Re: how

On 6/12/2024 4:33 PM, WM wrote:

Le 12/06/2024 à 20:54, Jim Burns a écrit :

On 6/12/2024 2:27 PM, WM wrote:

Le 12/06/2024 à 20:18, Jim Burns a écrit :

ℕ\{0,1,2,…}  =  ∅

>
So there are no followers?

>
there is no j ∈ ℕ after ℕ

>
i.e., after all natural numbers.

there is no
natural number after (≥) all natural numbers.

That means:
If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.

I have placed proposal 2 (sets: E A X) and
proposal 3 (meta.sets: E A X C MX) before you.
They have different consequences.
| 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
Under proposal 2,
each natural number (finite von Neumann ordinal)
has a finite sequence of adjunctions
which proves that it exists.
∃₂z=x∪{y}
Under proposal 3,
the meta.set ℕ of all and only natural numbers
has a description P(y) of a natural number
which proves that ℕ exists.
∃₃Z={y:P(y)}

If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.

If each finite von Neumann ordinal
is deleted from
the set of finite von Neumann ordinals,
nothing is left in the result.
We finite beings cannot perform a supertask,
which is what it takes to count all
finite von Neumann ordinals.
However,
it is not a supertask to describe them with P(y)
and explore their properties by
making finite claim.sequences which are
only not.first.false.
That is a distinction which we make
infinitely.prior to dealing with ℕ
"In principal" we can count to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ,
but we know very well we can't,
not in our 13.7 Gyo universe.
The tiny.ness of our universe doesn't stop us from
reasoning about all the numbers which we would count,
if we could.
In this tiny universe, we reason
about numbers bigger than we can count to
and about infinite sets
with tiny.length claims and
tiny.length sequences of claims.
We don't need to count to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ

ℕ is the minimal inductive meta.set.

>
ℕ is all natural numbers.
Not more and not less.

ℕ = {y: ∀₃X⤾⁺¹₀:X∋y}
X⤾⁺¹₀  ⇔  X∋0 ∧ ∀₂y∈X:X∋y⁺¹

for each j ∈ ℕ
there are ℵ₀.many followers in ℕ

>
Is ℕ more than every j ∈ ℕ?
What is it more?

∀₂j ∈ ℕ: ℕ ⊃ j
∀₂j ∈ ℕ: j⁺¹ ∈ ℕ\j
∀₂j ∈ ℕ: j⁺¹ ∉ ℕ\ℕ