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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Tue, 7 Jan 2025 18:30:39 -0500
Organization: i2pn2 (i2pn.org)
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On 1/7/25 7:51 AM, WM wrote:
> On 07.01.2025 13:35, Richard Damon wrote:
>> On 1/7/25 5:20 AM, WM wrote:
>>> On 07.01.2025 02:36, Ross Finlayson wrote:
>>>> On 01/06/2025 02:43 PM, Jim Burns wrote:
>>>
>>>>> It would be great if you (WM) did NOT
>>>>> find lemma 1 weird,
>>>>> but it is what it is.
>>>
>>> It is not weird. But your conclusions are weird.
>>>
>>>> The inductive set being covered by
>>>> initial segments is an _axiom_ of ZF.
>>>
>>> And the existence of the set ℕ is also an axiom of ZF. Therefore ZF 
>>> is incompatible with mathematics.
>>
>> No, ZF doesn't have as an axiom that the set of Natural Numbers exist.
> 
> AoI: There exists an infinite set S.

Which isn't that the NATURAL NUMBERS are an infinite set.

Only that such an infinite set will exist.

Or, do you think there is no such thing as an infinite set?

>>> All FISONs stay below the threshold |ℕ|/100, or in other words, 
>>> multiplication of any FISON by 100 is insufficient to cover |ℕ|.
>>>
>>
>> So? The union of an infinite set of them can have properites different 
>> that any set that is a union of only a finite number of them. That is 
>> a nature of infinity.
> 
> It is the nature of fools to believe every nonsense.

Sure is, and you prove it.

>>
>>> Every union of FISONs {1, 2, 3, ..., n} which stay below this 
>>> threshold stays below this threshold too.
>>
>> But not the union of *EVERY* FISON, the FULL INFINITE set of them. 
> 
> All are below 1 %.

No, your logic is all below some definite boundry (n) that is below 1%

So, all you prove is that a finite subset of an infinite set doesn't 
cover all of the infinite set.

Your logic can't handle *EVERY* FISON at once.

> 
>> SOething your "logic" can't handle,
> 
> Luckily.

So, you think you are "lucky" to be ignorant?

I guess that just shows how your logic works.

> 
> Regards, WM
>