Re: how

Le 24/06/2024 à 19:18, Jim Burns a écrit :

On 6/23/2024 4:04 PM, WM wrote:

Le 23/06/2024 à 19:58, Jim Burns a écrit :

 

U{FISON}  is the union of the set of
both necessary and unnecessary FISONs

>
The unnecessary FISONs can be removed.

 Your definition of 'unnecessary' is that
an unnecessaryᵂᴹ FISON when removed
leaves the union of remaining FISONs unchanged.

Yes. And that can be decided for every FISON.

One unnecessaryᵂᴹ FISON can be removed.
One.

Every one. By induction.

 {FISON} is inductive and well.ordered.

Yes. Every one can be removed without changing the union. Potential infinity.

 Because well.ordered,
Any necessaryᵂᴹ FISON in {FISON}
is last in {FISON}

Yes. But there is no last one in a potentially infinite sequence. With F(n) also F(n^n) is present. There are oo many bright FISONs.

 Because inductive,
No FISON is last in {FISON}

But every FISON has oo bright and ℵo dark successors.

 Any _one_ FISON can be removed and
leave the union of remaining FISONs unchanged.
Any _one_ FISON is unnecessaryᵂᴹ
 Removing all of those which
singly don't change the union
changes the union, ⋃{}≠ℕ
It is not a contradiction.

The contradiction is this: For every FISON we can determine whether it is necessary. Therefore the collection of neccessary FISONs is well-defined and, if being a set, has a smallest element. If its union could be ℕ, the necessary FISONs must be a set and have a smallest element. That is not the case.

 Compare to:
Suppose I have 5 elephants, and
an unnecessaryᴶᴮ elephant is one when removed
leaves 4 elephants.
 Any _one_ elephant can be removed and
leave 4 elephants.
Any _one_ elephant is unnecessaryᴶᴮ

First enumerate them. Then remove 1, 2, 3, 4. Stop.

 Removing all which
singly leave 4
doesn't leave 4
It is not a contradiction.
It is arithmetic.

Don't evade. Apply the logic:
If the union of FISONs is ℕ, then all unnecessary FISONs can be dropped and the remaining FISONs must be a set with union ℕ.
If they are a set they have a smallest element.
That is not the case. Hence they are not a set.

 

The union cannot be larger than all its FISONs.

>
The union cannot be smaller than any FISON.

>
Agreed.

 

Each FISON is smaller than another FISON,

 

Therefore the union is a FISON.

 Therefore the union is NOT a FISON.

Observe logic.
The union cannot be larger than all its FISONs.
The union cannot be smaller than any FISON.
Not larger and not smaller means ...?

It runs through ℕ
but never covers a substantial part.

 Related:
∀ᴿx>0: NUF(x) = ℵ₀

Yes, that is as much violating logic.
But I do not accept that.
Regards, WM