Sujet : Re: Replacement of Cardinality
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.logic sci.mathDate : 13. Aug 2024, 18:02:40
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <45ad1007-b1a7-49d0-a650-048f02738226@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 8/13/2024 10:21 AM, WM wrote:
Le 12/08/2024 à 19:44, Jim Burns a écrit :
On 8/12/2024 9:50 AM, WM wrote:
Le 11/08/2024 à 19:56, Jim Burns a écrit :
What causes an exception: nₓ ∈ ℕ:
⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?
>
The end of the positive axis.
>
There is no ⅟nₓ before the end of the positive axis
without ⅟(nₓ+1) before the end of the positive axis.
>
You cannot see it. It is dark.
The function NUF(x) is a step-function. It can increase from 0 at x = 0 to greater values,
0 isn't a unit.fraction.
either in a step of size 1
or in a step of size more than 1.
But increase by more than 1 is excluded by
the gaps between unit fractions.
0 isn't a unit.fraction.
(Note the universal quantifier there,
quantified over unit.fractions
according to which never –
in no limit and in no accumulation point –
two unit fractions occupy the same point x.)
Each unit fraction has GLB β > 0
which other unit.fractions are at least as far as.
0 is not a unit.fraction.
0 does not have GLB β > 0
which unit fractions are at least as far as.
Otherwise,
½⋅β is lower.bound and not.lower.bound.
Therefore
the step size can only be 1,
...at a unit.fraction.
0 isn't a unit.fraction.
resulting in a real coordinate x with NUF(x) = 1.
INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
NUF(INVNUF(1)) > 1
Contradiction.
…