Re: Contradiction of bijections as a measure for infinite sets

Le 03/04/2024 à 23:48, Jim Burns a écrit :

On 4/3/2024 9:32 AM, WM wrote:

Le 02/04/2024 à 17:51, Jim Burns a écrit :

 

If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.

 My trick proves that there is no bijection.

 See below,
for each k ∈ ℕ  its own iₖ/jₖ ∈ ℚᶠʳᵃᶜ
for each i/j ∈ ℚᶠʳᵃᶜ  its own kᵢⱼ ∈ ℕ
 Your trick gives incorrect results.

My trick unveils that there are mor fractions than indices.

Bijecting n and n/1 do not "destroy"
bijections between ℕ and ℚᶠʳᵃᶜ

Mapping n/1 in the fractions shows that no bijection is possible.
Regards, WM