Re: Contradiction of bijections as a measure for infinite sets

On 3/24/24 4:11 PM, WM wrote:

 Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can be added?
 If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that both sets have the same number of elements. 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.
 Regards, WM
 

Yep, because "infinity" just doesn't obey the logic you are used to, and insist on using, even in cases it can't be appled to, making your logic system just all blown up to smithereens.
Turns out that aleph0 = aleph0 - 1, = aleph0 + 1 = aleph0 / 2 =
    aleph0 * 2 = aleph0 ^ 2 = square root (aleph0)
despite that breaking most of you concepts of the logic of numbers.