Re: Cantor's cardinality finally refuted.


zuhair wrote:
Rupert wrote:
zuhair wrote:
Hi All,

Herein a refutation of Cantor's cardinality.

Aim: to illustrate what existed in Cantor's
thought, and then I will pinpoint were cantor's definitions becomes
contradictive or at best biased.

Cantor always related bijectability to equality of set size, the later
he called "cardinality".
he had this in his mind: Ef(f:A->B, f is bijective)<-> card A = card B.

Now the problem was that for some cases of A and B , we cannot find
such a bijective function though they look to be equal in size, for
example sets N={0,1,2,3,4,...}
and P={2,3,5,7,11,.....} , i.e. P is the set of all prime numbers.

Yes we can. Let f(n) be the (n+1)-th prime. Then f is a bijection N->P.

Here came Cantor-Bernsteing-Shroider Theorum ( I am not sure of the
naming ) ,

Cantor-Schroeder-Bernstein.

as a salvage, this thoerum states that if there exist an
injection from A to B and an injection from B to A, then a bijective
function between A and B exists, so if this bijective function is not
known we can call it the implicit bijective function.

But we do know it, because the proof explicitly constructs it.

Cantor asked himself, how can we relate this to size comparison of A
and B.
He concluded the following.

Ef(f:A->B,f is injective)<-> card A <= card B.

And Cantor Generalized this reasoning to Both finite an infinite sets.
so if we demonstrate injection from A to B then card A <= card B, And
if we
demonstrate an injection from B to A then card A >= card B. so we have
card A <= card B And card A >= card B, this leads to card A = card B.


Their is no problem with the reasoning above, except that it doesn't
tell the whole story, when it come to compare infinite sets that has
bijection between them.


I have demonstrated in a very nice manner in a previous post that the
two most important arguments Cantor's reasoning is built upon leads to
another arguments he didn't pay much attention to. as below;

Cantor's arguments are basically the followings:-
1)Ef(f:A->B,f is bijective)<-> card A = card B.
2)Ef(f:A->B,f is injective)<-> card A<= card B.

I demonstrated using consistent logical arguments the following:

1 ) -> Ef(f:A->B,f is injective but not surjective)<-> card A < card B.


Wrong.

2 ) -> Ef(f:A->B,f is injective but not surjective)<-> card A < card B.


Isn't that the same thing?

1) -> Ef(f:A->B,f is not injective but surjective)<-> card A > card B.


Wrong.


The details of this are as below:

The proof that 2 ) -> Ef(f:A->B,f is injective but not surjective)<->
card A < card B

we know that : Ef(f:A->B,f is injective)<->Ef(f:A->B,f is injective and
surjective
v f:A->B,f is injective and not surjective).

subtituting this in 2) above we have.

3)Ef(f:A->B,f is injective and surjective v
f:A->B,f is injective and not surjective) <-> card A <= card B.

From 1) we know that Ef(f:A->B,f is bijective)<-> card A = card B.
then dropping that from 3) should lead to

4)Ef(f:A->B,f is injective and not surjective)<-> card A < card B.


No, just because you have (A or B)<->(C or D) and B<->D you cannot
infer A<->C.

Therefore 2)->4).

The proof that 1 ) -> Ef(f:A->B,f is injective but not surjective)<->
card A < card B.

By negating both sides of 1) we have the following:

1' )~Ef(f:A->B,f is bijective)<-> card A <> card B.

we know that:~Ef(f:A->B,f is bijective)<-> Ef(f:A->B,f is injective and
not surjective v f:A->B,f is surjective and not injective)

subtituting this in 1' ) we have.

2' ) Ef(f:A->B,f is injective and not surjective v f:A->B,f is
surjective and not injective)<-> card A <> card B.

since it is clear that Ef(f:A->B, f is surjective and not injective)<->
card A > card B.

drop that from 2' ) we will get:

3' ) Ef(f:A->B,f is injective and not surjective)<-> card A < card B.


Same problem.

Cantor's fault is when he generalized his reasoning outlined above to
the case of infinit sets.

to demonstrate this lets take any two infinite sets A and B.

It can be proved using Dedekind definition of an infinite set, that
there between any two infinite sets A and B there exist the following
kinds of injections.

For A and B were A is infinite and B is infinite all the following are
true:-
1)Ef(f:A->B,f is bijective)
2)Ef(f:A->B,f is injective but not surjective)
3)Ef(f:B->A,f is injective but not surjective)

This mean that for any two infinite sets A and B the following is
true.

card A = card B and card A < card B and card A > card B..


No.

A striking contradiction .


