Re: how

On 6/14/2024 12:39 PM, WM wrote:

Le 14/06/2024 à 16:56, Jim Burns a écrit :

On 6/14/2024 4:34 AM, WM wrote:

Le 13/06/2024 à 20:25, Jim Burns a écrit :

On 6/13/2024 10:55 AM, WM wrote:

Every number has ℵo successors.

>
Yes.

>
No.
Here is the context:
WM:
If every number is subtracted,
then no successors remain.
FF:
Eine wirklich bemerkenswerte Erkenntnis!
WM:
It contradicts your claim that
every number has ℵo successors.
FF:
No,
WM (Proof by contradiction):
Every number has ℵo successors.
If every number is subtracted
the successors remain.
>

You:

Every number has ℵo successors.

>
In a proof by contradiction.

Every number in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
has ℵ₀ successors in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
{i:∀₃Xᴬ⤾⁺¹₀:X∋i}  is
the minimal inductive meta.set
Xᴬ⤾⁺¹₀  ⇔  X∋0 ∧ ∀₂j∈X:X∋j⁺¹
----
The composition of 1.to.1 functions is 1.to.1.
sⱼ(k) = k⁺¹
is a 1.to.1 function
sⱼ:{j<} → {j⁺¹<}: 1.to.1
{j<}  j.followers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
{j⁺¹<}  j⁺¹.followers in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
There is NO FIRST 0.follower j⁺¹ ∈ {0<}
WITHOUT g:{0<} → {j⁺¹<}: 1.to.1
WITH f:{0<} → {j<}: 1.to.1
...because
sⱼ∘f:{0<} → {j⁺¹<}  is 1.to.1
There is NO 0.follower j⁺¹ ∈ {0<}
WITHOUT sⱼ∘f:{0<} → {j⁺¹<}: 1.to.1
∀₂j⁺¹ ∈ {0<}: |{0<}| ≤ |{j⁺¹<}|
∀₂j⁺¹ ∈ {0<}:  {0<}  ⊇  {j⁺¹<}
∀₂j⁺¹ ∈ {0<}: |{0<}| ≥ |{j⁺¹<}|
∀₂j⁺¹ ∈ {0<}: |{0<}| = |{j⁺¹<}|
0.follower j⁺¹ has as many j⁺¹.followers
as 0 has 0.followers
...because
sⱼ∘f:{0<} → {j⁺¹<}  is 1.to.1
Therefore
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