Re: Incompleteness of Cantor's enumeration of the rational numbers

On 11/22/2024 9:51 AM, WM wrote:

On 22.11.2024 13:32, joes wrote:

Then we can consider those sets
to have the same number.

>
That is the big mistake.
It makes you think that
the sets of naturals and of prime numbers
could cover each other.

ℙ is the set of prime numbers.
ℙ(n) is the set of the first n prime numbers.
max.ℙ(n) is the nᵗʰ prime number.
For each n ∈ ℕ
a set ℙ(n) of the first n prime numbers exists,
and, since ℙ(n) is a finite set,
max.ℙ(n) = pr(n) the nᵗʰ prime number exists.
Therefore,
n ↦ pr(n) ↦ n: one.to.one
ℙ covers ℕ, and ℕ covers ℙ
ℙ ≠⊂ ℕ
ℙ and ℕ are infinite sets.
----
For each n ∈ ℕ
a set ℙ(n) of the first n prime numbers exists,
⎛ Assume otherwise.
⎜ Assume n′ ∈ ℕ and ℙ(n′) doesn't exist.v
⎜
⎜ The ℕ.subset of numbers i
⎜  such that ℙ(i) doesn't exist
⎜ is not empty, and
⎜ holds a first element 𝔊+1 such that
⎜  ℙ(𝔊+1) doesn't exist  and
⎜  ℙ(𝔊) exists
⎜
⎜ max.ℙ(𝔊) = pr(𝔊) is the 𝔊ᵗʰ prime.
⎜
⎜ pr(𝔊)!+1 is not.divisible by each of ⟦2,pr(𝔊)⟧ᴺ
⎜
⎜ The ℕ.subset of numbers i
⎜  such that i is not.divisible by each of ⟦2,pr(𝔊)⟧ᴺ
⎜ is not empty  and
⎜ holds a first element p such that
⎜  p is not.divisible by each of ⟦2,pr(𝔊)⟧ᴺ  and
⎜  each of ⦅pr(𝔊),p⦆ᴺ is divisible by one of ⟦2,pr(𝔊)⟧ᴺ
⎜
⎜ None of ⦅pr(𝔊),p⦆ᴺ divides p  or else
⎜  one of ⟦2,pr(𝔊)⟧ᴺ dividing one of ⦅pr(𝔊),p⦆ᴺ
⎜  would divide p, which it doesn't.
⎜
⎜ None of ⟦2,p⦆ᴺ divides p
⎜ p is a prime number.
⎜ p = pr(𝔊+1) next prime after pr(𝔊)
⎜ ℙ(𝔊+1) = ℙ(𝔊)∪{p} exists
⎜
⎜ However,
⎜ ℙ(𝔊+1) doesn't exist.
⎝ Contradiction.
Therefore,
for each n ∈ ℕ
a set ℙ(n) of the first n prime numbers exists.
And
n ↦ pr(n) ↦ n: one.to.one
ℙ covers ℕ, and ℕ covers ℙ
ℙ ≠⊂ ℕ
ℙ and ℕ are infinite sets.