Re: The set of necessary FISONs

On 2/16/25 6:18 AM, WM wrote:

On 16.02.2025 00:52, Jim Burns wrote:

On 2/15/2025 7:49 AM, WM wrote:

 

Induction proves the existence of
the inductive set of all its members.

 

ℕ has an only.inductive subset ℕ

 The set of FISONs is also an inductive set. Compare v. Neumann's definition of finite cardinal numbers. https://www.hs-augsburg.de/ ~mueckenh/Transfinity/Transfinity/pdf, p. 46.
 

{F} has only one inductive subset: {F}

 Correct.

>
{F} is a set of omissible FISONs.
{F} isn't an omissible set of FISONs.
Without the leap, there is no conflict.

 But you are wrong. If all FISONs are omissible by induction, then the set of all FISONs is omissible.
 What should remain if all F(n) are omitted? Inserting smaller F(n) after larger one have been omitted as useless?

>
You (WM) imagine a last finite step,
into the infinite.

 No, I accept Zermelo's proof of an infinite set by the axiom of induction: { } and a ==> {a}. That creates an infinite set.
All elements can be omitted. The set can be omitted. What should remain? And why? Note that you must define a first element.

Since he didn't do that, you are just showing that you are basing your logic on falsehood.
Zermelo didn't "prove" that an infinite set exists, he presumed it in an axiom, and then constructed a particular example.
Your "omitting" is not an operation that his set theory defines, so your set isn't a set in that theory.

 Regards, WM
     There is no doubt that this proves the whole set Z without assuming a last element.
 Regards, WM