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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: 2N=E
Date: Mon, 4 Nov 2024 07:26:43 -0500
Organization: i2pn2 (i2pn.org)
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On 11/4/24 6:15 AM, WM wrote:
> 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.

Yes, and ALL are there, and that means without end.

>>
>> That is the problem with your finite logic, that it can't actualy DO 
>> things in actual infinity,
> 
> I assume that it is possible.

Then you are making a false assumption that is the cause of your logic 
blowing up to smithereens, taking you with it.

IF you assume the cat you have is a dog, you won't be able to learn 
about what dogs are.

> 
>> 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.

And they are, and that means they go without end.

You can't have all, and have an end.

THAT is the contradiction that blows your system up.

> 
> Regards, WM
>