Re: Contradiction of bijections as a measure for infinite sets

On 4/2/2024 3:36 AM, WM wrote:

Le 02/04/2024 à 01:03, Richard Damon a écrit :

On 4/1/24 11:37 AM, WM wrote:

The different sets are ℕ and ℚ.
The bijection with the first column does not change that.

>
Yes it does,
as you are not "moving" the O out of the set of Q.
That makes a difference.

>
It does not make a difference.
It only shows that there is no bijectio.

If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.
k ∈ ℕ
k ⟼ iₖ/jₖ
sₖ = max{h: (h-1)(h-2/2 < k }
iₖ = k-(sₖ-1)(sₖ-2)/2
jₖ = sₖ-iₖ
iₖ/jₖ ∈ ℚᶠʳᵃᶜ
i/j ∈ ℚᶠʳᵃᶜ
i/j ⟼ kᵢⱼ
sᵢⱼ = i+j
kᵢⱼ = (sᵢⱼ-1)(sᵢⱼ-2)/2+i
kᵢⱼ ∈ ℕ
sᵢⱼ = sₖ
kᵢⱼ = k  ⟺  i/j = iₖ/jₖ
There is a bijection ℕ ⇄ ℚᶠʳᵃᶜ
Your assumption is wrong.