Sujet : Re: Contradiction of bijections as a measure for infinite sets
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 09. Apr 2024, 21:27:37
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <c0f7ebc9-ef92-465f-aba2-dfc76895dffb@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 4/9/2024 8:22 AM, WM wrote:
Le 09/04/2024 à 01:54, Jim Burns a écrit :
On 4/8/2024 9:55 AM, WM wrote:
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
>
For visible numbers only.
>
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
>
Not correct if
there are all natural numbers such that
no further one exists below ω.
You seem to be saying:
| Not correct if
| not all natural numbers have
| a further one below ω
In other words [1]:
| Not correct if
| ω is finiteⁿᵒᵗᐧᵂᴹ.
[1] is correct because
ω is infiniteⁿᵒᵗᐧᵂᴹ.
Infiniteⁿᵒᵗᐧᵂᴹ and infiniteᵂᴹ are different.
Infiniteᵂᴹ is hyper.humongous, large compared to
602214076000000000000000⁶⁰²²¹⁴⁰⁷⁶⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰
but still finiteⁿᵒᵗᐧᵂᴹ.
Any element of
{k: k=0∨∃⟦0,k⦆:∀i∈⟦0,k⦆:i⁺¹∈⦅0,k⟧ } := ℕ
is finiteⁿᵒᵗᐧᵂᴹ.
602214076000000000000000⁶⁰²²¹⁴⁰⁷⁶⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰
is finiteⁿᵒᵗᐧᵂᴹ.
⟦0,ω⦆ is the least upper bound of all finiteⁿᵒᵗᐧᵂᴹ ⟦0,k⟧
∀k < ω: k is finiteⁿᵒᵗᐧᵂᴹ
∀k > ω: k is not finiteⁿᵒᵗᐧᵂᴹ
∀k < ω: k⁺¹ < ω
[2]
∀k,m < ω: k+m < ω
[3]
∀k,m < ω: k⋅m < ω
[4]
[2]
∀k < ω: k⁺¹ < ω
|
| Assume otherwise.
| Assume, for k < ω,
| ∀i < k: i⁺¹ < ω
| but ω ≤ k⁺¹
|
| ∃⟦0,k⦆:∀i∈⟦0,k⦆:i⁺¹∈⦅0,k⟧
| ¬∃⟦0,k⁺¹⦆:∀i∈⟦0,k⁺¹⦆:i⁺¹∈⦅0,k⁺¹⟧
|
| However,
| for ⟦0,k⁺¹⦆ = ⟦0,k⦆∪{k}
| ∀i∈⟦0,k⁺¹⦆:i⁺¹∈⦅0,k⁺¹⟧
| k⁺¹ < ω
| Contradiction.
[3]
∀k,m < ω: k+m < ω
|
| Assume otherwise.
| Assume, for k,m < ω
| ∀i < m: k+i < ω
| but ω ≤ k+m
|
| k+m = (k+m⁻¹)⁺¹
| k+m⁻¹ < ω
| By [2], (k+m⁻¹)⁺¹ < ω
| k+m < ω
| Contradiction.
[4]
∀k,m < ω: k⋅m < ω
|
| Assume otherwise.
| Assume, for k,m < ω
| ∀i < m: k⋅i < ω
| but ω ≤ k⋅m
|
| k⋅m = (k⋅m⁻¹)+k
| k⋅m⁻¹ < ω
| By [3], (k⋅m⁻¹)+k < ω
| k⋅m < ω
| Contradiction.
Therefore, [2][3][4]
∀k < ω: k⁺¹ < ω
∀k,m < ω: k+m < ω
∀k,m < ω: k⋅m < ω
Multiplication by 2
creates numbers beyond ω, or
there would be numbers immune to multiplication.
All the numbers needed for
successors, sums, and products of numbers before ω
exist before ω
It is arithmetic.