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

On 04/19/2024 09:36 AM, Ross Finlayson wrote:

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, ....
>
>

1 / phi = phi - 1
x^x = x x
x = 1, 2
x^x = x + x
x^2 = 2x
x = 2
(x + x )^ (1/x) = x
x = 2
x - 1 = x x
x^2 - x = 1
x - 1 = 1/x
x = phi
1 - x = xx
x^2 + x = 1
x + 1 = 1/x
x = phi - 1
Then, there's that x^2, is (5 - 2root5 +1) / 4 = (3-root5)/2, ....
e^ P e^ -R = e^ P-R
A P, R
ln P/R = ln P - ln R
?
https://www.google.com/search?q=ln+P%2FR+%3D+ln+P+-+ln+R
https://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation
https://en.wikipedia.org/wiki/Gas_constant