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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?
Date: Mon, 7 Oct 2024 19:08:44 -0400
Organization: i2pn2 (i2pn.org)
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On 10/7/24 4:05 PM, Chris M. Thomasson wrote:
> On 10/7/2024 4:20 AM, Richard Damon wrote:
>> On 10/7/24 5:51 AM, WM wrote:
>>> On 07.10.2024 11:36, FromTheRafters wrote:
>>>> WM formulated the question :
>>>>> On 06.10.2024 19:03, FromTheRafters wrote:
>>>>>> A set is a collection of well-defined objects, meaning we must be 
>>>>>> able to determine if an element belongs to a particulr set.
>>>>>
>>>>> But you can't determine the smallest unit fraction although it is a 
>>>>> singleton set, a point on the real axis.
>>>>
>>>> There is no smallest unit fraction.
>>>
>>> If there are only fixed points, then there is a point such that 
>>> between it and zero there is no further point.
>>>
>>> Regards, WM
>>>
>>
>> Nope, not if you have an INFINITE set of fixed points.
>>
>> The problem is we can't have an infinite set of fixed points, as we 
>> are finite.
> 
> If each one of these is "fixed" in your line of thinking:
> 
> 1/1, 1/2, 1/3, 1/4, 1/5, ...
> 
> Well, there are infinitely many of them... ;^)
> 
> 

The problem with his "Actual infinity" is you need to get rid of the ... 
(as that is generative) and put in the full list of the number.

Because you have a full list of the numbers, he think that means there 
is a last one to that list.

The problem is that the full list can not be presented to a finite 
being, and thus, that "Actual infinity" can't be comprehended by it.

You need to b infinite, to actually have a fully generated infinite.

We can handle infinities by generative processes (his potential 
infinity) because we don't need to actually get to the end, we just know 
the process can continue forever and there will be no end, except after 
infinite work is done.

> 
>>
>> So, your "actual infinity" is something beyond what we can have, so it 
>> doesn't exist for us, and logic that assumes it is just breaks.
>