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

MoeBlee wrote:

On May 5, 2:39 pm, Phil <toob-head...@xxxxxxxxxxxxx> wrote:

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!


No, I told you that it is PROVEN from the axioms. I told you that it
was one of the many things that is proven before we even get to the
proof of the equinumerousnes of w and the set of even numbers. I told
you long ago that when I give you a set theory proof, then usually
there are premises in that proof that will have been previously proven
and that therefore the rational way for you to understand such proofs
is for you to START with the axioms and see all the proofs of along
the way toward the proofs of the premises used in such proofs as the
one I gave.

YOU KEEP BYPASSING THAT ISSUE.

One more time, since I've said it so many times, but you still don't
get it:

YOU KEEP BYPASSING THAT ISSUE.

You are mostly correct, except that everyone claims that the ESSENCE of an infinite set is that it has the same cardinality as a proper subset of itself. Well, I am pointing out that that is true ONLY when that infinite set is combined with the proper axioms. And in my response to Jesse I gave an example of an infinite set, and a valid method of using or viewing that infinite set, in which n+1 is NOT in the infinite set! Come on, Moe, do you ACTUALLY BELIEVE that when mathematicians marvel at the fact that N and E are equinumerous, that they attribute this "fact," NOT to the fact that N is an infinite set, but rather to the AXIOMS that are used with natural numbers, real numbers, etc? An infinite set is DEFINED as a set that is equinumerous with a proper subset of itself! Where are the footnotes that say that this is actually due to coexisting axioms, and that there may be another branch of mathematics in which N and E are NOT equinumerous?

Using the example I gave in my (latest) response to Jesse, when we cross the INFINITE set of halfway points from the middle of the room to the door, only we cross them in reverse order, from the door to the middle of the room, once we reach the middle of the room, how many more halfway points in this INFINITE SET of points can we cross? The halfway point at the "1/4 point," halfway between the middle of the room and the door, can be referred to as point n. Right? There's no "mystical reason" that prevents us from referring to this point. And n+1 is another point in the set. But n+2 is not, period. Does that make the axioms that prove that N and E are equinumerous false? Well, actually it might. But initially, we HAVE to say that it merely represents an ALTERNATIVE branch of mathematics, one where N+1 does NOT always exist, and one which has not been explored yet by mathematicians.

Phil


Do you think that if n is in N, then so
is 2n? That's an axiom; prove otherwise Jesse!


I told you I wouldn't prove for you the premises since that leads to a
regress where I have to prove the premises of the proof of the
premises of the proof of the premises...until I finally end at the
axioms. Thus, the rational approach is not to work BACKWARDS like
that, but rather for you just to read a good set theory textbook that
shows you all the proofs in FORWARD. But since you're being such a
stubborn ignoramus and won't do the reasonable thing of reading the
proofs in a textbook, here's the proof of the premise:

The operation of multiplication on natural numbers is proven to exist
by using the defintion by recursion theorem for natural numbers (which
is itself another premise that is proven from previously proven
premises, finally reaching back to the axioms), and one of the clauses
proven in the proof of the operation is that the operation is into w,
as that follows immediately from the definition by recursion theorem.
If you want to know these proofs, then you can look them up in many of
the common textbooks on set theory. I especially recommend Enderton's
'Elements Of Set Theory' for his excellent explanations and
presentations of the definition by recursion theorem and the theorem
of the existence of the operation of multiplication on natural
numbers.


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!


You don't even know what you're saying. "Twice as many" as an infinite
set? For an infinite set x, I can only take that to mean 2 times
card(x), where 'times' is cardinal multiplication. Well, in the case
of w, we have 2 times w = w, so you're CORRECT that we can't prove
what you're asking to be proved, which is GOOD, since a proof of it
would be a proof that set theory is inconsistent, and meanwhile the
lack of such a proof does NOT entail that we have not also proven that
for any natural number n, we have that 2n is a natural number.

[...]


Indeed, an infinite set is usually DEFINED as being a set which is
equinumerous with a proper subset of itself.


That is the Dedekind definition, and it is used by many authors, but a
different defintion is also common: n is infinite <-> ~ n is finite
(where 'finite' is previously defined not in the sense of Dedekind
finite). And that is the definition I use. So the Dedekind infinite
definition has NO BEARING on the proof I gave.


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!


No, ZF (which has the axiom of infinity) entails that there exists an
infinite set, but in ZF we do NOT take as an axiom nor do we even
prove that every infinite set ('infinite' defined as I mentioned, not
defined as Dedekind infinite) is equinumerous with a proper subset of
itself. Rather in ZFC (i.e., adding the axiom of choice), we proof
that every infinite set is equinumerous with a proper subset of
itself. That is, with the non-Dedekind definitions of 'finite' and
'infinite', in ZF, we do NOT prove that every infinite set is Dedekind
infinite, but rather we we add the axiom of choice (actually, we can
do it with the countable axiom of choice) which proves that every
infinite set is Dedekind infinite.


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.


NO, you are using UNDEFINED terminology when you say "half as many"
regarding INFINITE sets. There is no operation of taking "half as
many" for INFINITE cardinals. Just CLAIMING, without PROVING how do
make such a definition is typical CRANKSPEAK.

And you go on and on in your post. I've addressed enough for now. You
can either learn about the subject in a rational way, or you can keep
going on and on with your claims based in ignorance and confusion
about the subject. That's your choice.

MoeBlee


.

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