Sujet : Re: Replacement of Cardinality
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.logic sci.mathDate : 02. Aug 2024, 17:24:50
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <e7f793fb1fd82c90442679bfbcd2a334f10b7b13@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 8/2/24 12:05 PM, WM wrote:
Le 02/08/2024 à 17:59, Richard Damon a écrit :
On 8/2/24 7:38 AM, WM wrote:
Le 01/08/2024 à 18:04, joes a écrit :
Am Thu, 01 Aug 2024 12:27:27 +0000 schrieb WM:
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separated from 0 by any eps. Therefore your claim is wrong.
No. There is ALWAYS an epsilon.
>
Failing to separate almost all unit fractions.
Don't claim the contrary. Define (separate by an eps from 0) all unit fractions. Fail.
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Improperly revesing the conditionals.
Not at all! Recognizing that eps must be chosen. You cannot choose a eps that separates more than few unit fractions. That is why most cranks claim ∀x > 0: NUF(x) ≥ ℵ₀. It holds or all x that can be chosen. How should there be always an epsilon smaller than every x > 0 which fails? ? ?
Regards, WM
But you have the condition backwards.
The claim is that there is a finite difference that seperates any two specific unit fractions,
Not that there is a single finite difference that seperates any two arbirtry unit fractions.
Not understanding the order of the arguement just blow your logic up.
For your example, for any x, there is an eps = 1/(ceil(1/x)+1) that provides a unit fraction smaller than x, and thus NUF(x) can not be 1 for any finite x.
The fact that we get a different eps for every x is not a problem, it is just a property of the unbounded nature of the numbers we are using.
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