Re: The set of necessary FISONs

Am Sun, 16 Feb 2025 12:18:28 +0100 schrieb WM:

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.
 

{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.

That is simply false. You may remove one or another, but not at the
same time.

What should remain if all F(n) are omitted? Inserting smaller F(n) after
larger one have been omitted as useless?

Babbage: „I am not able to rightly apprehend the kind of confusion that
would provoke such a question.”
But you hint at the problem: you can’t omit all segments, because then
nothing remains.

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.

More precisely: Every single element can be omitted, but not more
than one.

The set can be omitted.

No, the set is not an element of itself that can be omitted.

What should remain?
And why? Note that you must define a first element.

Indeed, why should something remain if you remove everything? If you
only remove a finite number however, inf. many remaint.

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

WDYM „prove a set”? For every *element* you can prove the statement
„This, and only this segment can be omitted.” Unfortunately there
is no element that encompasses all numbers.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.