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

On May 16, 12:28 am, Phil <toob-head...@xxxxxxxxxxxxx> wrote:

I am pointing out that the CLAIM that N = E is an INHERENT PROPERTY of
any infinite set -- i.e., that the belief that IF set X has infinitely
many elements, then that fact BY ITSELF, with no need for additional
axioms, makes it possible for a proper subset of X to have the same
cardinality as X -- is false.

Who made such a claim?

I have been mentioning all along that we prove N = E from axioms.

Also, that every infinite set (in the sense of not being equinumeruous
with a natural number) is equinumerous with a proper subset of itself
(i.e. that every infinite set is Dedekind infinite) is also proven
from axioms (and the axiom of denumerable choice is used for this in
addition to just the axioms of ZF).

Neither you nor anyone else has proven,
using only the properties that are INHERENT TO ANY infinite set, that
any infinite set has the same cardinality as a proper subset of itself,

No, we prove "Every infinite set is Dedekind infinite" from axioms.

and THAT is what Cantor both tried and claimed to prove!

Cantor did not work in a formal set theory. We've come a long way
since Cantor.

Go ahead! Do
what no one else has done, and prove, WITHOUT using axioms that do NOT
necessarily follow from the mere fact that a set has infintely many
elements, that N = E!

Just to be clear here, a proof of "N = R" such as I mentioned makes no
use of any premise that every infinite set is equinumerous with a
proper subset of itself. On the other hand, "N = E" entails that there
is at least one infinite set that is equinumerous with a proper subset
of itself; but "N = E" doesn't entail that every infinite set is a
proper subset of itself. Anyway, the proof I mentioned of "N = E" is
from axioms and those axioms do not include "Every infinite set is
equinumerous with a proper subset of itself" and those axioms do not
even entail "Every infinite set is equinumerous with a proper subset
of itself".

Remember, the INFINITE set of halfway points from the middle of the room
to the door, when traversed in reverse, leaves, once we travel any
finite distance into the room whatsoever, a FINITE number of points
remaining in the set, meaning that even n + 1, let alone 2 * n, does NOT
always exist! Of course, you could claim that "we cannot address points
that are preceded by infinitely many points, and followed by finitely
many points," but that means that basic mathematics cannot handle
something as simple as walking into a room, and I neither buy that, nor
am I willing to accept that. What this means, whether you have the
intellectual honesty to admit it or not, is that it is not NECESSARILY
the case that in an infinite set, whenever n is in the set, so is 2n. Of
course, two times 0.75 = 1.5, and yet the number of points in the
segment [0,1] is infinite, so the fact that whenever n is an element in
an infinite set, that 2n may or may not be an element of the set,
shouldn't be a complete surprise to anyone.

In other words, the "fact" that if n is an element of infinite set X,
then so is 2n, is actually not a fact, but an axiom,

Who said "if n is an element of an infinite set X then so is 2n"?

No, what we prove as a THEOREM from AXIOMS is that if n is a member of
the set of natural numbers then 2n is a member of the set of natural
numbers. Not just any set infinte set X, since we also prove that
there ARE infinite sets X such that n is in X but 2n is NOT in X.
Rather, the infinite set that is the set of natural numbers is the
PARTICULAR infinite set we're talking about. And we prove that THAT
particular infinite set has the property that if n is a member of it
then 2n is a member of it.

one which does NOT
necessarily follow from simply from the fact that set X is infinite.

Who said that anything like that?

And
yet, everyone CLAIMS that because set X is infinite, that it IS equal to
a proper subset of itself.

No, I already told you that it is NOT the case that everyone claims
that. For example, what I claim is that in Z set theory with the axiom
of denumerable choice added, it is a theorem that every infinite set
is equinumerous with a proper subset of itself.

Now, if they claimed that when set X is
infinite AND we happen to choose an axiom that states that when n is in
X, then so is 2n, THEN they would have a valid proof that set X and any
infinite proper subset of X have the same cardinality.

That is ALL COMPLETELY mixed up. What we show is that from certain
axioms it follows that if n is a member of the set of natural numbers
then 2n is a member of the set of natural numbers. We don't claim that
if n is a member of just any infinite set X then 2n is a member of X.
And the proof that any infinite set X is equinumerous with a proper
subset of itself is a separate matter and is proven from axioms that
include some weak or full form of the axiom of choice.

But that would
mean that we could also choose an alternative axiom, whereupon this is
NOT true, and no one would be stupid enough to claim that N = E follows
simply because N is infinite.

It is consistent with ZF to have an axiom "It is not the case that
every infinite set is equinumerous with a proper subset of itself".
But it is INconsistent with Z (or with ZF) to have an axiom "No
infinite set is equinumerous with a proper subset of itself". And,
anyway, the proof of "N = E" is not very much related to that, since
"N = E" is provable in Z set theory alone, and the question of whether
every infinite set is Dedekind infinite does not affect the proof of
"N = E".

You are EXTREMELY mixed up about all of this. The only hope for you
have of being able talk coherently about this subject is to actually
study it while paying close attention to the PRECISE formulations in
those studies.

MoeBlee


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