Sujet : Re: Replacement of Cardinality
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.logic sci.mathDate : 08. Aug 2024, 18:58:27
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <73ce89ca-d6ce-4eb5-827a-00e5c8f75c80@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 8/8/2024 4:20 AM, WM wrote:
Le 07/08/2024 à 23:29, Jim Burns a écrit :
On 8/7/2024 3:01 PM, WM wrote:
Le 07/08/2024 à 20:29, Jim Burns a écrit :
The only part of your argument which you've shared is
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
>
That is the decisive part.
Never two or more unit fractions are added to NUF.
>
Arithmetic says:
⅟n >
⅟(n+1) >
⅟(n+2) >
⅟(n+3) >
⅟(n+4) >
...
>
NUF(0) = 0.
Never more than one unit fraction can be
added simultaneously to NUF(x).
==> Exists x with x = INVNUF(1).
INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
NUF(INVNUF(1)) > 1
There are two contradicting arguments.
One of them must be wrong.
Or actual infinity is wrong.
⎛ A finite order is trichotomous and
⎜ each non.{} subset with min and max
⎜
⎜ An infinite order is trichotomous and
⎜ has one or more non.{} subsets without min or max.
⎜
⎝ No set has both a finite and an infinite order.
A subset of a set with a finite order
has only non.{} subsets with min and max
has a finite order.
A superset of a set with an infinite order
has a non.{} subset without min or max
has an infinite order.
An actuallyᵂᴹ infinite set
( a finite.ordered set with an infinite.ordered subset
does not exist.