Re: how

On 6/6/2024 4:14 PM, WM wrote:

Le 06/06/2024 à 20:35, Jim Burns a écrit :

Therefore,
each k ∈ ℕ⁺ is ℵ₀.followed.in.ℕ⁺

>
You define some individual number and
many are following.

I give a description of an individual number j in ℕ⁺
-- a description which does not distinguish between
different numbers in ℕ⁺
⎛ Each number in ℕ⁺ has a successor.
⎜ Each nonzero number in ℕ⁺ has a predecessor.
⎝ Each nonempty subset of ℕ⁺ holds a first number.
⎛ The successor relation  x+1=y  is 1.to.1
⎜ The image.set S(X) = {y: ∃x ∈ X: x+1=y} is
⎜ the same size/cardinality as set X
⎝ |X| = |S(X)|
⎛ A superset is at least as large as
⎜ any of its subsets.
⎝ X ⊇ Y  ⇒  |X| ≥ |Y|
⎛ {j<}  =  ℕ⁺\{1,2,3,...,j}
⎜  is followers.of.j  for j ∈ ℕ⁺
⎝ {j=<}  is j and followers.of.j
1.
|{j=<}| = |{j<}|  for j ∈ ℕ⁺
proof
  {j=<}  ⊇  {j<}
|{j=<}| ≥  |{j<}|
  S{j=<}  ⊆  {j<}
|S{j=<}| ≤ |{j<}|
|{j=<}| = |S{j=<}|
  |{j=<}| ≤ |{j<}|
|{j=<}| = |{j<}|
2.
|{0<}| = |{k<}|  for k ∈ ℕ⁺
proof
The set {k∈ℕ⁺:|{0<}|≠|{k<}|}
of counter.examples to my claim
is empty or not.empty.
If {k∈ℕ⁺:|{0<}|≠|{k<}|} is empty
then
|{0<}| = |{k<}|  for k ∈ ℕ⁺
On the other hand,
if {k∈ℕ⁺:|{0<}|≠|{k<}|} is not.empty
then
{k∈ℕ⁺:|{0<}|≠|{k<}|}  holds first number j
|{0<}| ≠ |{j<}|
|{0<}| = |{j-1<}|
However,
|{j-1<}| = |{j=<}|
by (1.)
|{j=<}| = |{j<}| ≠ |{0<}|
|{0<}| ≠ |{0<}|
Contradiction.
Therefore,
|{0<}| = |{k<}|  for k ∈ ℕ⁺
|{0<}| = |{k<}|  for k ∈ ℕ⁺
|{0<}| = ℵ₀
{k<}  =  ℕ⁺\{1,2,3,...,k}
∀k ∈ ℕ⁺: |ℕ⁺\{1,2,3,...,k}| = ℵ₀

They are following even upon
the last number that you ever have defined. Therefore they are undefined by you.
That is the difference between ℕ_def and ℕ.

I see that you (WM) aren't denying that
∀k ∈ ℕ⁺: |ℕ⁺\{1,2,3,...,k}| = ℵ₀
⎛ Each number in ℕ⁺ has a successor.
⎜ Each nonzero number in ℕ⁺ has a predecessor.
⎝ Each nonempty subset of ℕ⁺ holds a first number.
Of being defined, nothing is said.
There is nothing to be denied with respect to
being defined or not defined.

It is not Cantor's actual infinity.

>
It is our familiar arithmetic.

>
Of course.

Do you agree that
your (WM's) claims are not about
our familiar arithmetic?

Most natural numbers are undefined
by you and be anyone else.

⎛ Each number in ℕ⁺ has a successor.
⎜ Each nonzero number in ℕ⁺ has a predecessor.
⎝ Each nonempty subset of ℕ⁺ holds a first number.