viable options.

Yes, I know that above you CLAIM that a very technical proof exists that proves that if n is in N, then so is 2n, but if this is to have any meaning, then it must be a "true" proof, meaning that the alternative CANNOT lead to a viable branch of mathematics. I will tell you now, for the record, you are going to find that any such "proof" ASSUMES, somewhere, one of the logical equivalents of N and E are equinumerous. Bet your ass on it. Because it is NOT an inherent feature of infinite sets, any more than the "fact" that 2 * 0.75 can be found on the segment [0,1] is an INHERENT feature of the set/line segment [0,1].

Phil

If you agree to (1), (2) and (3), we can see that 2n is in N whenever
n is in N. This is not the usual proof, since the usual proof
requires a digression into recursion. But perhaps this simpler
argument will convince.

.