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Subject: Re: how
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Date: Mon, 24 Jun 24 18:41:31 +0000
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From: WM <wolfgang.mueckenheim@tha.de>
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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