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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: 2N=E
Date: Mon, 4 Nov 2024 12:15:35 +0100
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On 03.11.2024 22:21, Richard Damon wrote:
> On 11/3/24 12:00 PM, WM wrote:
>> 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. You must be 
>> clairvoyant if you knew in advance whether something is done at all.

> The problem is that if you need to do them in "order" you can't complete 
> the infinite task.

Cantor says that all are there and can be paired with all fractions, for 
instance. That is what I accept for a moment.
> 
> That is the problem with your finite logic, that it can't actualy DO 
> things in actual infinity,

I assume that it is possible.

> We don't need to be clairvoyant to understand what WILL happen with a 
> deterministic operation.

Either all numbers are there before - or not. These are the only 
alternatives. You must switch to and fro.

Regards, WM