Re: Contradiction of bijections as a measure for infinite sets

Le 24/03/2024 à 22:02, Dieter Heidorn a écrit :

WM schrieb:

 Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
can be added?
 

Sure

Of course.

 But you can do the following described by Cantor:

Irrelevant. Important is only this: You cannot add a natural number to the set ℕ.

 

If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove
that both sets have the same number of elements.

 Infinite sets don't have a "number of elements".

This cannot be denied: A bijection, if really existing, proves that one of both sets has not one element more nor less than the other!

 And indeed: there is a bijection from the set of natural numbers ℕ
to the set of even natural numbers 𝔼 = {2, 4, 6, ..}.
     f: ℕ → 𝔼 ,  n ↦ 2n
 This function is both injective (or one-to-one) and surjective (or
onto), thus it is bijective.

If so, that would result in: The set 𝔼 has not one element more nor less than the set ℕ.

 

Then the completion of 𝔼 resulting in E = {1, 2, 3, 4, 5, 6, ...}
would double the number of its elements. Then there are more natural
numbers than were originally in ℕ.

 Rubbish. The cardinality of an infinite set is described by an
transfinite cardinal number and not by a finite "number of elements".

Here we do not use the rubbish of cardinality but the definition of bijection proving that one of both sets has not one element more or less than the other!

Your problem is: You try to apply facts,

I apply logic which is universally valid.
If the set 𝔼 = {2, 4, 6, ..} has not one element more or less than the set ℕ = {1, 2, 3, ...}, then adding an element to 𝔼 destroys this state.

that hold for finite sets,
on infinite sets. That doesn't work.

Your problem is you deny logic which is universally valid.
Regards, WM