Re: Cantor's cardinality finally refuted.


Rupert wrote:
zuhair wrote:
Rupert wrote:
zuhair wrote:
zuhair wrote:
Example:

For the FINITE sets X and Y.
Let f:X->Y
A= f is injective and not surjective.
B= f is injective and surjective.
C= (card X = card Y)
D= (card X < card Y).

((Ef(A) xor Ef(B)) <->(C xor D)) /\ (Ef(B)<->D)) -> (Ef(A)<->C).

Is that correct.

Rupert wrote:
Yes.

Zuhair

In reality for the finite sets X and Y and all the above. even the
following statement is true.

Ef((A) xor (B)) <->(C xor D)) /\ (Ef(B)<->D)) -> (Ef(A)<->C).

Now what I want to show is the following :

1: Ef(f:X->Y,f is injective)<->card X <= card Y. This is a Cantorian
statement.

Ef(f:X->Y, f is injective)<-> Ef(f:X->Y, f is injective and surjective
v f:X->Y, f is injective and not surjective)

subtituting this in 1.

Ef(f:X->Y, f is injective and surjective v f:X->Y, f is injective and
not surjective) <->
card X <= card Y.

You say that since Ef(f:X->Y, f is injective and surjective) and
Ef(f:X->Y,f is injective and not surjective)are not mutually exclusive
( which is correct), then even though Ef(f:X->X,f is bijective)<->card
X = card Y ,it doesn't follow that Ef(f:X->Y,f is injective and not
surjective)<->card X < card Y.

But what it follows then.

Let me use the abbreviated notation again.

For the ANY sets X and Y.
Let f:X->Y
A= f is injective and surjective.
B= f is injective and not surjective.
C= (card X = card Y)
D= (card X < card Y).

You are saying that the following statement is incorrect.

Ef((A) or (B)) <->(C xor D)) /\ (Ef(A)<->C)) -> (Ef(B)<->D).

But what follows then. I have these probabilities in my mind.

1)Ef(A)<-> C xor D and Ef(B)<-> C xor D

I don't follow you here. It's not true that Ef(A)<->(C xor D).

Read the whole thread, I already said that.

Oh right. Well the problem with (4) is that Ef(A) and Ef(B) are not
mutually exclusive. So if you adopted it then sometimes you'd have both
C and D, i.e. you'd have card X < card Y and card X = card Y at the
same time. Most people agree that shouldn't happen, so that's why we
don't use definition (4).

Hmmm...., This is not enough cause to refuse it, it is only enough to
say that it is incomplete determination of size comparison, since we
can apply restrictions on the definition of size to prevent this. Of
course when I say that card X < card Y I should apply restirction to
prevent card X = card Y, by finding a fair way to select which of these
two functions reflect the true size comparison.. Anyhow, now I
understand matters, appearantly mathematicians do not like these
restrictions, that's why they prefer Cantor's cardinality which work
without them.

Zuhair



2)Ef(A)<-> C xor D and Ef(B)<-> D
3)Ef(A)<-> C and Ef(B)<-> C xor D
4)Ef(A)<-> C and Ef(B)<-> D
5)Ef(A)<-> C and Ef(B) -> C xor D.

Now 1) , 2) , 3) leads to contradictions.

what remains is 4) and 5). Obviouselly Cantor chose 5).

My question is that chioce Cantor did a bias or not?

Why I say it is a bias, because there is no logical evidence against
4).

IF it is a bias, then I can also make a bias and hold that 4) is true
not 5).since it works for both finites and infinites, and it represent
a more extentional concept of set size from finites to infinite, I mean
set size as defined after 4) would be the same concept for finites and
infinites. While Cantorian selection of 5) would lead to a size that is
conceptually different from that of finites.

Why not develop a fair methodology that enable us to choose among 4) or
5) according to the sets X and Y. themselfs, i.e. let X and Y choose
among 4) and 5) and not our biased preferences.


Zuhair

.

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