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

On 10/14/2024 2:07 PM, WM wrote:

On 14.10.2024 19:13, Jim Burns wrote:

On 10/14/2024 10:43 AM, WM wrote:

On 13.10.2024 18:28, Jim Burns wrote:

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} .

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Each γ≠0 preceding ω is (our) finite.

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There are two alternatives:
Either doubling creates natnumbers,
  then they are not among those doubled,
  then we have not doubled 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
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⎛ 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)

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Correct.
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⎜ For each j such that 0 < j ≤ 2⋅(𝔊+1)
⎜  j-1 exists.
⎜ 2⋅(𝔊+1) < ω

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Mistake.

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ω is the first not.finite ordinal.
Each finite ordinal is before ω

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Yes.

Your 'yes' is really disagreement with us
because you have your private idea of
what 'finite' means.

If 2⋅𝔊 is countable.to from 0
then 2⋅(𝔊+1) is countable.to from 0

>
Mistake.

lemma.
2⋅(𝔊+1) is countable.to from 2⋅𝔊
Proof:  ⟨ 2⋅𝔊 2⋅𝔊+1 2⋅(𝔊+1) ⟩
----
If 2⋅𝔊 is countable.to from 0
and 2⋅(𝔊+1) is countable.to from 2⋅𝔊  [lemma]
then 2⋅(𝔊+1) is countable.to from 0

If 2⋅𝔊 is finite
then 2⋅(𝔊+1) is finite

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Mistake.
>

If 2⋅𝔊 < ω
then 2⋅(𝔊+1) < ω

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Mistake.
>
Try again
considering the darkness of most numbers.

⎛ A set S of ordinals holds first.S or is empty.
⎜
⎜ Ordinal ξ has successor ξ+1 = ξ∪{ξ}
⎜
⎜ _Finite_ non.0 ordinal k
⎜  has predecessor k-1:  (k-1)+1 = k  and
⎜ each prior non.0 ordinal j < k
⎝  has predecessor j-1:  (j-1)+1 = j
Whether k or ξ or any ordinal is dark
is irrelevant to that description.