Re: how

On 6/16/2024 10:17 AM, WM wrote:

Le 14/06/2024 à 20:52, Jim Burns a écrit :

every number in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
has ℵ₀ successors in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
>

Note that
only the numbers with successors are deleted,
the successors remain by definition.

>
sⱼ∘f:{0<} → {j⁺¹<}  is 1.to.1

minimal inductive meta.set {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
0.followers {0<}  =  {i:∀₃Xᴬ⤾⁺¹₀:X∋i ∧ 0<i}
j⁺¹.followers {j⁺¹<}  =  {i:∀₃Xᴬ⤾⁺¹₀:X∋i ∧ j⁺¹<i}
Non.empty subset S ⊆ {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
holds a first element k₁  with
predecessor j₁ = k₁⁻¹  unless k₁ = 0
j₁ ∉ S  ∧  j₁⁺¹ ∈ S
Define
∃f:{0<}⇉{j<}  ⇔
exists f:{0<} → {j<}: 1.to.1
{j<} holds at least as many as {0<}
|{0<}| ≤ |{j<}|
Define
S  =  {i:∀₃Xᴬ⤾⁺¹₀:X∋i ∧ ¬∃f:{0<}⇉{i<}}
S is the set of numbers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
without ℵ₀.many followers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
S is empty or nonempty.
If S is empty
then
NO number in {i:∀₃Xᴬ⤾⁺¹₀:X∋i} is
WITHOUT ℵ₀.many followers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
Deleting all numbers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
with ℵ₀.many followers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
deletes all numbers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
If S is nonempty
then
S holds a first element k₁  with
predecessor j₁ = k₁⁻¹  unless k₁ = 0
j₁ ∉ S  ∧  j₁⁺¹ ∈ S
∃f:{0<}⇉{j₁<}  ∧  ¬∃g:{0<}⇉{j₁⁺¹<}
However,
sⱼ₁:{j₁<}⇉{j₁⁺¹<}:1.to.1
sⱼ₁∘f:{0<}⇉{j₁⁺¹<}:1.to.1
∃g=sⱼ₁∘f:{0<}⇉{j₁⁺¹<}:1.to.1
Contradiction.
Therefore,
S the set of numbers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
without ℵ₀.many followers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
is empty
and
deleting all numbers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
with ℵ₀.many followers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
deletes all numbers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}

If only numbers having ℵo successors are removed
and only as long as ℵo successors remain,
then ℵo successors remain and
every definable number is removed.
You cannot defina a number that remains.
But ℵo successors remain

If all numbers having ℵ₀ followers are removed
then no numbers remain.
There is no first number remaining.
There is no number remaining.
We can _define_ a number that remains; however,
the _existence_ of the number defined
leads to self.contradiction