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

On 4/19/2024 11:05 AM, WM wrote:

Le 19/04/2024 à 00:09, Jim Burns a écrit :

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  = X
|  |  |  |  |||    | 0, 2, 4, 6, ..., w*2  = Y
>

Please answer the question.

>
| Set X fits set Y

| 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′)

Yes both sets have the same number of elements.
But the interval covered by Y is twice as large as
that covered by 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

 The question was: Where in the second line sits ω?
>
Sorry, I could not find the symbol ω in your text.

You (WM) misunderstand ω
I thought I would try answering without ω

I assume you deny to answer because
the answer would prove you wrong.

I answer and you refuse to see it because
ω isn't what you (WM) want ω to be.
"Infinite" doesn't mean "humongous".
A type I set is same.sized as nearby sets
S⁻ˣ [=] S [=] S⁺ʸ
A type F set is not.same.sized as nearby sets
S⁻ˣ [<] S [<] S⁺ʸ
Only type I sets are nearby type I sets.
Only type F sets are nearby type F sets.
Sets are only type I or type F.
Each ordinal κ has a prequel.set ⟦0,κ⦆
A type I ordinal has a type I prequel.set.
A type F ordinal has a type F prequel.set.
Only type I ordinals are nearby type I ordinals.
Only type F ordinals are nearby type F ordinals.
Ordinals are only type I or type F.
ω is first.upper.bound of type F ordinals.
ω bounds every type F ordinal.
Nothing before ω bounds every type F ordinal.
ω+1 is not bounded by ω
ω+1 isn't type F
ω+1 is type I
ω is nearby ω+1 (type I)
ω is type I
If ω-1 exists
then
ω-1 is nearby ω (type I)
ω-1 is type I
ω-1 bounds everything which ω bounds except ω-1
ω-1 isn't type F
ω-1 bounds every type F ordinal.
ω-1 is before ω.
ω isn't first.upper.bound of type F ordinals.
If ω-1 exists
then
ω isn't first.upper.bound of type F ordinals.
If ω is first.upper.bound of type F ordinals
then
ω-1 not.exists
visibleᵂᴹ or darkᵂᴹ, ω-1 not.exists
not ∃κ: κ < ω ≤ κ⁺¹
if ∃κ,μ: κ,μ < ω ≤ κ+μ
then ∃κ,λ: κ+λ < ω ≤ κ+λ⁺¹
κ+λ⁺¹ = (κ+λ)⁺¹
not ∃κ,λ: κ+λ < ω ≤ (κ+λ)⁺¹
not ∃κ,μ: κ,μ < ω ≤ κ+μ
if ∃κ,μ: κ,μ < ω ≤ κ⋅μ
then ∃κ,λ: κ⋅λ,κ < ω ≤ κ⋅λ⁺¹
κ⋅λ⁺¹ = (κ⋅λ)+κ
not ∃κ,λ: κ⋅λ,κ < ω ≤ (κ⋅λ)+κ
not ∃κ,μ: κ,μ < ω ≤ κ⋅μ

The question was: Where in the second line sits ω?

0, 1, 2, 3, ...,   w  = X
|  |  |  |  |||    | 0, 2, 4, 6, ..., w*2  = Y

ω is first.upper.bound of type F ordinals.
not ∃κ: κ < ω ≤ κ⁺¹
not ∃κ,μ: κ,μ < ω ≤ κ+μ
not ∃κ,μ: κ,μ < ω ≤ κ⋅μ
ω isn't in Y