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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: How many different unit fractions are lessorequal than all unit
 fractions? (infinitary)
Date: Sat, 12 Oct 2024 16:47:38 -0400
Organization: i2pn2 (i2pn.org)
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On 10/12/24 1:49 PM, WM wrote:
> On 10.10.2024 21:54, joes wrote:
>> Am Thu, 10 Oct 2024 20:53:07 +0200 schrieb WM:
>>> On 10.10.2024 20:45, Alan Mackenzie wrote:
>>>> WM <wolfgang.mueckenheim@tha.de> wrote:
>>>
>>>>> If all natnumbers are there and if 2n is greater than n, then the
>>>>> doubled numbers do not fit into ℕ.
>>>> For any finite n greater than zero, 2n is greater than n.  The same
>>>> does not hold for infinite n.
>>> There are no infinite n = natural numbers.
>> Exactly! There are furthermore no infinite doubles of naturals (2n).
> 
> But the doubles are larger. Hence after doubling the set has a smaller 
> density and therefore a larger extension on the real line. Hence not all 
> natural numbers have been doubled.

The doubles are larger than the element they replace, but that value was 
always in the set to begin with, so it never creates a "new" term.

>>
>>>>>>> Numbers multiplied by 2 do not remain unchanged.
>>>>> Either doubling
>>>>> creates new natural numbers. Then not all have been doubled. Or all
>>>>> have been doubled, then some products fall outside of ℕ.
>>>> No.  Not even close.
>>> Deplorable. But note that all natural numbers are finite and follow this
>>> law: When doubled then 2n > n. If a set of natural numbers is doubled,
>>> then the results cover a larger set than before.. 
>> Additionally: if n is finite, so is 2n. It cannot go beyond w.
> 
> Then there is no complete set. The doubling can be repeated and 
> repeated. Always new numbers are created. Potential infinity.

You got it! The complete set of infinity is not available to finite 
beings to directly observe and handle, because it is just too big for 
use to work on.

That was the conclusion that Aristotle came up millennia ago.



> 
> Regards, WM
>>
>