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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: Contradiction of bijections as a measure for infinite sets
Date: Sat, 6 Apr 2024 16:03:59 -0400
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On 4/6/24 3:40 PM, WM wrote:
> Le 06/04/2024 à 15:58, Richard Damon a écrit :
>> On 4/6/24 9:55 AM, WM wrote:
> 
>>> That mapping is Cantor's proposal. But for every other mapping, the 
>>> O's would also remain. All O's! It is th lossless exchange which 
>>> proves it.
>>
>> Cantor's proposal is between members of two distinct sets.
> 
> No. He does not specify that. And there is no reason to do so, except 
> that it can be used to contradict the ridiculous nonsense that there are 
> as many fractions as prime numbers.y

But he DOES, as he talks about the two SETS of numbers that are matched up.

And yes, the size of the set of all fractions is EXACTLY of the same 
size as the set of all Prime Numbers, and that size is Aleph_0.

That you can't understand that is YOUR problem, because your mind just 
can't understand things bigger than it.

> +
>>> It is Cantor's famaous mapping, more than a century believed to be a 
>>> bijection.
>>
>> But HIS does work, when you do it right.
> 
> No, his bijection works only for potential infinity applying the "...". 
> I show that his mapping does not work for the complete actually infinite 
> sets. He uses "and so on". Why does any intelligent mind believe that? I 
> show that the remainder will never decrease. There is no belief 
> required. It is provable fact.

Excpet that he doesn't need to use "..." as he can just list the formula 
that maps any k to n,d and back, and show that each n.d generates a 
unique k, and each k list a unique n,d. That is the bijection.

Only if you want to try to LIST all the members of the mapping, do you 
need to use ..., and that is because the set is INFINITE in length, and 
thus unlistable.

You logic is just INVALID when applied to infinite sets, and totally 
blows up and becomes inconsistant when you try to do so.

You seem to be stuck in "Dead Man Walking" mode.

>>
>>> If it operates, it must operate within one set too.
>>
>> Why?
> 
> Because there are as many naturals in ℕ as in ℚ. Precisely as many. But 
> only my approach shows that they are less than the fractions.

Since Q is the set of ALL RATIONAL numbers, what "fraction" isn't a 
Rational Number?

Note, the fact that you can map the Natural numbers N to a subset of Q 
(the elements that have the same value as a natural number) doesn't mean 
that their can not be also a mapping from N to ALL of Q.

You are just showing you don't understand that nature of infinite sets, 
because you brain just can't handle something bigger than it.

> 
> Regards, WM
> 
>