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

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Sujet : Re: because g⤨(g⁻¹(x)) = g(y) [2/2] Re: how
De : ross.a.finlayson (at) *nospam* gmail.com (Ross Finlayson)
Groupes : sci.math
Date : 19. Apr 2024, 17:36:51
Autres entêtes
Message-ID : <BX6dnRXuToLrAL_7nZ2dnZfqn_qdnZ2d@giganews.com>
References : 1 2 3 4 5 6 7 8 9 10 11 12
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On 04/18/2024 03:09 PM, Jim Burns wrote:
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)
>
>
That's kind of interesting.
1/phi = phi - 1, ....
You know they say that Fibonacci numbers and the Golden Ratio phi
show up a lot, ....
Convolution is a very usual setting, ....

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