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

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.

| Set X fits set Y
| X [≤] Y
  is a claim that
  a function exists 1.to.1.from.to:
| ∃f: X → Y: ¬∃x≠x′:f(x)=f(x′)
| X and Y are same.sized
| X [=] Y
  is a claim that
  a function exists 1.to.1.onto.from.to:
| ∃f: X → Y: ¬∃x≠x′:f(x)=f(x′) ∧ f(X)=Y
X [≤≥] Y  ⟺  X [=] Y  (Cantor–Schröder–Bernstein)
X [≤≱] Y  ⟺  X [<] Y
X [≰≥] Y  ⟺  X [>] Y
X [≰≱] Y  (forbidden by Axiom of Choice)
X ⊆ Y  ⟹  X [≤] Y
X [≤] Y [≤] Z  ⟹  X [≤] Z
X [<≠≯] Y  ∨  X [≮=≯] Y  ∨  X [≮≠>] Y
 From the existence of an injection between sets,
we can deduce
the existence of an injection between nearby sets.
S⁻ˣ  and  S⁺ʸ  are sets nearby S
S⁻ˣ  and  S⁺ʸ  abbreviate
S\{x}:x∈S  and S∪{y}:y∉S
S⁻ˣ  and  S⁺ʸ  are not.quite.operations.
They exist, but x and y are any member or non.member.
S⁻ˣ  and  S⁺ʸ  are not uniquely.valued.
| Assume
| ∃f: S → S⁻ˣ: ¬∃s≠s′:f(s)=f(s′)
|
| Define f↗(y) = x
| otherwise f↗(s) = f(s)
| f↗: S⁺ʸ → S: ¬∃s≠s′:f↗(s)=f↗(s′)
Concisely,
S⁻ˣ [≥] S  ⟹  S⁻ˣ [≥] S [≥] S⁺ʸ
| Assume
| ∃g: S⁺ʸ → S: ¬∃s≠s′:g(s)=g(s′)
|
| Define g⤨(y) = x
| g⤨(g⁻¹(x)) = g(y)
| otherwise g⤨(s) = g(s)
| g⤨: S → S⁻ˣ: ¬∃s≠s′:g⤨(s)=g⤨(s′)
Concisely,
S [≥] S⁺ʸ  ⟹  S⁻ˣ [≥] S [≥] S⁺ʸ
And always
S⁻ˣ [≤] S [≤] S⁺ʸ
because
S⁻ˣ ⊆ S ⊆ S⁺ʸ
Those results about injections.between.sets
can be concisely summarized as
S⁻ˣ [<] S [<] S⁺ʸ
  or
S⁻ˣ [=] S [=] S⁺ʸ
Those are the only options, because of the definitions
f↗(y) = x  and  g⤨(g⁻¹(x)) = g(y)
Your theory of darkᵂᴹ elements involves last.finites,
that is, some set S such that
☠  S⁻ˣ [<] S [=] S⁺ʸ
☠   or
☠  S⁻ˣ [=] S [<] S⁺ʸ
Sets like that do not exist,
because
f↗(y) = x  and  g⤨(g⁻¹(x)) = g(y)