Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)

On 10/12/2024 2:06 PM, WM wrote:

On 10.10.2024 22:47, Jim Burns wrote:

On 10/9/2024 11:39 AM, WM wrote:

{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .

>
Each γ≠0 preceding ω is (our) finite.

>
There are two alternatives:
Either doubling creates natnumbers,
 then they are not among those doubled,
 then we have not doubles all.
Or we have doubled all
 but then larger numbers have been created.

There are two alternatives:
Either 𝔊 exists such that 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
Or {2,4,6,...} ᵉᵃᶜʰ< ω
⎛ Assume 2⋅𝔊 < ω ≤ 2⋅(𝔊+1)
⎜
⎜ 2⋅𝔊 < ω
⎜ For each j such that 0 < j ≤ 2⋅𝔊
⎜  j-1 exists.
⎜ 2⋅𝔊 = (2⋅𝔊+1)-1 exists
⎜ 2⋅𝔊+1 = (2⋅𝔊+2)-1 exists
⎜ 2⋅𝔊+2 = 2⋅(𝔊+1)
⎜ For each j such that 0 < j ≤ 2⋅(𝔊+1)
⎜  j-1 exists.
⎜ 2⋅(𝔊+1) < ω
⎜
⎜ However,
⎜ ω ≤ 2⋅(𝔊+1)
⎝ Contradiction.
Therefore,
{2,4,6,...} ᵉᵃᶜʰ< ω

β < ω  ∧  γ < ω  ⇒  β+γ < ω
>
⎛ Assume a counterexample.
⎜ Assume
⎜ β < ω  ∧  γ < ω  ∧  β+γ ≥ ω

 Dark numbers have no discernible individuality.

Natural numbers are countable.to from 0.
⎛ Each non.0 natural number k
⎜ has predecessor k-1  and
⎜ each prior non.0 natural number j < k
⎜ has predecessor j-1
⎜
⎜ Each set S of natural numbers
⎜ holds first.in.S or is empty.
⎜
⎜ Each natural number k
⎝ has successor k+1 = k∪{k}

β < ω  ∧  γ < ω  ⇒  β×γ < ω
>
⎛ Assume a counterexample.
⎜ Assume
⎜ β < ω  ∧  γ < ω  ∧  β×γ ≥ ω

 Dark numbers have no discernible individuality.

Natural numbers are countable.to from 0.