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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
(extra-ordinary)
Date: Mon, 9 Dec 2024 23:13:56 +0100
Organization: A noiseless patient Spider
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On 09.12.2024 23:05, FromTheRafters wrote:
> WM presented the following explanation :
>> On 09.12.2024 22:45, FromTheRafters wrote:
>>> WM explained :
>>>> On 09.12.2024 21:55, Jim Burns wrote:
>>>>> On 12/9/2024 2:45 PM, WM wrote:
>>>>>> On 09.12.2024 18:20, Jim Burns wrote:
>>>>>
>>>>>>> Your two sequences as you have written them
>>>>>>> are equal, and have equal limits: the empty set.
>>>>>>>
>>>>>>> I suspect that it is the distinction between
>>>>>>> cardinality of limit #⋂{E(i):i} = #{} = 0 and
>>>>>>> limit of cardinalities ⋂{#E(i):i} = ⋂{ℵ₀:i} = ℵ₀
>>>>>>
>>>>>> The cardinality of the limit is
>>>>>> the cardinality of the limit set.
>>>>
>>>> By the way, we need no cardinality. We need only the sequence of
>>>> sets with the empty set in the limit.
>>>>
>>>>> The limit set is
>>>>> the set of numbers in common with each end.segment
>>>>> and isn't anything else.
>>>>
>>>> The limit set is the same for both sequences.
>>>> (E(1)∩E(2)∩...∩E(n)) and (E(n))
>>>> In order to stop tricksters we go without cardinality.
>>>
>>> But size matters, or so I've heard.
>>
>> In sequences of sets only sets matter.
>
> The why are you averse to the emptyset?
Not at all! It proves the existence of dark numbers.
Regards, WM