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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: 2N=E
Date: Mon, 4 Nov 2024 22:08:52 -0500
Organization: i2pn2 (i2pn.org)
Message-ID: <b9782cf210c9e0b67522205fa1991b29baaa7a2b@i2pn2.org>
References: <vb4rde$22fb4$2@solani.org> <vfoqi1$14lcd$6@dont-email.me>
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On 11/4/24 6:19 AM, WM wrote:
> On 03.11.2024 23:12, joes wrote:
>> 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.
> 
> It has only half of the reality of the natnumbers. But when doubling 
> them, their full reality is maintained.
>>
>>> You must be
>>> clairvoyant if you knew in advance whether something is done at all.
>> I know I will get even numbers.
> 
> But you will get larger even numbers than were multiplied.
> 
> Regards, WM
> 
> 

No, because that number was also in the set that was multiplied.

Since there is no highest number, it works, even if it blows your brains 
out because you can only handle finite sets.