Re: The set of necessary FISONs

Am Sat, 25 Jan 2025 12:02:39 +0100 schrieb WM:

On 24.01.2025 16:44, Jim Burns wrote:

On 1/24/2025 4:37 AM, WM wrote:

On 23.01.2025 16:18, Jim Burns wrote:

 
The union of all FISONs covers UF(n)

 
Simply contradicted by:
∀n ∈ UF(n): |ℕ \ {1, 2, 3, ..., n}| = ℵo

[assuming UF(n) means {1, 2, ..., n}]
That just says that N is not a single FISON.

Each FISON is a proper subset of another FISON. Each FISON is a proper
subset of UF(n). No FISON is UF(n)

That is potential infinity.

No, it’s just a factual description.

Whatever contains each FISON  contains UF(n)

Alas this is not a set but a (potentially in-) finite changing
collection.

No. Can you not conceive of an infinite set?

Otherwise Cantor's theorem would require the existence of a first
necessary FISON.

There is obviously no single FISON that is equal to N.

Do you agree that or every FISON the question whether it is necessary
can be answered?

Yes, in the negative. That does not imply the nonexistence of a
sufficient set.

Up to every FISON |ℕ \ {1, 2, 3, ..., n}| = ℵo.

N is not a FISON.

For any two FISONs {1,2,...,j} {1,2,...,k} their sum
{1,2,...,j,j+1,j+2,...,j+k} is a FISON

Therefore the union cannot be larger than a FISON.

The union is infinite, since each FISON adds a different number.

The infinite union is the infinite FISON. But there is no infinite FISON

The limit of FISONs is N.

⎛ Consider Bob such that,
⎜ before all FISON.end.swaps n⇄n+1 ⎜ Bob is in the first FISON.end 0 ⎜
⎜ If Bob is in FISON.end n ⎜ then ⎜ it is after n-1⇄n and before n⇄n+1
⎜ If it is after all FISON.end.swaps ⎜ then Bob is not.in any
FISON.end,
⎜ even though ⎜ no FISON.end.swap takes Bob ⎝ anywhere else.
 

Swaps cannot eliminate Bob. He remains but i the darkness.

An infinite sequence of swaps may not correspond to a single swap.

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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.