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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: 2N=E
Date: Thu, 24 Oct 2024 13:59:52 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
Message-ID: <b415bcc470029bd4814b226ea14a6a7753125dfd@i2pn2.org>
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Am Thu, 24 Oct 2024 15:49:48 +0200 schrieb WM:
> On 24.10.2024 14:10, joes wrote:
>> Am Thu, 24 Oct 2024 12:46:05 +0200 schrieb WM:
>>> On 23.10.2024 20:59, joes wrote:
>>>> Am Wed, 23 Oct 2024 20:03:48 +0200 schrieb WM:
>>>>> On 23.10.2024 13:37, Richard Damon wrote:
>>>>>
>>>>>> Your complete set of the Natural Numbers is not complete.
>>>>> I take what is given: the complete set of natural numbers. I double
>>>>> it an get, according to mathematics 2n > n greater numbers than were
>>>>> given to me.
>>>> Every single number is greater than its original, but for every
>>>> number in the mapped set there is a greater one in the former set.
>>> Impossible since all have been mapped, even the greater ones.
>> Yes, and those are mapped to even greater ones!
> The possibility of always even greater ones in natural numbers proves
> potential infinity.
You have it backwards. Surely the "complete" set should not be missing
those greater numbers that the "potential" set includes.

> The greater ones have not been doubled because doubling of a complete
> set creates a set covering a greater interval than covered before. (Half
> the density implies twice the extension.)
They have also been doubled, along with their doubles. The powers of 2 and
their multiples form a subset of the naturals. The "size" of this set
is omega, and 2w=w, regardless of "reality". 

-- 
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.