Re: how

On 6/11/2024 10:44 AM, WM wrote:

Le 10/06/2024 à 22:16, Jim Burns a écrit :

each natural number, defined or undefined,
has ℵ₀ followers.

>
No.

Proposal 2, sets.
(E) The empty set exists.
¬∃₂x ∈ ∅
(A) For existing x,y, adjunct x∪{y} exists.
∀₂x,y: ∃₂z=x∪{y}
(X) Two equi.membered sets are
the same set.
∀₂x,y: ∀₂u:u∈x⇔u∈y  ⇒  x=y
Proposal 3, meta.sets.
(C) If,  for predicate.on.sets P(x):
for each set x: P(x) xor ¬P(x)
then  meta.set {y:P(y)} exists
∀₂x: P(x) ⊻ ¬P(x)  ⇒  ∃₃Z={y:P(y)}
(MX) Two equi.membered meta.sets are
the same meta.set.
∀₃X,Y: ∀₂u:u∈X⇔u∈Y  ⇒  X=Y

each natural number, defined or undefined,
has ℵ₀ followers.

>
No.
ℕ \ {1, 2, 3, ...} = ?
Where are  the followers?

Define
(Presume it is true that, when I say … , I mean … )
x < y  ⇔  x ∈ y
0 = ∅
x⁺¹ = x∪{x}
y⁻¹ < z  ⇔  ∃₂x < z: x⁺¹ = y
{0,1,2,…}  =  {y: 0≤x ∧ ∀₂y≤x:( y⁻¹<x ∨ y=0 )}
X⤾⁺¹₀  "X is inductive"
X⤾⁺¹₀  ⇔  X∋0 ∧ ∀₂y∈X:X∋y⁺¹
ℕ = {y: ∀₃X⤾⁺¹₀:X∋y}
ℕ is the minimal inductive meta.set.

each natural number, defined or undefined,
has ℵ₀ followers.

>
No.
ℕ \ {1, 2, 3, ...} = ?
Where are  the followers?

{0,1,2,…} exists₃
{0,1,2,…} is inductive
{0,1,2,…}⤾⁺¹₀
ℕ exists₃
{0,1,2,…} is a superset of ℕ
{0,1,2,…} ⊇ ℕ
ℕ\{0,1,2,…}  =  ∅

each natural number, defined or undefined,
has ℵ₀ followers.

Define
{j<} "followers of j"
{j<} = {y: y∈ℕ ∧ j<y}
sⱼ(k) = k⁺¹
sⱼ: {j<} → {j⁺¹<}
----
for each k ∈ ℕ: ⟨k,k⁺¹⟩ = {{k},{k,k⁺¹}} exists₂
for each j ∈ ℕ:
{j<} {j⁺¹<} exist₃
sⱼ = {⟨k,k⁺¹⟩: k∈{j<}} exists₃
sⱼ is 1.to.1
The composition of 1.to.1 functions is
a 1.to.1 function.
There is NO FIRST j⁺¹ ∈ ℕ
WITHOUT 1.to.1 g:{0<} → {j⁺¹<}  and
WITH 1.to.1 f:{0<} → {j<}
...because
sⱼ∘f:{0<} → {j⁺¹<} is 1.to.1
There is NO j⁺¹ ∈ ℕ
WITHOUT 1.to.1 sⱼ∘f:{0<} → {j⁺¹<}
∀₂j⁺¹ ∈ ℕ:
|{0<}| ≤ |{j⁺¹<}|
∀₂j⁺¹ ∈ ℕ:
  {0<}  ⊇  {j⁺¹<}
|{0<}| ≥ |{j⁺¹<}|
|{0<}| = |{j⁺¹<}|
Each natural number, defined or undefined,
has ℵ₀ followers.