Re: Incompleteness of Cantor's enumeration of the rational numbers

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.
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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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