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

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

On 09.10.2024 17:11, Alan Mackenzie wrote:

WM <wolfgang.mueckenheim@tha.de> wrote:

No.
When we *in actual infinity*
multiply all |ℕ|natural numbers by 2,
then we keep |ℕ| numbers
but only half of them are smaller than ω,
i.e., are natural numbers.
The other half is larger than ω.

You (WM) are treating ω as though it is (our) finite.
ω is the first (our) transfinite ordinal: not finite.

Ha ha ha ha! This is garbage.
If you think doubling some numbers gives results
which are "larger than ω"
you'd better be prepared to give
an example of such a number.
But you're surely going to tell me that
these are "dark numbers

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

Each γ≠0 preceding ω is (our) finite.
Each γ≠0 preceding ω is predecessored  and
each β≠0 preceding γ is predecessored.
γ < ω  ⇔
⎛ for each β: 0 < β ≤ γ  ⇒
⎝ exists α: 0 ≤ α < β  ∧  α+1 = β
γ < ω  ⇔
∀β ∈ ⦅0,γ⟧: ∃α ∈ ⟦0,β⦆: α+1=β
----
⎛ If
⎜  γ≠0 preceding ω is predecessored and
⎜  each β≠0 preceding γ is predecessored,
⎜ then
⎜  γ+1≠0 is predecessored and
⎜  each β≠0 preceding γ+1 is predecessored, and
⎝  γ+1 precedes ω
Therefore,
γ < ω  ⇒  γ+1 < ω
----
⎛ 0+γ = γ
⎝ (β+1)+γ = (β+γ)+1
β < ω  ∧  γ < ω  ⇒  β+γ < ω
⎛ Assume a counterexample.
⎜ Assume
⎜ β < ω  ∧  γ < ω  ∧  β+γ ≥ ω
⎜
⎜ The nonempty set
⎜ {β < ω: γ < ω ∧ β+γ ≥ ω)
⎜ holds a minimum 𝔊+1 and
⎜ 𝔊+1 has a predecessor 𝔊 not.in the set.
⎜
⎜ 𝔊 < 𝔊+1 < ω  ∧  γ < ω
⎜ 𝔊+γ < ω  ∧  (𝔊+1)+γ ≥ ω
⎜
⎜ However,
⎜  γ < ω  ⇒  γ+1 < ω
⎜ 𝔊+γ < ω  ⇒  (𝔊+γ)+1 < ω
⎜ (𝔊+γ)+1 = (𝔊+1)+γ
⎜ (𝔊+1)+γ < ω
⎝ Contradiction.
Therefore,
β < ω  ∧  γ < ω  ⇒  β+γ < ω
----
⎛ 0×γ = 0
⎝ (β+1)×γ = (β×γ)+γ
β < ω  ∧  γ < ω  ⇒  β×γ < ω
⎛ Assume a counterexample.
⎜ Assume
⎜ β < ω  ∧  γ < ω  ∧  β×γ ≥ ω
⎜
⎜ The nonempty set
⎜ {β < ω: γ < ω ∧ β×γ ≥ ω)
⎜ holds a minimum 𝔊+1 and
⎜ 𝔊+1 has a predecessor 𝔊 not.in the set.
⎜
⎜ 𝔊 < 𝔊+1 < ω  ∧  γ < ω
⎜ 𝔊×γ < ω  ∧  (𝔊+1)×γ ≥ ω
⎜
⎜ However,
⎜  β < ω  ∧  γ < ω  ⇒  β+γ < ω
⎜ 𝔊×γ < ω  ⇒  (𝔊×γ)+γ < ω
⎜ (𝔊×γ)+γ = (𝔊+1)×γ
⎜ (𝔊+1)×γ < ω
⎝ Contradiction.
Therefore,
β < ω  ∧  γ < ω  ⇒  β×γ < ω

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

γ < ω  ⇒  γ+1 < ω
There is no first successor out of ⟦0,ω⦆
There is no successor out of ⟦0,ω⦆
β < ω  ∧  γ < ω  ⇒  β+γ < ω
There is no first sum out of ⟦0,ω⦆
There is no sum out of ⟦0,ω⦆
β < ω  ∧  γ < ω  ⇒  β×γ < ω
There is no first product out of ⟦0,ω⦆
There is no product out of ⟦0,ω⦆
{1,2,3,...} ᵉᵃᶜʰ< ω
{1,2,3,...}ᵉᵃᶜʰ×2  =  {2,4,6,...} ᵉᵃᶜʰ< ω

Should all places ω+2, ω+4, ω+6, ... remain empty?

Should ω be (our) finite?

Should the even numbers in spite of doubling
remain below ω?
Then they must occupy places not existing before.
That means
the original set had not contained all natural numbers.

ω is the first (our) transfinite ordinal.
∀γ:  γ ∈ ⟦0,ω⦆  ⇔
∀β ∈ ⦅0,γ⟧: ∃α ∈ ⟦0,β⦆: α+1=β