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From: Moebius <invalid@example.invalid>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Sat, 16 Nov 2024 02:10:15 +0100
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Am 16.11.2024 um 01:27 schrieb Moebius:

> Hint: Let's consider your claim: "an infinite set is never exhausted".
> 
> But IN \ {1} \ {2} \ {3} \ ... _should_ be {}, I'd say. After all, which 
> natural number would "remain" (=be) in the set
> 
>                IN \ {1} \ {2} \ {3} \ ...
> 
> ? :-P
> 
> Yeah, slightly "paradoxical". IN \ {1} is infinite, IN \ {1} \ {2} is 
> infinite, IN \ {1} \ {2} \ {3} is infinite, etc. Actually, for each and 
> everey natural number n: IN \ {1} \ ... \ {n} is infinite (in THIS sense 
> your "never" is true). But what's about IN \ {1} \ {2} \ {3} \ ...? 
> WHICH natural number would be in this set? :-P
> 
> Be aware of the infinite!
> 
> Remember:
> 
>   > You (WM) are misinterpreting the infinite as
>   > ☠( just like the finite, but bigger.
> 
> .
> .
> .

Actually, I'd prefer to consider a related problem (for certain 
technical reasons).

Consider the (infinitely many) unions:

           {1} u {2}, {1} u {2} u {3}, ...

Here you might claim: "an union of finite sets will never be infinite", 
after all for each and every n e IN:

           {1} u ... u {n}

is finite (namely {1, ..., n}). (Hint: Thats WM's "position".)

But actually,

           {1} u {2} u {3} u ...

IS infinite, namely {1, 2, 3, ...} = IN.

Technically, in set theory there's a certain union operation U which 
allows to "unite" infinitely many sets. There we would write:

           U{{1}, {2}, {3}, ...} = {1, 2, 3, ...} .

[If we want to be PRECISE: U{{n} : n e IN} = IN.}

Hope this helps.

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