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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Tue, 17 Dec 2024 07:34:09 -0500
Organization: i2pn2 (i2pn.org)
Message-ID: <75921cc1f17cdb691969a99e666f237cd09c0b09@i2pn2.org>
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On 12/17/24 4:13 AM, WM wrote:
> On 17.12.2024 00:52, Richard Damon wrote:
>> On 12/16/24 3:30 AM, WM wrote:
>>> On 15.12.2024 21:21, joes wrote:
> 
>>>>> Therefore we use all [1, n].
>>>> Those are all finite.
>>>
>>> All n are finite.
>>
>> But N isn't, so the sets [1, n] aren't what the bijection is defined on.
> 
> Every element is the last element of a FISON [1, n]. ℕ is the set of all 
> FISONs. I use all FISONs. ∀n ∈ ℕ: f([1, n]) =< 1/10.
> Ever heard of the effect of the universal quantifier?


But your logic can't deal with ALL Fisons.

Note, the mapping isn't in your [1, n] but in N.

Your logic that if it holds for all FISONs, it holds for N, is what 
shows that 0 == 1, so we see that logic is broken when it is applied to 
truly infinite things.

> 
>>> All intervals do it because there is no n outside of all intervals 
>>> [1, n]. My proof applies all intervals.
>>
>> And all the intervals are finite, and thus not the INFINITE set N, 
>> which is where the bijection occurs.
> 
> According to Cantor the "bijection" uses all n and nothing more.

Right, but no FISON uses contains ALL n.

> 
> Regards, WM
>