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

On 11/23/2024 5:30 AM, WM wrote:

On 22.11.2024 22:50, Jim Burns wrote:

ℙ covers ℕ, and ℕ covers ℙ

>
Let every unit interval on
the infinite real axis
be coloured white. Cover the unit intervals of prime numbers
by red hats.
It is impossible to shift the red hats

Yes, because we are finite beings,
and there are infinitely.many red hats.

It is impossible to shift the red hats
so that all unit intervals of
the whole real axis get red hats.
There are too few prime numbers.

No.
⎛ Assume that that is so.
⎜ Assume that there are
⎜ enough red hats for the first 𝔊 numbers
⎜ but not enough for the 𝔊+1ᵗʰ
⎜
⎜ Before shifting red hats,
⎜ the 𝔊ᵗʰ red hat was at the 𝔊ᵗʰ prime pr(𝔊)
⎜ and there were no more primes after pr(𝔊)
⎜
⎜ However,
⎜ the number pr(𝔊)!+1 is not.divisible.by
⎜ each number in sequence ⟦2,pr(𝔊)⟧
⎜ ⟦2,pr(𝔊)⟧ ᵉᵃᶜʰ~⃒ pr(G)!+1
⎜
⎜ Non.empty.set {i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i} of numbers
⎜  not.divisible.by each in ⟦2,pr(𝔊)⟧
⎜ holds least min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i}
⎜
⎜ min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i} is prime
⎜
⎜⎛ Each of ⟦2,pr(𝔊)⟧
⎜⎜  not.divides min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i}
⎜⎜
⎜⎜ For each of ⦅pr(𝔊),min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i}⦆
⎜⎜  one of ⟦2,pr(𝔊)⟧ divides it,
⎜⎜ otherwise,
⎜⎜( min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i} isn't least,
⎜⎜ which it is.
⎜⎜
⎜⎜ Each of ⦅pr(𝔊),min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i}⦆
⎜⎜  not.divides min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i}
⎜⎜ Otherwise,
⎜⎜⎛ one of ⟦2,pr(𝔊)⟧ which divides it
⎜⎜⎝ also divides min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i}
⎜⎝ which it doesn't.
⎜
⎜ Each of ⟦2,min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i}⦆
⎜  not.divides min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i}
⎜
⎜ min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i} is prime
⎜ min.{i∈ℕ:⟦2,pr(𝔊)⟧ᵉᵃᶜʰ~⃒i} > pr(𝔊)
⎜
⎜ Therefore,
⎜ there is another prime after pr(G)
⎝ Contradiction.

There are too few prime numbers.

No,
there being too few primes
leads to contradiction.