Re: Contradiction of bijections as a measure for infinite sets

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.
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

Or could you explain why
first bijecting n and n/1 should destroy
an existing bijection?

For finite sets,
one bijection implies
no non.bijection (no not.onto injection).
one non.bijection implies no bijection.
ℕ and ℚᶠʳᵃᶜ aren't finite sets.
For ℕ and ℚᶠʳᵃᶜ
one non.bijection not.implies no bijection.
Bijecting n and n/1 do not "destroy"
bijections between ℕ and ℚᶠʳᵃᶜ