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 Set theory debate : cantor 1st / Virgil argument. tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 12/08/2015, 11:08 PM I had to comment on virgil's / cantor's so called proof that there are more reals than integers. This is NOT the diagonal argument. I had set theory debates before as most of you probably know. Quote 22:23timmy1729 Virgil wrote : Virgil In article <760eb651-bb0e-4c13-9950-154ce5f66ccf@googlegroups.com>, WM wrote: > Without actual infinity, i.e., without having "all" digits, there is > no defined diagonal number, therefore Cantor's diagonal argument > fails. It shows at most that his assumption of finished infinity, > taken from holy bible and St. Augustin, is as inconsistent, as it > sounds. Of course without different cardinals there is no continuum > hypothesis to decide. Any alleged failure in Cantor's second proof in no way impacts on his first proof: Cantor's First Proof (of his theorem that the set of reals cannot be enumerated) revisited and simplifed. It states that there cannot be any surjection from the set of all naturals to the set of all reals). Note: If one assumes every real has been indexed with a different natural, it would otherwise follow from the argument below that some real must be indexed by a natural larger than each of infinitely many other naturals. Which is impossible! Lemma: Assume every real has been indexed with a different natural. Then every open real interval will contain two reals of lowest possible index for that original open interval, and the interior points of this new interval will necessarily have higher indices than its endpoints. Proof of lemma: obvious! Proof of theorem: Iterate the lemma to produce a nested sequence of such closed intervals, each a subset of its predecessors and interior points of each such interval having all indices greater than those of its own endpoints and the endpoints of all prior such intervals. Such an infinite sequence of nested closed real intervals is known to have a non-empty intersection, at least one point interior to all those intervals. But any such inside point must have a natural number index larger than all the infinitely many different natural numbers indexing different endpoints of the infinite sequence of intervals containing it. Which is impossible in proper mathematics! Thus any assumption that one can surject the naturals ONTO the reals is proven false and the theorem is true, and the set of reals is UNcountable!!! -- Virgil "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller) --- No ! False proof. Here is why You say infinitesimals do not occur on the real line. Then you take a limit ( without infinitesimals ?!? = dubious ) And end Up with An open interval A containing one point. And a closed interval B containing one point. Guess what : neither A or B exist - without infinitesimals - and its just a point. Because it is a point, the whole argument of intervals failed. Example : give the interval containing only 0. Give a closed and open example that are distinct. A point is not An interval. [- 0,0000...1 , 0] is not allowed. Problem ? You just proved there are An infinite amount of reals. That is all. Regards Tommy1729 ---- Im aware of "clopen " and neither open nor closed. But as stated the proof is incorrect imho. Hope my objection is clear. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 12/08/2015, 11:14 PM By analogue we can show any open interval with endpoint 2 consecutive integers contains infinitely many fractions. This SUGGESTS that there is no bijection from integers to fractions. But there is. This proof strategy fails. Regards Tommy1729 « Next Oldest | Next Newest »

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