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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: The reality of sets, on a scale of 1 to 10 [Was: The
 non-existence of "dark numbers"]
Date: Sun, 13 Apr 2025 13:35:20 +0200
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On 13.04.2025 10:25, joes wrote:
> Am Thu, 03 Apr 2025 22:28:44 +0200 schrieb WM:
>> On 03.04.2025 21:10, Alan Mackenzie wrote:

>>> Yes, and aleph_0^2 = aleph_0.  There are as many positive fractions as
>>> natural numbers.
>> This is easily contradicted by observing that 1/2 is not a natural
>> number while all natural numbers are fractions.
> No,

Yes.
> 
>>> This was proven by Cantor.  That you don't understand the proof is your
>>> problem, not ours.
>> I understand that you are duped. And I have explained why. Every pair of
>> the bijection has almost all elements as successors.
> Bijections aren’t ordered.

Bijections with ℕ are ordered by the well-ordered set ℕ.
> 
>>>> The cardinality is the same because it counts only the first elements.
>>> That's a meaningless concatenation of words.
>> It is a pity that you can't understand. Every natural number that you
>> can use in a bijection has finitely many predecessors but infinitely
>> many successors which will never be used.
> No, a bijection can be an infinite set of pairs - must be, if the sets
> are infinite.

Every pair of the bijection has almost all elements as successors.
> 
>>> "So that each element of the set stands at a definite position of this
>>> sequence."  That has no relevance to anything at issue here.  In
>>> particular, it has no relevance to the loss of your favourite set
>>> element caused by an infinite sequence of transpositions.
>> Just this is excluded. Only definite positions are admitted. No evasion
>> into the infinite.
> Then you cannot talk about the limit.

So it is. Bijections with ℕ are well-ordered and have no limit but only 
all definable terms.

Regards, WM
>