Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary, effectively)

Am 05.01.2025 um 23:39 schrieb Chris M. Thomasson:

On 1/5/2025 11:47 AM, WM wrote:

On 05.01.2025 18:39, Richard Damon wrote:

On 1/5/25 12:26 PM, WM wrote:

On 05.01.2025 13:47, Richard Damon wrote:

On 1/5/25 5:31 AM, WM wrote:

On 04.01.2025 11:59, joes wrote:

Am Sat, 04 Jan 2025 09:52:08 +0100 schrieb WM:

For all FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Not even wrong.
What he meant to express is most likely:
       For all FISONs F: |ℕ \ F| = ℵo.

But not for the union.

Right: |ℕ \ U{F : F is a FISON}| = |ℕ \ ℕ| = |{}| = 0.

What should make the union larger than all FISONs?

Holy shit! What an idiotic/nonsensical "question".
A reasonable question might be:
      "What should make the union larger than each and every FISON?"
That fact that the union of ALL FISONs "comprises" ALL natural numbers, while each and every FISON only "comprises" finitely many numbers.

Every union of FISONs which stay below a certain threshold stays belown that threshold.

Nope. For all FISONs F: F c_proper ℕ, while U{F : F is a FISON} = ℕ.

Because you never actually USED *ALL* FISONs [for the union(s)],

Right.

You said that I never used all FISONs. But I do.

No, you (don't and) didn't.
See: https://www.youtube.com/watch?v=wxrbOVeRonQ&t=76s

All are insufficient.

Nope, they are "sufficient".
Hint: U{F : F is a FISON} = ℕ.

but your claim doesn't match your conclusion, as the union of *ALL* FISONs will reach the size of the [set of all] Natural Numbers, even though no [union of] finite[ly many FISONs] reaches [...] it.

Right.

Do one or more FISONs grow during the union process? (WM)

1. Sets (and hence FISONs) DON'T "grow" and
2. there is no "union process",
you silly asshole full of shit!
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