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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: 2N=E
Date: Sat, 2 Nov 2024 12:20:56 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Fri, 01 Nov 2024 18:08:45 +0100 schrieb WM:
> On 01.11.2024 13:37, FromTheRafters wrote:
>> WM laid this down on his screen :
>>> On 01.11.2024 11:59, FromTheRafters wrote:
>>>> WM used his keyboard to write :
>>>>> On 31.10.2024 22:53, FromTheRafters wrote:
>>>>>> WM pretended :
>>>>>
>>>>>>>> Our sets do not change.
>>>>>>> Multiplication of all infinitely many fractions of the open
>>>>>>> interval (0, 1) results in some fractions in (1, 2).
>>>>>>> Multiplication of all infinitely many numbers of the open interval
>>>>>>> (0, ω) result in some numbers in (ω, ω*2).
>>>>>> No, there are no finite numbers in the transfinites.
>>>>> Numbers like ω + 4 are in the infinite. But if all natural numbers
>>>>> are doubled, then numbers in the infinite are produced.
>>>> What makes you think so?
>>> Simplest mathematics. Doubling increases the value. If all natnumbers
>>> are existing, then the greatest existing natnumber is existing too,
>> That is not the nature of the set of natural numbers. There is no last,
>> that is for finite sets only.
> But there are all. No point can escape doubling.
>>> Actual infinity is not about variable sets!
>> Sets don't change!
> Therefore not all doubled elements can be absorbed by the set.
Why "absorbed"? Do you think some multiple of a power of 2 is not natural?

-- 
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.