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

In article <463CF9A1.8030903@xxxxxxxxxxxxx>,
Phil <toob-headman@xxxxxxxxxxxxx> wrote:

Jesse F. Hughes wrote:

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!

Not so.

Definition of N:
N is a set with a a function succ: N -> N such that
(1) 1 is a member of N
(2) for every n in N succ(n) in N is such that
(a) 1 is not succ(n) for any n in N
(b) for all m, n in N, succ(m) = succ(n) iff m = n
(3) if K is a subset of N such that 1 is a member of K and
for every k in K succ(k) is in K then K = N.

Definition of addition, (+), on N:
(+): NxN --> N: <m,n> |--> m+n is a function such that
(1) for all n in N, n + 1 = succ(n)
(2) for all m,n in N, m + succ(n) = succ(m + n)

It follows from the definition of N, that m + n is defined and is a
member of N for all m and all n in N.

Definition of 2n: 2n = n + n.

This competes the proof that 2n is a member of N for all n in N.
E is, by definition, the set of all 2n's for n in N.

There is no circularity here.




Do you think that if n is in N, then so
is 2n?

Since all the major steps of the PROOF are laid out above, we think it
is PROVED that if n is in N then so is 2n.



That's an axiom;

Things that are proved are called theorems, which this is.
Things that are assumed without proof are axioms, which this is not.

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!

No needs to do that in order to prove
"If n is a member of N then so is 2n".

The rest of Phil's rant is irrelevant to the above demonstration of
total absence of any circularity at all.
.

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