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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: 2N=E
Date: Sun, 3 Nov 2024 22:12:23 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
Message-ID: <1c8871dbcef7cfa4085c2947d52f9ce08412ceb3@i2pn2.org>
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Am Sun, 03 Nov 2024 18:00:18 +0100 schrieb WM:
> On 03.11.2024 16:55, joes wrote:
>> Am Sun, 03 Nov 2024 12:56:48 +0100 schrieb WM:
>>> On 03.11.2024 09:50, joes wrote:
>>>> pparently you do think that there is a natural n such that 2^n is
>>>> infinite.
>>> If all naturals are there, then no further one is available. But
>>> doubling all yields a greater number than all.
>>> In actual infinity there is no way to avoid this.
>> We don't need any further ones because we ALREADY HAVE ALL OF THEM,
>> even including the doubles.
> But you have not what is done to all of them afterwards.
Yes I have. The set of even numbers is a subset.

> You must be
> clairvoyant if you knew in advance whether something is done at all.
I know I will get even numbers.

-- 
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.