Re: 2N=E

On 10/31/2024 4:27 PM, WM wrote:

On 31.10.2024 20:25, Jim Burns wrote:

On 10/31/2024 2:30 PM, WM wrote:

On 31.10.2024 19:07, Jim Burns wrote:

'Infinite' means 'not finite'.

>
Potentially infinite means
finite but variable without bound.

>
Our sets do not change.

>
Then:
Multiplication of
all infinitely many fractions of the open interval (0,1)
results in some fractions in (1, 2).

No.
_Of course_ they don't.
a/b,c/d ∈ (0,1) ⇒
a < b, c < d  ⇒
a⋅c < b⋅c < b⋅d  ⇒
(a/b)⋅(c/d) = (a⋅c)/(b⋅d) ∈ (0,1)
Did you mean "addition" instead of "multiplication"?

Multiplication of
all infinitely many numbers of the open interval (0,ω)
result in some numbers in (ω, ω*2).

No.
_Of course_ they don't.
⎛ Assume otherwise.c
⎜ Assume j,k ∈ ⟦0,ω⦆  ∧  j×k ∉ ⟦0,ω⦆
⎜
⎜ The set {i∈⟦0,ω⦆: j×i∉⟦0,ω⦆} of
⎜ finite.ordinals with infinite products
⎜ holds k, is not empty, and
⎜ holds least element ψ+1 ∈ ⟦0,ω⦆
⎜ j×(ψ+1) = j+(j×ψ) ∉ ⟦0,ω⦆
⎜ j×ψ ∈ ⟦0,ω⦆
⎜
⎜ The set {i∈⟦0,ω⦆: j+i∉⟦0,ω⦆} of
⎜ finite.ordinals with infinite sums
⎜ holds j×ψ, is not empty, and
⎜ holds least element φ+1 ∈ ⟦0,ω⦆
⎜ j+(φ+1) = (j+φ)+1 ∉ ⟦0,ω⦆
⎜ j+φ ∈ ⟦0,ω⦆
⎜
⎜ The set {i∈⟦0,ω⦆: i+1∉⟦0,ω⦆} of
⎜ finite.ordinals with infinite successors
⎜ holds j+φ, is not empty, and
⎜ holds least element ξ+1
⎜ ξ+1 ∉ ⟦0,ω⦆
⎜ ξ ∈ ⟦0,ω⦆
⎜
⎜⎛ However,
⎜⎜ the definition of ω is not
⎜⎜ what you (WM) think it is.
⎜⎜ k ∈ ⟦0,ω⦆  :⇔
⎜⎝ ∀j ∈ ⦅0,k⟧: ∃i ∈ ⟦0,j⦆: i+1=j
⎜
⎜ ξ ∈ ⟦0,ω⦆
⎜ ∀j ∈ ⦅0,ξ⟧: ∃i ∈ ⟦0,j⦆: i+1=j
⎜ Also,
⎜ ∀j ∈ {ξ+1}: ∃i ∈ ⟦0,j⦆: i+1=j
⎜ because
⎜ ξ ∈ ⟦0,ξ+1⦆  and  ξ + 1 = ξ+1
⎜
⎜ Thus,
⎜ ∀j ∈ ⦅0,ξ⟧∪{ξ+1}: ∃i ∈ ⟦0,j⦆: i+1=j
⎜ ∀j ∈ ⦅0,ξ+1⟧: ∃i ∈ ⟦0,j⦆: i+1=j
⎜ ξ+1 ∈ ⟦0,ω⦆
⎜
⎜ However
⎜ ξ+1 ∉ ⟦0,ω⦆
⎝ Contradiction.
Therefore,
for each two j,k ∈ ⟦0,ω⦆:  j×k ∈ ⟦0,ω⦆

Multiplication of
all infinitely many numbers of the open interval (0,ω)
result in some numbers in (ω, ω*2).

No.
_Of course_ they don't.

Reducing the density increases the interval.

'Infinity' does not mean what you want it to mean.