because g⤨(g⁻¹(x)) = g(y) [2/2] Re: how

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Sujet : because g⤨(g⁻¹(x)) = g(y) [2/2] Re: how
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.math
Date : 18. Apr 2024, 23:09:49
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <6f99874e-8275-482c-be2b-a6433b1bb6e2@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 4/18/2024 10:54 AM, WM wrote:
Le 18/04/2024 à 00:50, Jim Burns a écrit :
On 4/16/2024 10:33 AM, WM wrote:
Le 15/04/2024 à 23:06, Jim Burns a écrit :
On 4/15/2024 7:59 AM, WM wrote:

That is wrong if
all natnumbers are present already such that
no further natnumbers fits below ω.

It is enough if you explain
where ω is in the lower line:
>
0, 1, 2, 3, ...,   w
|  |  |  |  |||    | 0, 2, 4, 6, ..., w*2
>
Please answer the question.
Consider an ordinal as a prequel.set:
the set of all ordinals before that ordinal.
κ = ⟦0,κ⦆
⟦κ,μ⦆ = {ordinal λ: κ ≤ λ < μ }
We have gotten the result for a set S
S⁻ˣ [<] S [<] S⁺ʸ
  or
S⁻ˣ [=] S [=] S⁺ʸ
because we can define
f↗(y) = x  and  g⤨(g⁻¹(x)) = g(y)
A prequel.set ⟦0,κ⦆ is a set.
Thus,
⟦0,κ⦆⁻¹ [<] ⟦0,κ⦆ [<] ⟦0,κ⦆⁺¹
  or
⟦0,κ⦆⁻¹ [=] ⟦0,κ⦆ [=] ⟦0,κ⦆⁺¹
We can define, for each ordinal κ
a unique successor κ⁺¹
for an operation κ+1
where non.unique S⁺ʸ is not.
κ+1 = ⟦0,κ+1⦆ = ⟦0,κ⦆∪{⟦0,κ⦆}
Consider the set
{ordinal λ:  ⟦0,λ-1⦆ᴲ [<] ⟦0,λ⦆ [<] ⟦0,λ+1⦆ }
ᴲif λ-1 exists.
Note that κ+1 always exists.
| Assume
| κ ∈ {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}
|
| ⟦0,κ-1⦆ [<] ⟦0,κ⦆ [<] ⟦0,κ+1⦆
| not( ⟦0,κ⦆ [=] ⟦0,κ+1⦆ [=] ⟦0,κ+2⦆ )
| ⟦0,κ⦆ [<] ⟦0,κ+1⦆ [<] ⟦0,κ+2⦆ )
| κ+1 ∈ {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}
Therefore,
∀κ ∈ {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}:
κ+1 ∈ {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}
The set {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}
is closed under the successor.operation.
Its closure has nothing to do with humongosity.
Closure ultimately follows from the definitions
f↗(y) = x  and  g⤨(g⁻¹(x)) = g(y)
We build on that result.
Addition.closure from successor.closure
because  κ+μ⁺¹ = (κ+μ)⁺¹  and no.first.exception.
Multiplication.closure from addition.closure
because  κ⋅μ⁺¹ = (κ⋅μ)+κ  and not.first.exception.
∀κ,μ ∈ {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}:
κ+μ,κ⋅μ ∈ {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}
The meaning of ω is the least.upper.bound of
this set: {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}
Its meaning is NOT connected to some vague humongosity.
{ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆} is
the prequel set of ω  so we can write
⟦0,ω⦆ = {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}
which is a nice improvement in conciseness.
Even when we write
0, 1, 2, ..., ω, ω+1, ω+2, ...
⟦0,ω⦆ = {ordinal λ: ⟦0,λ-1⦆ᴲ[<]⟦0,λ⦆[<]⟦0,λ+1⦆}
ω is NOT _humongous_
ω is _infinite_ which has properties different from finite.

It is enough if you explain
where ω is in the lower line:
>
0, 1, 2, 3, ...,   w
|  |  |  |  |||    | 0, 2, 4, 6, ..., w*2
∀κ < ω: k⋅2 < ω
∀κ ≥ ω: k⋅2 > ω
∀κ <≥ ω: k⋅2 ≠ ω
...because
f↗(y) = x  and  g⤨(g⁻¹(x)) = g(y)

Date Sujet#  Auteur
6 Apr 24 * Re: Contradiction of bijections as a measure for infinite sets1217WM
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