Sujet : Re: how
De : wolfgang.mueckenheim (at) *nospam* tha.de (WM)
Groupes : sci.mathDate : 13. Jun 2024, 11:55:45
Autres entêtes
Organisation : Nemoweb
Message-ID : <doQDrJl4xLXY-Ml9z444aw3p-Jg@jntp>
References : 1 2 3 4 5 6 7 8 9 10
User-Agent : Nemo/0.999a
Le 12/06/2024 à 23:12, Jim Burns a écrit :
On 6/12/2024 4:33 PM, WM wrote:
Le 12/06/2024 à 20:54, Jim Burns a écrit :
On 6/12/2024 2:27 PM, WM wrote:
Le 12/06/2024 à 20:18, Jim Burns a écrit :
On 6/11/2024 10:44 AM, WM wrote:
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
>
ℕ\{0,1,2,…} = ∅
>
So there are no followers?
>
there is no j ∈ ℕ after ℕ
>
i.e., after all natural numbers.
There is no natural number after (≥)
all natural numbers.
But there are many after all discernible natural numbers.
That means:
If every number is subtracted,
then no successors remain.
If every number is deleted,
then every number is deleted.
But if every discernible number is deleted, then ℵo natural numbers are not deleted.
If only definable numbers are subtracted, then successors remain.
Only if some natural number is undefinable.
Yes. Obviously most are undefinable.
If any natural number is undefinable, then
the first undefinable has a definable predecessor.
That is your error. The definable numbers are definable and have definable successors. You will never get into the dark numbers by counting or defining.
No undefinable has a definable predecessor.
Right.
No natural number is undefinable.
Wrong. For the difference between definable and dark numbers see above: All definable numbers have ℵo successors. All dark numbers have no successors. Otherwise you could not delete all numbers.
After all definables are deleted from ℕ
no successors (no anything) remain in ℕ
Wrong. Easy to falsify: Delete any definable that has no successors remaining. Fail.
By 'ℕ' I mean the minimal inductive meta.set.
That is the set that contains all definables. There is none without ℵo dark successors.
Regards, WM
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