Re: The set of necessary FISONs

On 2/7/25 5:24 AM, WM wrote:

On 07.02.2025 09:59, joes wrote:
 

N is exactly the set of all, and only the natural numbers.
It does not matter that you call it N_def.

 It is described by the axioms 1 ∈ ℕ and n ∈ ℕ ==> n+1 ∈ ℕ. This is an infinite set containing all natnumbers.

Which isn't the "induction" axiom.

 The set M of all FISONs is described by the axioms F(1) ∈ M and F(n) ∈ M ==> F(n+1) ∈ M . This is an infinite set containing all FISONs.
 

Your extension of
that set is unclear, since you have not provided any axioms.

 Cantor claims that *a fixed quantity greater than all natural numbers* exists.
An increasing sequence of natural numbers is never greater than all natural numbers.

So?, What Cantor claims is that there must exist another fixed quantity, that isn't part of the Natural Numbers.
Cantor is pointing out that the Set of Natural Numbers, as an infinite set, has properties that can't be described with Natural Numbers. Things like the size of an infinite set, can not be a finite number, and thus Aleph_0 is created as a non-finite value to describe the size of that set.

 

If the set M is described as the smallest set satisfying F(1) ∈ M and
F(n) ∈ M ==> F(n+1) ∈ M then M contains all FISONs which can be
subtracted from U(Fn)) without changing the assumed result ℕ.

Wrong. M cannot be finite.

 Inductive sets are infinite sets, according to set theory.

Right, but you need to use *ALL* the properties. Note, your problem seems to be that you don't understand what axioms are, and which axioms are which.
For instnace, that there exists a 0, and for every n in the set S(n) is also in the set, does not by itself generate just the natural numbers, but can also generate a lot of other number sets. You need the additional properties like for n that are not 0 there is an m such that S(m) = n, that there is no n such that S(n) == 0, and that if S(m) == S(n) then m == n.
The Natural Numbers come from a set that meets ALL those rules. The Transfinite numbers come from a relaxation of them.

 Regards, WM