Re: how

On 4/12/2024 9:18 AM, WM wrote:

Le 11/04/2024 à 19:36, Jim Burns a écrit :

On 4/11/2024 8:07 AM, WM wrote:

Which ones were unused by e = 2*n?

>
Those of {1, 2, 3, ...} with less than ℵo successors.

>
{1,2,3,…} is
the least.upper.bound of all finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k}

>
Yes,

tl;dr
usedᵂᴹ or not.usedᵂᴹ
not.exists
k, n in {1,2,3,…}  and
k⁺¹ not.in {1,2,3,…} or
n⁺¹ not.in {1,2,3,…} or
k+n not.in {1,2,3,…} or
k⋅n not.in {1,2,3,…}
That is arithmetic.
There are reasons arithmetic is like that.
not.exists
finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k} and
not.finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k,k⁺¹}
not.exists
k in {1,2,3,…} and
k⁺¹ not.in {1,2,3,…}
[!]
not.exists
k+m in {1,2,3,…} and
(k+m)⁺¹ not.in {1,2,3,…}
define.
k+m⁺¹ = (k+m)⁺¹
not.exists
k+m in {1,2,3,…} and
k+m⁺¹ not.in {1,2,3,…}
if exists
   k, n in {1,2,3,…} and
   k+n not.in {1,2,3,…}
then exists first
   m⁺¹ in {1,2,3,…,n}
   k+m in {1,2,3,…} and
   k+m⁺¹ not.in {1,2,3,…}
   (which not.exists)
not.exists
k, n in {1,2,3,…} and
k+n not.in {1,2,3,…}
[!]
not.exists
k⋅m, k in {1,2,3,…} and
k⋅m+k not.in {1,2,3,…}
define
k⋅m⁺¹ = k⋅m+k
not.exists
k⋅m, k in {1,2,3,…} and
k⋅m⁺¹ not.in {1,2,3,…}
if exists
   k, n in {1,2,3,…} and
   k⋅n not.in {1,2,3,…}
then exists first
   m⁺¹ in {1,2,3,…,n}
   k⋅m in {1,2,3,…} and
   k⋅m⁺¹ not.in {1,2,3,…}
   (which not.exists)
not.exists
k, n in {1,2,3,…} and
k⋅n not.in {1,2,3,…}
[!]

but many natnumbers will never be used.

usedᵂᴹ or not.usedᵂᴹ
not.exists
k, n in {1,2,3,…}  and
k⁺¹ not.in {1,2,3,…} or
n⁺¹ not.in {1,2,3,…} or
k+n not.in {1,2,3,…} or
k⋅n not.in {1,2,3,…}
It is arithmetic.
Those are the reasons arithmetic is like that.

For each m ∈ {1,2,3,…}:
( for each k ∈ {1,2,3,…}
m+k is a successor of m in {1,2,3,…} )

>
Consider the set {1, 2, 3, ..., ω}

|​̲1|​̲2|​̲3|​̲4|​̲5|​̲6| ... |​̲ω ω+1 ω+2 ... ω+ω ... | ω₁ ...
The ordinals.of.different.sizes
have been separated with |'s
Nearest.ordinals to a _finite_ ordinal
are different sizes.
ω is first among ordinals with
same.sized nearest.ordinals.
Infinite doesn't mean
humongous.with.different.sized.nearest.ordinals.

and multiply every element by 2
with the result {2, 4, 6, ..., ω*2}.

|1|​̲2|3|​̲4|5|​̲6| ... |ω ω+1 ω+2 ... ​̲ω​̲+​̲ω ... | ω₁ ...

What elements fall between ω and ω*2?

ω+1 ω+2 ...

What size has the interval between ℕ*2 and ω*2?

I take you to be asking how many ordinals are
after each of {2,4,6,...} and before ω+ω
Those ordinals are
ω ω+1 ω+2 ...
The set of them is the same size as the set of
|1|​̲2|3|​̲4|5|​̲6| ...

If you were right, then no elements would fall between ω and ω*2

No doubled ordinal is between ω and ω+ω

but all new elements
(larger than all in {1,2,3,…})

There is nothing in {1,2,3,…} which is
larger than all in {1,2,3,…}
Not even larger than all.but.one in {1,2,3,…}
All doubled finite.ordinals are finite.ordinals.
ω > k ⟼ k+k < ω

would stay in {1,2,3,…}
while the intervall between ω and ω*2
would be infinite and empty.

Your logicᵂᴹ says that
ordinals ≥ ω should be like ordinals < ω
However,
if ordinals ≥ ω were like ordinals < ω
they would be ordinals < ω
This is the same kind of claim as that
squares should have three corners.
If squares had three corners, they'd be triangles.