Yes, and since the contradiction arises from your reasoning it would
seem most reasonable to assume that something has gone wrong with your
reasoning.

Yea I discovered what is wrong with my reasoning, but it was not what
you mentioned,

Yes, it was.

in reality it was a pointed out by Virgil in another
thread.

Cantor sayes that 1)Ef(f:A->B, f is bijective ) <-> card A = card B.

Now negating both sides we get.

Af(f:A->B,f is ~bijective)<-> card A <>card B.
this leads to the following

Af(f:A->B, f is injective and not surjective v f:A->B, f is surjective
and not injective)<-> card A <>card B.

This leads to the followings:

Af(f:A->B, f is injective and not surjective)<-> card A < card B
and
Af(f:A->B, f is surjective and not injective)<-> card A > card B.

in reality it was Virgil who refuted my alleged refutation of Cantor,
thanx to him.


If Virgil helped you to understand the concepts better, that's great.
But the above doesn't show what is wrong with your reasoning. I pointed
out the error in your reasoning.

But still my trial to find another definition of size stands a chance
but on other grounds.

My feeling is in the following manner:

I tend to think that if we KNOW the surjective state and the injective
state of any function between A and B then we should arrive at a strict
quantitative relation between A and B, as dictated of course by this
function. in symboles I mean the following:

Ef(f:A->B, f is injective and surjective) <-> (size A)_f = (size B)_f
Ef(f:A->B, f is injective and not surjective)<->(size A)_f < (size B)_f
Ef(f:A->B, f is surjective and not injective)<->(size A)_f > (size B)_f

while if we only know the injective xor the sujrective state of a
function between A and B, then we can only determine a non strict
quantitative relation between A and B, as dictated by this function. in
symboles I mean the following:

Ef(f:A->B, f is injective ) <-> (size A)_f <= (size B)_f
Ef(f:A->B, f is surjective)<-> (size A)_f >= (size B)_f

Now the absolute size comparison of A and B depends on weather all
injective functions between A and B determiens the same size comparison
per function. If so then the absolute size comparison is the same as
any functional size of the set.

so the following exist in my head,

Af(f:A->B, f is injective and not surjective )<-> size A < size B.
Af(f:A->B, f is injective and surjective)<-> size A = size B.
Af(f:A->b, f is sujrective and not injective)<-> size A > size B.

But what about A and B, in which not every f is of the same type.i.e
some are injective and not sujrective others are injective and
surjective, etc...

In this condition , I think size of A and the size of B should be
determined by one and only one function , i call it the size
determining injective function"s.d.i.f". But till now I didn't manage
to form a rigorous definition of this function. though I know if f(x)=x
is injective between A and B then it is the s.d.i.f. , so the size
comparison determined by this function reflects the true size
comparison between A and B.

Anyhow I think my approach will be proove to be the right one at the
end, not Cantor's,Cantor's approach is a compromise. In reality it is
biased bo bijection, and this will lead to a lot of false positive
equality of set size. Days will prove that I was right.


I see. Great.


Either that or identify a problem with Cantor's reasoning.
If there's no problem with either, then ZFC is inconsistent.

Though I am sure that ZFC is inconsistent, but it is not due this
reason.


Why are you sure? Have you found a proof of a contradiction in ZFC?

Zuhair

.

Relevant Pages

  • Re: Cantors cardinality finally refuted.
    ... injection from A to B and an injection from B to A, ... And Cantor Generalized this reasoning to Both finite an infinite sets. ... so if we demonstrate injection from A to B then card A <= card B, ...
    (sci.math)
  • Re: Cantors cardinality finally refuted.
    ... injection from A to B and an injection from B to A, ... And Cantor Generalized this reasoning to Both finite an infinite sets. ... so if we demonstrate injection from A to B then card A <= card B, ...
    (sci.math)
  • Re: Cantors cardinality finally refuted.
    ... injection from A to B and an injection from B to A, ... And Cantor Generalized this reasoning to Both finite an infinite sets. ... so if we demonstrate injection from A to B then card A <= card B, ...
    (sci.math)
  • Re: Cantors cardinality finally refuted.
    ... injection from A to B and an injection from B to A, ... And Cantor Generalized this reasoning to Both finite an infinite sets. ... so if we demonstrate injection from A to B then card A <= card B, ...
    (sci.math)
  • Re: Cantors cardinality finally refuted.
    ... injection from A to B and an injection from B to A, ... And Cantor Generalized this reasoning to Both finite an infinite sets. ... so if we demonstrate injection from A to B then card A <= card B, ...
    (sci.math)