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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Tue, 19 Nov 2024 16:27:52 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Tue, 19 Nov 2024 16:24:52 +0100 schrieb WM:
> On 18.11.2024 23:40, FromTheRafters wrote:
>> WM wrote on 11/18/2024 :
>>> On 18.11.2024 22:58, FromTheRafters wrote:
>>>> on 11/18/2024, WM supposed :
>>>>> On 18.11.2024 18:15, FromTheRafters wrote:
>>>>>> WM brought next idea :
>>>>>
>>>>>>>> |ℕ| - |ℕ| = 0 because if you subtract one element from ℕ then you
>>>>>>>> have no longer ℕ and therefore no longer |ℕ| describing it.
>>>>> If you remove one element from ℕ, then you have still ℵo but no
>>>>> longer all elements of ℕ.
>>>> But you do have now a proper subset of the naturals the same size as
>>>> before.
>>> It has one element less, hence the "size" ℵo is a very unsharp
>>> measure.
>> Comparing the size of sets by bijection. Bijection of finite sets give
>> you a same number of elements, bijection of infinite sets give you same
>> size of set.
> Why? Because only potential infinity is involved. True bijections pr5ove
> equinumerosity.

What is a "true" bijection?

>>>>> If |ℕ| describes the number of elements, then it has changed to |ℕ|
>>>>> - 1.
>>>> Minus one is not defined.
>>> Subtracting an element is defined. |ℕ| - 1 is defined as the number of
>>> elements minus 1.
>> Nope!
> The number of ℕ \ {1} is 1 less than ℕ.
And what, pray tell, is Aleph_0 - 1 ?

>>>>> If you don't like |ℕ| then call this number the number of natural
>>>>> numbers.
>>>> Why would I do that when it is the *SIZE* of the smallest infinite
>>>> set.
>>> The set of prime numbers is smaller.
>> No, it is not.
> It is, because 4 and 8 are missing.
It is a subset.

>  > There is a bijection.
> Only between numbers which have more successors than predecessors,
All of them do.

> although it is claimed that no successors are remaining.
-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.