Sujet : Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 14. Oct 2024, 18:13:28
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <68b8be64-7fe8-47e7-a991-7adf14713af5@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
User-Agent : Mozilla Thunderbird
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} .
>
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 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)
>
Correct.
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⎜ For each j such that 0 < j ≤ 2⋅(𝔊+1)
⎜ j-1 exists. ⎜ 2⋅(𝔊+1) < ω
>
Mistake.
ω is the first not.finite ordinal.
Each finite ordinal is before ω
If 2⋅𝔊 is countable.to from 0
then 2⋅(𝔊+1) is countable.to from 0
If 2⋅𝔊 is finite
then 2⋅(𝔊+1) is finite
If 2⋅𝔊 < ω
then 2⋅(𝔊+1) < ω
⎜ However,
⎜ ω ≤ 2⋅(𝔊+1)
⎝ Contradiction.
>
No, your mistake.
You (WM) think that
infinite is just like finite,
but more.
However,
finite is countable.to from nothing, and
infinite is otherwise, is not countable.to
Whatever is countable.to
is countable.past, to more countable.to
No last finite successor.
𝔊 < ω ⇒ 𝔊+1 < ω
Sums are defined from successors.
k+(𝔊+1) = (k+𝔊)+1
No last finite sum.
k+𝔊 < ω ⇒ (k+𝔊)+1 < ω ⇒ k+(𝔊+1) < ω
Products are defined from sums.
k×(𝔊+1) = k+(k×𝔊)
No last finite product.
k×𝔊 < ω ⇒ k+(k×𝔊) < ω ⇒ k×(𝔊+1) < ω
And so on, staying within the finites.
k^(𝔊+1) = k×(k^𝔊)
k^𝔊 < ω ⇒ k×(k^𝔊) < ω ⇒ k^(𝔊+1) < ω
k⇈(𝔊+1) = k^(k⇈𝔊)
k⇈𝔊 < ω ⇒ k^(k⇈𝔊) < ω ⇒ k⇈(𝔊+1) < ω
...
ω is the first not.finite ordinal.
Finite does NOT mean less.than.unreasonably.large.
Finite means countable.to.from.0
ω is the first not.countable.to.from.0 ordinal.
Everything before ω is countable.to.from.0
ω and everything after ω is otherwise.